strand 1 of 5 - dublin academy of education

6th Year

Maths

Higher Level

Strand 1 of 5 Topics:

Statistics

Probability

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©The Dublin School of Grinds Page 1

Your Leaving Cert paper is based on the syllabus published by a gang called the NCCA.

The NCCA have the syllabus broken into 5 sections, which are called the 5 strands.

Strand 1 is worth 17% to 21% of The Leaving Cert.

It appears on Paper 2.

Contents Statistics

Theory .................................................................................................................................................................................................... 2

1) Measures of centre ................................................................................................................................................................. 5

2) Measures of variation ............................................................................................................................................................ 8

3) Stem and leaf .......................................................................................................................................................................... 13

4) Histograms and distribution curves ............................................................................................................................ 16

5) Scatter plots and correlation........................................................................................................................................... 24

6) Linear regression ................................................................................................................................................................. 30

7) Measures of relative standing ........................................................................................................................................ 34

8) Inferential Statistics ............................................................................................................................................................ 48

9) Solutions to Statistics ......................................................................................................................................................... 63

Probability Theory ................................................................................................................................................................................................. 79

1) And/ or ..................................................................................................................................................................................... 79

2) Relative frequency ............................................................................................................................................................... 83

3) Probability rules ................................................................................................................................................................... 84

4) Binomial distribution (also known as Bernoulli trials) ...................................................................................... 90

5) Expected value ...................................................................................................................................................................... 98

6) Factorial/ arranging/ choosing/ hangman ............................................................................................................ 101

7) Solutions to probability ................................................................................................................................................... 107

Past and probable exam questions

1) Section A questions ........................................................................................................................................................... 114

2) Section B questions ........................................................................................................................................................... 133

3) Solutions to past and probable exam questions ................................................................................................... 155


Statistics We can break Statistics into 2 sections:

Theory

Problems

Theory

There are some questions asked for which we must know definitions ‘off by heart’. These are listed below. There are other questions which will ask theory about general problems. To answer these we need to understand the problems in this chapter inside out. (Note: The following definitions have been minimised as much as possible for easy learning. These are full mark definitions).

Types of Studies:

Observational study:

An observational study is a study in which a researcher observes behaviour without influence. Example: Traffic on a road. Advantage: Can be inexpensive. Disadvantage: Reliance on subjective measurement (ie: the researcher may be biased).

Designed experiment:

A designed experiment is an experiment in which a treatment is applied and the researcher observes the effects. Example: Testing of drugs. Advantage: Researcher can control variables. Disadvantage: Problem of ethics.

Sample survey:

A sample survey is a study that obtains data from a sample of a population in order to estimate attributes of the population. Example: Opinion poll on government parties. Advantage: Can be inexpensive. Disadvantage: May include sampling error.

Census -v- Sample Survey

A census is a survey of the whole population in question. A sample survey is a study that obtains data from a sample of a population in order to estimate attributes of the population. A sample takes less time but may include sampling error. Note: There is a difference between ‘sampling error’ and ‘bias’. The ‘unofficial’ way to remember it is:

Sampling error comes from the number of things. Bias comes from the type of things.

Sampling error can be predicted, calculated and accounted for. Bias can’t. (More on how to predict, calculate and

account for sampling error later).

Population/Sampling frame/Sampling units

The population is the entire group being studied, eg. students of UCD. (Note: The word population doesn’t just mean the number of people in a country) The sampling frame is the actual list of sampling units from which the sample is selected, eg. Those on the secretaries computer in the science building. The sampling units is the set of elements chosen from the sampling frame, eg. The actual ones chosen to be surveyed.


Population parameter -v- Sample statistic

A population parameter is a numerical measure from the entire population. A sample statistic is a numerical measure from a sample of the population. Sampling Distribution: A sampling distribution is a distribution of statistics obtained through a large number of same size samples drawn from a specific population. Types of sample surveys:

Simple Random Sampling:

A simple random sample gives each member of the population an equal chance of being chosen. Example: Random names can be pulled from a hat. Advantage: Can be highly representative. Disadvantage: Expensive, as those sampled may be spread over a large area. Stratified Random Sampling:

The population is divided into subgroups, so that subjects within each subgroup have a common characteristic. Then a

simple random sample is drawn from each subgroup according to their proportion of the population, to make the full

sample.

Example: Which team is going to win the All-Ireland this year? Here we would break the country into county subgroups and take a simple random sample from each. Advantage: Eliminates potential bias. Disadvantage: May be difficult in identifying appropriate strata. Systematic Random Sampling:

This is random sampling with a system. From the sampling frame, a starting point is chosen at random, and thereafter at regular intervals. Example: Suppose you want to sample 8 houses from a street of 120 houses. 120/8=15, so every 15th house is chosen after a random starting point between 1 and 15. If the random starting point is 11, then the houses selected are 11, 26, 41, 56, 71, 86, 101, and 116. Advantage: Easier to conduct than a simple random sample. Disadvantage: The system may interact with some hidden pattern in the population, e.g. every third house along the street might always be the middle one of a terrace of three.

Cluster Sampling:

The population is divided into clusters, and some of these are then chosen at random. Then units are chosen within each cluster. Example: Households in the same street. Advantage: Saving of travelling time, and consequent reduction in cost. Disadvantage: Units close to each other may be very similar and so less likely to represent the whole population.

Quota Sampling:

In quota sampling the selection of the sample is made by the interviewer, who has been given quotas to fill from specified sub-groups of the population. Example: An interviewer may be told to sample 50 females between the age of 45 and 60. Advantage: Quick and cheap to organise. Disadvantage: Because the sample is non-random, there could be bias.

Convenience Sampling:

This is a non-probablity sampling method where subjects are chosen in the most convenient way. Example: Just ask family members. Advantage: Quick and cheap. Disadvantage: Could be very biased.


Other Definitions:

1. Descriptive statistics: use of graphs, tables, charts and various measurements/calculations to organise and summarise information.

2. Inferential statistics: a sample or portion of the population is taken and then conclusions are made about the entire population.

3. Variable: A numerical characteristic of interest in each element of the sample.

4. Observations: value of a variable for ONE PARTICULAR element of the sample.

5. Data set: value of all observations of a variable for the elements of the sample.

6. Data capture: process by which data is transferred from paper copy eg: questionnaire, to an electronic file eg: a computer.

7. Primary data: data collected by the organisation/person who is going to use it.

8. Secondary data: data already available not collected by the person/organisation who is going to use it. (newspapers, books, historical, records, internet)

9. Univariate data: only one item of information is collected from each member of a group eg: height or annual salary or eye colour.

10. Bivariate data: two items of information are collected from each member of a group eg, height and weight of a person or starting salary and numbers of years in education.

11. Outlier: very high or very low value not typical of the other values in the data.

12. Explanatory variable: this is the controlled variable whose effects on the response variable we wish to study.

13. Response variable: this is the variable whose changes we wish to study.

14. Control group: split a group into 2 parts to test a drug. Both think they are getting the drug but the second group are getting an inactive substance called a placebo. This group is called the control group.

15. Quantitative data (numerical): data that can be counted or measured. 2 types exist. Discrete: only have certain values in a given range eg: number of goals in a match, shoe

size, number of tails when tossing a coin 30 times. Continuous: can take any value in a given range. eg: height, weight, time, rainfall

measurements.

16. Qualitative data (categorical): data that cannot be counted or measured or answered in numbers. 2 types exist.

Ordinal: data has an obvious order (pain level: none/low/moderate/severe, or restaurant service: poor/good/excellent, or exam results A/B/C/D/E/F/NG).

Nominal data: data that cannot be ordered i.e. blood type A, B, O, AB or hair colour or favourite football team.

17. Questionnaires: set of questions designed to obtain data from a population.

18. Respondent: People who answer questionnaires.

19. Good qualities of a questionnaire: (a) be as brief as possible (b) use clear and simple language. (c) not cause embarrassment or offend (d) no leading questions (e) accommodate all possible answers (f) contains tick boxes


Problems: There are 8 types of problems in Statistics:

1) Measures of centre

There are 3 measures of centre:

1.

Mean - The sum of values divided by the number of values (on page 33 of The Formulae and Tables

Booklet you will see this explained as 𝜇 =Σ𝑥

𝑛).

Advantage: Uses all the data. Disadvantage: Affected by outliers.

2. Mode - The value that occurs the most. Advantage: Easy to find. Disadvantage: Not necessarily unique (may be more than one answer).

3. Median - The middle value when values are listed in order. Advantage: Not affected by outliers. Disadvantage: Can be tedious to arrange a large list of numbers in order.

Example 1 Find the mean, mode and median of the following list of numbers: 2,3,5,3,7,4,3,9,5,3

Mean = 2+3+5+3+7+4+3+9+5+3

10

= 4.4 Median = list the data in order: 2,3,3,3,3,4,5,5,7,9

Since the line doesn’t go through a number we get the mean of the numbers on either side:

=> median = 3+4

2

= 3.5 Mode = the one that occurs the most = 3


It is important to know when to use each measure. Consider the following example:

Example 2 Daithi is interested in finding the typical weight of trophies in the local sports club’s cabinet. He weighs 10 random trophies as follows (in kgs): 2.7, 2.9, 3.1, 3.4, 3.7, 4.1, 4.3, 4.7, 4.7, 40.8 If we calculate the mean we get 7.44kg. This figure has clearly been affected by the extreme outlier of 40.8, so is not a suitable measure of this scenario. The mode is 4.7kg. Again this is not very representative of the data. The median is calculated as follows: 2.7, 2.9, 3.1, 3.4, 3.7, 4.1, 4.3, 4.7, 4.7, 40.8

median = 3.7+4.1

2

= 3.9kg This is the best representation of the weight of trophies in this case.

Question 1.1 Twenty five students each measure and record a particular angle of elevation, in degrees, each using his or her own home made clinometer

24 20 22 15 70 15 16 15 16 15 18 16 21 21 73 16 20 12 18 20 18 18 14 22 18

Find what you consider to be the best estimate of the true value of the angle, explaining your reasoning.


You can also be asked to calculate the mean of a ‘frequency distribution’. This is calculated by getting the sum of the frequencies by the corresponding values and dividing by the sum of the frequencies.

(On page 33 of The Formulae and Tables Booklet you will see this explained as 𝜇 =Σ𝑓𝑥

Σ𝑓).

Example 3 Isabella walked up a street to find out how many parking spaces each house had. Here are her results:

Parking spaces No. of houses 1 15 2 27 3 8 4 5

Calculate the mean number of parking spaces, correct to 2 decimal places. Solution:

mean = Σ𝑓𝑥

Σ𝑓

= (15)(1)+(27)(2)+(8)(3)+(5)(4)

15+27+8+5

= 2.05 Hint: If you aren’t sure which list you should divide by, it is always the “other one”. In the last example we were looking for the mean number of parking spaces so we divided by the “other one”, ie: the number of houses.

You may also be given a grouped frequency distribution. To calculate the mean we must first find mid-interval values (which is the same thing as finding the mean).

Example 4 The ages of players in the Arsenal football squad in 2014 are given in the table below. Calculate the mean age of an Arsenal player, to the nearest year.

Ages No. of players 17-21 12 21-26 15 26-31 7 31-36 4 36-40 0

Note: 17-21 means 17 is included, but not 21, and so on.

In this type of question there will always be a little note at the end of the table (see above). You can ignore this. Firstly, list out the mid interval values by getting the mean of each of the groupings:

Ages No. of players 19 12

23.5 15 28.5 7 33.5 4 38 0

Now, just like the last example: 𝜇 =Σ𝑓𝑥

Σ𝑓

=(12)(19)+(15)(23.5)+(7)(28.5)+(4)(33.5)+(0)(38)

12+15+7+4+0

=794

38

= 20.894 ≈ 21 𝑦𝑒𝑎𝑟𝑠


2) Measures of variation

There are 3 measures of variation (also known as measures of spread or measures of dispersion)

1. Range: (𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 − 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒)

Advantage: Easy to calculate. Disadvantage: Can be affected by outliers.

2. Interquartile range: (𝑈𝑝𝑝𝑒𝑟 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 − 𝐿𝑜𝑤𝑒𝑟 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝑣𝑎𝑙𝑢𝑒)

Advantage: Not affected by outliers. Disadvantage: Ignores some data.

Note: There is some confusion on how to calculate the upper quartile value and lower quartile value. Let’s clear this

up…

Lower quartile rule:

List the values in order.

Mulitply 1

4 by the number of values.

If you get a decimal, then round up and use this value.

If you get a whole number, then add this value and the next value, then divide by 2.

Example 1 Find the lower quartile value of the following lists: List A: 2, 3, 6, 7,10, 12, 13, 13, 17

1

4× 9 = 2.25

ie: a decimal ⇒ round up to 3 ⇒Lower quartile value = 6 List B: 2, 5, 6, 7, 9, 10, 11, 18

1

4× 8 = 2

ie: a whole number ⇒ add the 2nd and 3rd value, then divide by 2.

⇒ Lower quartile value =5+6

2

= 5.5

Note: There are other ways to do this that sometimes give different answers. This method is used by the State

Examinations Commission.

Upper quartile rule:

This is the same as the lower quartile except we multiply by 3

4 rather than

1

4.


Question 2.1

Find the range and the interquartile range of the following times taken to complete a race:

Name Time (seconds)

Richard 37.6 Stephanie 20.3

Brian 42.9 Joe 50.1 Simon 44.2 Daragh 21.7

3. Standard deviation: Measures the spread of the data from the mean. Advantage: Uses all the data. Disadvantage: Can be affected by outliers.

For example, take the following lists of data:

List 1: 12, 15, 11, 13, 13, 14, 14, 11, 12, 15 List 2: 1, 16, 20, 26, 31, 2, 8, 2, 20, 4 If you were to calculate the mean of both lists you would find they are both 13. However, it is clear by just looking at the

lists, that list 2 is more ‘spread out’. This means it has a greater standard deviation.


To calculate the standard deviation of a list of numbers, you can use the formula from page 33 of The Formula and

Tables Booklet or use your calculator. The NCCA have confirmed that students can not be required to use the formula

alone. This means that if you just learn the calculator steps you are perfectly fine. Some of the text books and some

teachers are unaware of this. The instructions for using your calculator are as follows:

Example 2 Find the standard deviation of the following list of numbers: 8, 8, 9, 11, 14. How to do it on a CASIO fx-83MS:

=

How to do it on a Sharp EL - W531: Clear your calculators memory first by pressing 2nd function, then MODE (CA). Then type the following:

=

Be careful: After using the steps above, your calculator is now in ‘Statistics Mode’. To change it back to “Normal Mode’,

follow the below instructions:

Casio:

Press: MODE (SETUP), then press 1.

Sharp:

Press the RESET button on the back of your calculator.


You can also be asked to find the standard deviation of a ‘frequency table’. Again, we use our calculator as follows:

Example 3 Find the standard deviation of the following frequency table:

Score 12 16 20 24 28 Frequency 13 28 26 21 12

How to do it on a CASIO fx-83MS:

=

How to do it on a Sharp EL - W531: Clear your calculators memory first by pressing 2nd function , then MODE (CA). Then type in the following:

=

3


Question 2.2 What is the standard deviation for the numbers: 75, 83, 96, 100, 121, 125?

Question 2.3 Jack did a survey of the number of pets owned by his classmates, with the following results.

Number of pets Frequency 0 4 1 12 2 8 3 2 4 1 5 2 6 1

What was the standard deviation?

Question 2.4


3) Stem and leaf

To draw a stem and leaf:

Make sure each number is in increasing order out from the stem.

Write in the key.

Example 1 Complete a stem and leaf plot of the following exam results obtained by students in Economics and Geography. Economics(%): 90, 72, 14, 22, 63, 63, 54, 57, 72 Geography(%): 88, 87, 63, 41, 41, 99, 6, 41, 53

Note: In higher level it is very unlikely you will be asked to draw a stem and leaf plot (although it is on the

syllabus).


Question 3.1


4) Histograms and distribution curves

Histograms are used to represent data in categories.

Example 1 The following frequency distribution gives the number of marks obtained by students in an examination.

(i) Represent the data with a histogram. (ii) Name the modal class. (iii) In which class interval does the median lie? (iv) On your histogram, indicate clearly where the median lies and write down your value for the median. Solution: (i)

(ii) The modal class is the interval with the most students in it, ie: 80 − 100 (iii) The median is the middle student. Since there are 71 students the middle one will be the 36th student. (Think about it, if we listed out 71 numbers we would have 35 students to the left, 35 to the right and the 36th number in the middle). So, since there are 29 students in the 0 - 40 interval and 37 students in the 0 - 60 interval, the 36th student must be in the 40 - 60 interval. (iv) Well there are 29 students from 0 - 40 marks. So if we’re looking for the 36th student we need another 7. But

there are 8 in the 40 - 60 interval, so the median must be 7

8 along the 40 - 60 interval.

This interval is 20 so 7

8 of 20 = 17.5

Therefore the median is approx 40 + 17.5 = 57.5

Shapes of Histograms: Histograms can have certain shapes…

Normal (also known as symmetrical or bell shaped)


Left skewed (also known as negatively skewed)

Right skewed (also known as positvely skewed)

Bi modal: (if it has two peaks)

Multimodal: (if it has more than 2 peaks)

Linear

When we have continuous data we get curved lines which make distribution curves: For example, a left skewed distribution curve would look as follows:

Note: You must know the difference between ‘continuous data’ and ‘discrete data’. An unofficial rule is as follows:

Continuous data is data which has decimals (eg: heights, weights etc).

Discrete data is data which hasn’t got decimals (eg: number of airplanes in the sky, amount of people at a match).

Note: Shoe size is discrete even though it has halves.


Empirical Rule:

When a curve is normally distributed we have an important rule called the Empirical Rule, as follows:

68% of all data lies within mean ± standard deviation.

95% of all data lies within mean ± 2 standard deviations.

99.7% of all data lies within mean ± 3 standard deviations.

Let’s explain this with examples…

Question 4.1 Suppose it takes you 30 minutes on average to travel to the city centre with a standard deviation of 6 minutes. Given the distribution of these journey times is normal, then using the Empirical Rule, find the percentage of journeys you will arrive in the city centre taking: i) Less than 36 minutes ii) More than 36 minutes iii) Less than 18 minutes iv) More than 18 minutes


Question 4.2 The height of men is normally distributed with a mean of 172.5cm and a standard deviation from the mean of 7cm. Tom the statistician has designed a house with doorway high enough to allow all men, except the tallest 2.5%, to pass through without bending. By using the Empirical Rule, decide what doorway height Tom has used.

Question 4.3 The scores of secondary school students taking the verbal section of an aptitute test are normally distributed with mean 490 and standard deviation 100. Using the Empircal Rule, answer the following questions: a) What percentage of students scored between 390 and 590 on the test?

b) One student scored 795 on the test. What can you say about this student?

c) A certain course in Dublin University only admits students who are among the highest 2.5% of the scores. What score

would a student need to get on the test to get into this course?

Question 4.4 In an exam you scored 73%. If the mean is 65% would you prefer the standard deviation of the scores to be 8% or 16%? Why? (You can assume the scores are normally distributed). Use the Empirical Rule to make your decision.


Another important fact you must know is how the shape of a distribution (histogram or distribution curve) affects the

mean, median and mode:

Normal:

mean = median = mode

Left skewed:

mean < median < mode

Right skewed:

mode < median < mean

Note how the mode is always the highest one, the median is always the middle one and the mean is always the other

one. Also note that when we say less than we mean along the horizontal axis, not the vertical axis.

Question 4.5


Question 4.6

As the standard deviation of a normal distribuition increases, it gets ‘fatter’ and ‘lower’. This is because the data

is more spread from the mean. However, a change in standard deviation causes no affect to the area under the curve.

Question 4.7


Shifting data: Point 1: When we shift data by adding (or subtracting) When we add (or subtract) a constant to all the data values, all measures of centre (the mean, mode and median) are increased (or decreased) by that constant, while measures of variation (the range, interquartile range and standard deviation) are unchanged. Point 2: When we shift data by multiplying (or dividing) When we multiply (or divide) all the data values by a constant, all measures of centre (mean, mode and median) and measures of variation (range, interquartile range and standard deviation) are multiplied (or divided) by that same constant.

Example 2 Three friends are aged 6, 8 and 9.

Mean =6+8+9

3

= 7.6 Range = 9 − 6 = 3 Now add 10 years onto their ages, so they are now aged 16, 18 and 19.

Mean =16+18+19

3

= 17.6 (ie: old mean +10) Range = 19 − 16 = 3 (ie: same as old range) ⇒ This illustrates Point 1 above. Instead, what if we multiplied their ages by 10, so they are now aged 60, 80 and 90.

Mean =60+80+90

3

= 76.6 (ie: old mean x 10) Range= 90 − 60 = 30 (ie: old range x 10) ⇒ This illustrates Point 2 above.


Question 4.8 If 𝑑 is the standard deviation and 𝑚 is the mean of the data 𝑥, 𝑦, 𝑧 , what is the standard deviation and mean of 𝑥 + 5, 𝑦 + 5, 𝑧 + 5?

Question 4.9 If 𝑑 is the standard deviation and 𝑚 is the mean of the data 𝑥, 𝑦, 𝑧 what is the standard deviation and the mean of 𝑥

8 ,

𝑦

8 ,

𝑧

8 ?

Question 4.10 The lifetimes of a certain type of batteries have a mean of 20 hours and standard deviation of 5 hours. Find the mean and standard deviation of the total lifetime of a pack of 6 batteries?

Question 4.11 If the Government decided to add a tax of €2 to all smart phone covers, how would this affect the summary statistics (minimum, maximum, mean, standard deviation, median, IQR)?

Note on shapes of stem and leaf diagrams: The Examiner can ask you to comment on the shape of stem and leaf diagrams.

For the right hand side of a stem and leaf turn your head clockwise. For the left hand side of a stem and leaf turn your head anti-clockwise and give the opposite to what you see!

Consider the following back-to-back stem and leaf diagram:

3 6, 2

7, 4, 1 8, 5, 5, 3, 2

9, 9, 9, 7, 4, 4, 2, 1 9, 9, 9, 6, 6, 3, 2

8, 7, 2

0 1 2 3 4 5 6

7 3, 6 3, 4, 7 2,4, 5, 6 1, 2, 2, 2, 6, 7, 9, 9, 9 3, 4, 7, 9, 9, 9, 9, 9 1, 4

We can see the right hand side is left skewed.

The left hand side seems right skewed but we must give the opposite to what we see, meaning it is actually left skewed.

The reason this happens is because the numbers are listed backwards when we draw the left hand side of a stem and

leaf diagram.


5) Scatter plots and correlation

We use scatter plots to investigate the relationship between 2 sets of data.

Note: If one of the two ‘things’ being compared causes the other, then we put the ‘cause’ on the horizontal

axis, ie: the x-axis.

Since the ‘number of years’ causes the ‘distance travelled’ to change we put this on the horizontal axis.

Sometimes we don’t know which thing causes the other so it doesn’t matter which axis we use for each thing

(eg: arm length -v- leg length).

The correlation coefficient (denoted by the letter ‘r’) measures the strength of the relationship between two

sets of data. It is always between -1 and 1.

We can have: Positive correlations (uphill) Negative correlations (downhill) No correlation (all over the gaf)

Use the following descriptions for correlations: perfect/very strong/moderately strong/strong/very weak/moderately

weak/weak.

Some examples are as follows:

1) Perfect positive correlation (𝑟 = 1)

eg: Number of tickets sold -v- number of full seats.

2) Strong positive correlation (𝑟 ≈ 0.8)

eg: Years of education -v- annual income


3) Moderately strong positive correlation (𝑟 ≈ 0.6)

eg: Hours study -v- exam result

4) Strong negative correlation (𝑟 ≈ −0.7)

eg: Time spent in shopping mall -v- money left in pocket

5) Weak negative correlation (𝑟 ≈ −0.3)

eg: Weight -v- kilometres ran

6) No correlation (𝑟 ≈ 0)

eg: Hours study -v- goals scored


As well as estimating the correlation coefficient, the Examiner can ask you to calculate the correlation coefficient.

For this you use your calculator.

Example 1 Calculate the correlation coefficient of the following data:

How to do it on a CASIO fx-83MS

You should get an answer of 0.9228479016 which you can round off as required.

How to do it on a Sharp EL-W531 Clear your calculators memory first by pressing 2nd function, then MODE (CA). Then type in the following. Mode – 1- 1 20 (x, y) 3150 DATA 110 (x, y) 8800 DATA 290 (x, y) 9310 DATA RCL (which is above number 8) r (which is actually the ÷ button) You should get an answer of 0.922847901 which you can round off as required.

Be careful: After using the steps above, your calculator is now in ‘Statistics Mode’. To change it back to “Normal Mode’,

follow the below instructions:

Casio:

Press: MODE (SETUP), then press 1.

Sharp:

Press the RESET button on the back of your calculator.


Question 5.1

Question 5.2 Match a correlation coefficient to the description below:

Description Correlation coefficient No correlation Very weak negative correlation Perfect negative correlation Strong negative correlation Strong positive correlation Very weak positive correlation


Question 5.3


What about this weird question?

Question 5.4

Important note: Just because two variables are correlated does not necessarily mean they cause each other.

Consider the following diagram, which was created when research was done into the number of shark attacks and the

number of ice-creams sold over the course of 15 years.

It can be seen from the scatter plot that there is a strong positive correlation. But the amount of ice-creams sold don’t

cause shark attacks, or vice-versa. The thing that is causing both these to increase is temperature. This is called the

‘lurking variable’ because it is hiding (lurking) in the background.

Question 5.5 Explain with the aid of an example, what is meant by the statement:

“Correlation does not imply causality”


6) Linear regression

When we have a scatter plot, the Examiner can ask us to draw a ‘regression line’ (also known as ‘the line of best

fit’). We simply do this by eye.

If he asks us to find the equation of the line, we use the formula for the equation of a line from Page 18 of our

Formulae and Tables Booklet:

𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1)

This requires the slope which we find by using the slope formula on the same page:

𝑚 =𝑦2 − 𝑦1

𝑥2 − 𝑥1

Warning: When choosing your 2 points for the slope formula, make sure to take them off the line you drew and not points which aren’t on the line.

Important: The slope of the line of best fit always represents the expected change in the 𝑦 value for each additional 𝑥

value.

Lets say the following line of best fit had a slope of 1.8

This would mean we expect a 1.8% increase in grade for every extra hour studied.


Question 6.1


Question 6.2


7) Measures of relative standing

There are two measures of relative standing (also known as measures of position) examinable on your Leaving Cert:

Section 1: 𝑍 - scores

Section 2: percentiles

Section 1: 𝑍 - scores (also known as z-values)

Before we explain what z – scores are we must know how to deal with them.

To do this we use the table on pages 36 and 37 of the log tables.

Example 1 Find 𝑃(𝑍 ≤ 0.84) Answer from the table: 0.7995

Note: If the 𝑍 - score is not in the table, you will get full marks by rounding off to the nearest value.

Example 2 Find 𝑃(𝑍 ≤ 1.857) Answer from the table by rounding to 𝑃(𝑍 ≤ 1.86): 0.9686

There are 3 rules you must know about z – scores: Rule1: If we are asked for the probability of a value being greater than a 𝑍 - score we calculate:

1 – probability that its less than or equal to the z-score

Question 7.1

Calculate 𝑃(𝑍 > 1.49)

Rule 2: If we are asked for the probability of a value being less than a minus value we use:

1 – probability that it’s less than or equal to the positive value

Question 7.2

Calculate 𝑃(𝑍 ≤ −0.46)


Now let’s mix these 2 rules together:

Question 7.3

Calculate 𝑃(𝑍 > −0.06)

Rule 3: To find the probability that a z-score is between two values, you find: 𝑃(𝑧 ≤ 2𝑛𝑑 value) − 𝑃(𝑧 ≤ 1𝑠𝑡 value)

Question 7.4

𝑃(1.7 ≤ 𝑍 ≤ 2.4)

So you may be wondering: why do we use z-scores?

Think about this: A scientist might want to know the probability of a cell being less than a certain size, a car manufacturer may want to know the probability of a car weighing less than a certain weight, a gardener might want to know the probability of a plant being less than a certain height, and so on … But rather than having loads of probability tables for cells, cars, plants etc., we just ‘standardize’ all these to z – scores,

and use one probability table.

The formula for 𝑍-scores is given on Page 34 of the Formulae and Tables Booklet:

𝑧 =𝑥 − 𝜇

𝜎

where 𝑥 = 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑖𝑛 𝑞𝑢𝑒𝑠𝑡𝑖𝑜𝑛 𝜇 = 𝑡ℎ𝑒 𝑚𝑒𝑎𝑛 𝜎 = 𝑡ℎ𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛

In the syllabus, The Examiner only requires us to deal with 𝑍-scores in normal distributions (as seen by the diagram at

the top of Page 36 in The Formulae and Tables Booklet).

Look out for The Examiner’s key words:

normal

mean

standard deviation

When you see all these words in a question you will know it must be either a 𝑍-score question or an Empirical Rule

question. (z-scores are more accurate than the Empirical Rule, but take more time. Also the Empirical Rule only has 3

parts to it, so it doesn’t work all the time. Therefore, the question will determine which method to use)


Question 7.5

A random varialble 𝑥 follows a normal distribution with mean 150 and standard deviation 20. Find 𝑃(𝑥 ≤ 190).

Question 7.6

A random variable 𝑥 follows a normal distribution with mean 180 and standard deviation 8.

Find 𝑃(168 ≤ 𝑥 ≤ 184.5).


The Examiner has the scope in the syllabus to put a story behind the question:

Question 7.7

The mean length of boats in a large harbour is 151m with a standard deviation of 15m. Assuming that the lengths are

normally distributed, find what percentage of boats are:

a) less than 160m in length

b) greater than 180m in length

Question 7.8 A coffee company knows that the weights of the packets of one of their brands are normally distributed with a mean of 300 grams and a standard deviation of 6 grams. How many of 1000 packets, selected at random, can be expected to weigh: a) less than 295 grams b) between 296 grams and 307 grams


In questions, such as the previous 2 examples, we are given a value (such as length, weight, height etc.) and asked to find a probability. Sometimes the Examiner actually gives you a probability and you have to work backwards to find a value (for length, weight, height etc): Question 7.9 The mean height of applicants for an army division was 180cm, with a standard deviation of 15cm. You can assume the heights are distributed normally. The shortest 60% of applicants are rejected. What is the minimum height an applicant must be to be accepted?

Question 7.10 (a) The mean height of students in a Leaving Cert class was 170cm with a standard deviation of 13cm. The heights were normally distributed. An airline has requirements for flight attendants for safety reasons to ensure that they can reach overhead safety equipment. Typically the acceptable heights are between 160cm and 185cm tall. What is the probability that a student chosen at random from this class complies with these typical requirements?


(b) The Teacher checks with the airline and is told that 4% of the class are too tall to be flight attendants. What is the

maximum flight attendant height used by the airline?


Question 7.11


Question 7.12 A normal distribution has a mean of 120 and a standard deviation of 20. For this distribution: i) What score separates the top 40% of the scores from the rest?

ii) What range of scores would form the middle 60% of this distribution?


Question 7.13 To enter a particular college course, candidates must complete an aptitude test. In 2013 the mean score was 490 with a standard deviation of 100. The distribution of the scores on the aptitute test is a normal distribution. i) What percentage of candidates scored between 390 and 590 on this aptitute test?

ii) One student scored 795 on this test. How does this student’s score compare to the rest of the scores?


iii) The college only admits students who were among the highest 16% of the scores on this test. What score would a

student need on this test to be qualified for admission to this college?

iv) Alice is preparing to sit the aptitude test in 2014. She heard that a score of over 650 would guarantee her a place on

the course. She knew 20 people who were going to take the test. Based on the mean and standard deviation in 2013,

approximately how many of the people Alice knew were likely to get a score of above 650 and secure a place on the

course? Justify your answer.


Section 2: Percentiles

Percentiles divide a set of data into 100 equal parts. If you were in the 83rd percentile for an exam it would mean 83% of

people got less marks than you.

Note: I am just using exam scores as an example. We can use percentiles for any set of numerical data (height, weight

etc).

Always follow these steps when finding percentile numbers:

Step 1: Step 2: Step 3:

Put the data in order. Mulitiply the percentile by the number of values and then divide by 100. If you get a whole number value, add that value and the next value and then divide by 2.

If you get a decimal number value , round up to the next value.

The value you get locates the required percentile.

THIS IS BEST EXPLAINED WITH AN EXAMPLE:

Example 3 Lets say I gave you the following 40 levels of nicotine (in grams) in smokers and asked you to find the 30th percentile (ie: the nicotine level below which 30% of the data lies).

Lets follow the steps: Step 1: Put the data in order:

Step 2: Mulitiply the percentile by the number of values and then divide by 100:

=30 × 40

100

= 12 Step 3: Since this is a whole number value we add the 12th and 13th values and then divide by 2:

=103 + 112

2

= 107.5 𝑔𝑟𝑎𝑚𝑠


In the example above I gave you a percentile and asked you to find a nicotine level. What if I asked the question another

way around:

Example 4 One particular smoker has a nicotine level of 245. What percentile are they in?

Always follow this rule when finding percentile ranks: Put the number of values less than your number over the total number of values and then multiply by 100:

Using our example above we can see that 28 of the 40 values are less than 245 so according to our rule we have:

=28

40× 100

= 70 ie: The 70th percentile

Question 7.14 The following are weights (in kgs) of tools on a building site:

304 278 266 79 177 763 579 121 703 317 1253 417 614 139 591 639 76 231 741 289 122 973 231 50 274 140 190 467 958 273 128 32 266 691 523 251

i) Display the distribution on an appropriate graph.

ii) Describe the distribution

iii) Calculate the 20th percentile

iv) Calculate the 65th percentile

v) Explain the meaning of these percentiles, in the context of this question.

vi) Some of the workers claim that the tools are too heavy to be stored on a wooden pallet. The construction manager

replies:

“Ah sure about 90% of the tools are lighter than 700kg, get back to work, I’m off to the shop”.

Comment on the construction managers reply, using calculations where necessary.

More space for solution on the next page…


Question 7.15

The final exam test scores in a school were: 71, 78, 62, 94, 89, 87, 75, 84, 62, 66, 75, 85, 81, 83, 92, 93, 95, 99, 85, 89.

Find the percentile rank for a score of 85 on this test.

Question 7.16

The heights of students in inches in a maths class are 59, 64, 68, 72, 61, 63, 64, 55, 60, 59, 74, 69, 68, 65. Find the

percentile rank for a height of 61 inches.

Question 7.17

Which result of 18 exam results would be the 90th percentile?

z-scores actually tell you how many standard deviations you are above or below the mean.

Question 7.18

Which is relatively better, a score of 84 in a Maths test or a score of 48 in an Irish test? Scores on the Maths test have a

mean of 90 and standard deviation of 10. Scores on the Irish test have a mean of 55 and a standard deviation of 5. You

can assume the results of both tests are normally distributed.


8) Inferential Statistics

This section of the syllabus helps you draw conclusions about data, or in other words, it helps you infer conclusions

about data.

Understanding 𝑝 and �̂� (pronounced “𝑝 hat”)

Let’s say we want to find the proportion of people in Ireland who support Liverpool. We decide to do a sample survey

and 47 out of 200 people say they support Liverpool.

𝑝 is the population proportion (this is the value to be estimated).

�̂� is the sample proportion (this is 47

200 in this case).

Standard error and margin of error.

Using the example above, it is very unlikely that our sample survey would be perfectly accurate.

There is a thing called standard error (𝜎𝑝), (we will explain what this actually stands for in a minute). A formula is given

for this in the log tables on page 34:

𝜎𝑝 = √𝑝(1 − 𝑝)

𝑛

Note 1: 𝑛 is the size of the sample and works best when ≥ 30.

Note 2: The problem with this formula is that includes 𝑝 but sure we’re looking for 𝑝!!! So, what you actually do is just

use �̂�, since it is the next best thing!

Now, when we infer conclusions, we might want to be, let’s say 99% confident, especially if the study is really important

(for example a medical study). But sometimes we mightn’t care as much, so, let’s say, 90% confidence is sufficient.

However, on our syllabus the only level of confidence is 95%.

This 95% leads us to the following diagram:

Which leads us to the following diagram:

Which leads us to pages 36/37 of the log tables where we get 𝑧 = 1.96.

This leads us to a formula, called margin of error:

𝐸 = 1.96𝜎𝑝

i.e.:

𝐸 = 1.96√𝑝(1 − 𝑝)

𝑛

Or if we don’t know 𝑝, we use �̂� (as we said):

=> 𝐸 = 1.96√�̂�(1 − �̂�)

𝑛


So, you may be wondering what is the difference between standard error (𝜎𝑝) and margin of error (𝐸).

Well when you find �̂� (the sample proportion), it is a random variable, since it depends on who you sampled. In our

example we got �̂� =47

200 but if we sample another group we might get �̂� =

53

200 and so on. As a random variable, it has a

mean, median, mode, standard deviation etc.

The standard error (𝜎𝑝) is your best guess of the standard deviation, while the margin of error is the “half width of the

confidence interval”, as seen below:

The 95% confidence interval for a proportion

The above work leads us to a formula which we can expect to be asked to use very often in the Leaving Cert. It is called

the 95% confidence interval for a population:

Sample proportion

− Margin of error

≤ True

proportion ≤

Sample proportion

+ Margin of error

�̂� − 1.96𝜎𝑝 ≤ 𝑝 ≤ �̂� + 1.96𝜎𝑝

�̂� − 1.96√𝑝(1 − 𝑝)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

𝑝(1 − 𝑝)

𝑛

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

What this formula means is that if we did the survey numerous times, about 95% of these confidence intervals would

contain the true proportion (and the other 5% wouldn’t).

Note 1: When 𝑝 or �̂� are close to 1

2 a more rough estimate of 𝐸 is 𝐸 =

1

√𝑛 but this formula is more suitable for Ordinary

Level, or when we don’t have �̂�.

Note 2: Another estimate you may see is 𝐸 = 2√𝑝(1−𝑝)

𝑛 which is based on the Empirical Rule.

Note 3: As 𝑛 increases, 𝐸 decreases.

Note 4: As 𝑛 doubles, 𝐸 isn’t halved, as some silly people think.


Example 1

A politician claims that 90% of voters support his views on healthcare. Use a suitable confidence interval to decide whether 435 from 500 is an unusual proportion.

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

435

500− 1.96

√435500

(1 −435500

)

500≤ 𝑝 ≤

435

500+ 1.96

√435500

(1 −435500

)

500

0.841 ≤ 𝑝 ≤ 0.899

Since the 90% (or 0.9) claimed by the politician is outside the confidence interval, his claims seems unusual (spoofer!).

Example 2

Before a big election, newspapers often hold surveys so that they can report on the support for certain candidates. In one particular survey, 45% of the 700 voters indicate that they will vote for a new politician, Mark Buggy. (i) Find the standard error at the 95% confidence level. (ii) Mark Buggy contacts the newspaper who ran the survey and asks them to determine his level of support to

within ±1

2%. What sample size should the newspaper take at the 95% confidence level?

(i)

𝜎𝑝 = √𝑝(1 − 𝑝)

𝑛

= √�̂�(1 − �̂�)

𝑛

= √0.45(1 − 0.45)

700

= 0.01880349512 ≈ 1.9%

(ii) “To within ±1

2%" means that the margin of error should be less than

1

2% (or 0.005).

⇒ 𝐸 < 0.005

1.96√�̂�(1 − �̂�)

𝑛< 0.005

1.96√0.45(1 − 0.45)

𝑛< 0.005

÷ 1.96: √0.2475

𝑛<

1

392

𝑆𝑞𝑢𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠: 0.2475

𝑛<

1

153,664

38,031.84 < 𝑛 ⇒ 𝑛 = 38,032


Question 8.1 In a sample survey of 400 fans taken in the Aviva Stadium, Dublin, it was found that 136 of them thought that the queues for the food stalls were too long. (i) Estimate the percentage of all fans at the stadium who thought that the queues for food stalls were too long?

(ii) Calculate the 95% confidence limits for this estimate and explain briefly what these mean?

(iii) What sample size would have to be taken in order to estimate the percentage to within ±2%?

(iv) Describe how a 99% Confidence interval would compare with a 95% confidence interval.

(v) An earlier survey about fans satisfaction of the toilet queue lengths indicates that fans opinions are evenly split. The surveyors want to carry an accurate survey to within 1% accuracy. What sample size should they take at the 95% confidence level?


Hypothesis testing of �̂�

If someone makes a claim, for example:

“Only 20% of drivers pass their driving test”, we may want to test this. What if we surveyed 400 people and 98 of them

pass their driving test. This is 24.5%, so is the original claim a lie or what’s the craic? To find out, we do what is called a

hypothesis test. There are steps to follow:

Step 1) State the null hypothesis

The null hypothesis, often called 𝐻0, is that nothing is wrong with the claim. (“null” means “nothing”).

So, in our example about driving tests: 𝐻0: 𝑝 = 0.2

Step 2) List the alternative hypothesis

The alternative hypothesis, often called 𝐻1 or 𝐻𝑎 , simply means that 𝐻0 is wrong.

So, in our example about the driving tests: 𝐻1: 𝑝 ≠ 0.2

Note: On other courses you may have 𝑝 > or 𝑝 <, but on our course it’s only 𝑝 ≠.

Step 3) Find the confidence interval:

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

So, in our example about the driving tests:

0.245 − 1.96√0.245(1 − 0.245)

400≤ 𝑝 ≤ 0.245 + 1.96√

0.245(1 − 0.245)

400

0.203 ≤ 𝑝 ≤ 0.287

Step 4) Make a decision

If 𝐻0 is inside the confidence interval, we do not reject it.

If 𝐻0 is outside the confidence interval, we reject it.

So, in our example about the driving tests. 𝐻0 is outside the confidence interval so we reject 𝐻0.

It is important to note that “do not reject” is not the same as saying we “accept” the claim.

You might see this idea in court. Just because someone is “not guilty”, does not mean they are “innocent”. It just

means we couldn’t prove they were guilty.

Note: As with most questions in this section there are various alternative methods for your solution, so don’t

panic if you see other methods.

Example 1

A medical company has started manufacturing a new drug which they claim has an 80% success rate. However a lobby group doubts this claim and carries out a test, in which 1232 of 1600 patients say the drug works. Is this a result in tune with the company’s claim, at the 95% level of confidence? Carry out a hypothesis test to justify your answer. 𝐻0: 𝑝 = 0.8 𝐻1: 𝑝 ≠ 0.8

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

0.77 − 1.96√0.77(1 − 0.77)

1600≤ 𝑝 ≤ 0.77 + 1.96√

0.77(1 − 0.77)

1600

0.749 ≤ 𝑝 ≤ 0.791 𝐻0 is outside the interval so we reject 𝐻0.


Example 2

BBC sport claim that 55% of television viewers in Ireland tune into their station to watch the 6 Nations rugby tournament. To test the validity of this claim, RTE Sport survey some television viewers and admit that the claim by BBC sport cannot be rejected. (i) If the survey size had been 1,000 persons, determine the least and greatest number of viewers that could

have said they tune into BBC sport. (ii) If the outcome of the survey was that 52% of the sample said that they viewed BBC Sport, what is the

greatest possible size of the sample. (i)

�̂� − 1.96√𝑝(1 − 𝑝)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

𝑝(1 − 𝑝)

𝑛

𝑥

1000− 1.96√

0.55(1 − 0.55)

1000≤ 0.55 ≤

𝑥

1000+ 1.96√

0.55(1 − 0.55)

1000

𝑥

1000− 0.031 ≤ 0.55 ≤

𝑥

1000+ 0.031

𝑥

1000− 0.031 ≤ 0.55

𝑥

1000≤ 0.581

𝑥 ≤ 581

0.55 ≤𝑥

1000+ 0.031

0.519 ≤𝑥

1000

519 ≤ 𝑥 => Lowest = 519 Highest =581

(ii)

�̂� − 1.96√𝑝(1 − 𝑝)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

𝑝(1 − 𝑝)

𝑛

0.52 − 1.96√0.55(1 − 0.55)

𝑛≤ 0.55 ≤ 0.52 + 1.96√

0.55(1 − 0.55)

𝑛

0.52 − 1.96√0.2475

𝑛≤ 0.55 ≤ 0.52 + 1.96√

0.2475

𝑛

0.52 − 1.96√0.2475

𝑛≤ 0.55

√0.2475

𝑛≥ −

3

196

0.2475

𝑛≥

9

38416

(0.2475)(38416)

9≥ 𝑛

1056.44 ≥ 𝑛

0.55 ≤ 0.52 + 1.96√0.2475

𝑛

3

196≤ √

0.2475

𝑛

9

38416≤

0.2475

𝑛

𝑛 ≤(0.2475)(38416)

9

𝑛 ≤ 1056.44 => 𝑛 = 1056


Example 3

A crisp manufacturer claims that 8 out of 10 people prefer its brand, called ‘Crispy’. Some rival crisp manufacturers don’t believe this claim so they carry out a survey of n people. (i) What hypothesis should be made? (ii) If the hypothesis is rejected when the percentage of people who prefer ‘Crispy’ is below 78%, calculate n. (iii) Find the greatest possible value of n if a result of 76% is not significant. (i) 𝐻0: 𝑝 = 0.8 (ii) Difference between 0.8 and 0.78 is 0.02

Since it’s rejected when below 78%, then 𝐸 > 0.02

=> 0.02 > 1.96√𝑝(1 − 𝑝)

𝑛

0.02 > 1.96√0.8(1 − 0.8)

𝑛

÷ 1.96: 1

98> √

0.16

𝑛

Square both sides: 1

9604>

0.16

𝑛

𝑛 > (0.16)(9604) 𝑛 > 1536.64

=> 𝑛 = 1537

(iii) Difference between 0.8 and 0.76 is 0.04 If it’s not significant, this means we don’t reject it, so 𝐸 < 0.04

1.96√0.8(1 − 0.8)

𝑛< 0.04

÷ 1.96: √0.16

𝑛 <

1

49

Square both sides: 0.16

𝑛 <

1

2401

(0.16)(2401) < 𝑛 384.16 < 𝑛

=> 𝑛 = 385

Note: You may get full credits for using the simpler version of E (𝐸 =1

√𝑛), however the version we used above is

more accurate.


Question 8.2

A rugby manager has a hypothesis that young players born in the first three months of the year have an advantage over

players born later in the same year, of being selected to represent their country.

(i) Estimate the expected percentage of players born in the first three months of the year.

(ii) For a sample size of 40 players, find the margin of error, at the 95% level of confidence.

(iii) A survey of 400 players finds that 35% of those who represent their country were born in the first three

months of the year. Carry out a hypothesis test, at the 5% level of significance to decide whether there is

sufficient evidence to justify the manager’s claim. State the null hypothesis and state your conclusion.


The Central Limit theorem

Suppose we were to find the mean age of people attending a 1-Direction concert at Croke Park. The mean might be, let’s

say 16 years and the distribution would look as follows:

Now imagine that instead of asking everyone, you took a sample of, let’s say 50 people and calculate their mean age.

And you repeat this several times. If you do this, 3 things happen:

1) The sample means will be distributed normally.

Note: The Central Limit Theorem only works for large samples (30 or more).

2) The mean of the sample means will equal the mean of the population.

3) The standard deviation of the sample means will equal 𝜎

√𝑛.

So, consider the following:

Example 4

Strictly Come Dancing is a TV show in which celebrities partner with professional dancers to perform dancing acts. At the end of each dance, the duo are ranked from 1 to 10. It is known that the mean score is 7.4, which a standard deviation of 2.1. A statistician decides to carry out an analysis of the ratings. He takes numerous sample sizes of 40 duos. Describe the shape, the mean and the standard deviation of these sample means. Shape: Normal Mean = 7.4

S.D.: 𝜎

√𝑛.

=2.1

√40

= 0.332

Example 5

(i)

Shape: normal Mean = €22.05

S. D. = 𝜎

√𝑛

=10.64

√200

= 0.752


(ii)

𝑧 =𝑥 − 𝜇

𝜎

√𝑛

=23 − 22.05

0.752

= 1.26 => 𝑃(𝑧 > 1.26)

= 1 − 𝑃(𝑧 ≤ 1.26) = 1 − 0.8962

= 0.1038 × 1000

103.8 ≈ 103

Question 8.3

A game at a funfair involves throwing a hoop at 100 hooks. The hoop always lands on one of the hooks and is as likely to

land on any one number as any other.

i) Describe the distribution of the hooks landed on as the number of throws increases.

ii) The distribution of all possible selections has a mean of 50 and a standard deviation of 28.868 Samples sizes of 80 are taken from the funfair game and the mean of each sample is calculated. Describe the distribution of the sample means and calculate its mean and standard deviation.

iii) If sample sizes of 5 were taken, would you be able to describe the distribution of the sample means as normal? Justify your answer.


iv) One sample of size 80 is taken. Calculate the probability that its mean is between 52 and 58?

The 95% confidence interval for a population mean.

So far we have met confidence intervals for a proportion but this section is different:

If we are looking for a mean of a population, 𝜇, we would have to measure everyone. Since this is not practical, we will

take a sample (𝑛 ≥ 30), and find its mean, �̅�, and its standard deviation, 𝑠.

The confidence interval is as follows:

�̅� − 1.96𝜎

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝜎

√𝑛

However, the problem with this formula is that we do not know 𝜎 (which stands for the standard deviation of the

population), so instead we use the next best thing, which is ‘𝑠’ (which stands for the standard deviation of the sample):

�̅� − 1.96𝑠

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝑠

√𝑛

Example 6

A random sample of 64 workers was taken, and their annual sick leave was recorded. The mean number of sick leave days of these workers was found to be 14 days, with a standard deviation of 4.4 days. Form a 95% confidence interval for the mean number of sick leave days for all workers in the population.

Formula: �̅� − 1.96𝜎

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝜎

√𝑛

�̅� = 14 𝑠 = 4.4 𝑛 = 64

We use 𝑠 = 4.4 as the standard deviation, 𝜎, of the population is not available. Then the 95% confidence interval for the mean number of sick leave days taken by workers is given by:

�̅� − 1.96𝑠

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝑠

√𝑛

14 − 1.964.4

√64≤ 𝜇 ≤ 14 + 1.96

4.4

√64

12.922 ≤ 𝜇 ≤ 15.078


Question 8.4 A random sample of 100 plants from a garden had their heights measured. The mean was found to be 53cm with a standard deviation of 5cm. Form a 95% confidence interval for the mean height of all plants in the garden.

Hypothesis testing of �̅�

This section uses the same method as the section of ‘Hypothesis testing of �̂�’.

Example 7

- 𝐻0: 𝜇 = 3.5 The mean distance from students’ homes to school is 3.5km. 𝐻1: 𝜇 ≠ 3.5 The mean distance from students’ homes to school is different to 3.5km.

�̅� − 1.96𝑠

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝑠

√𝑛

3.7 − 1.960.5

√60≤ 𝜇 ≤ 3.7 + 1.96

0.5

√60

3.573 ≤ 𝜇 ≤ 3.827 𝐻0 is outside confidence interval so we reject 𝐻0. So the principal’s claim is false.

Notice the way we were asked to “clearly state” things. When the examiner says this you must explain in plain English.

Question 8.5

The mean number of peas in a certain brand of pea packet has been determined from ongoing analysis as 366. The

factory manufacturing these packets purchases a new machine and from it 150 sample packets are taken. These have a

mean of 360 peas, with a standard deviation of 24 peas. Is there sufficient evidence to say that the new machine

produces pea packets with a different mean peas per packet than before, at the 5% level of significance?


𝑝-Values

Instead of using hypothesis tests of �̅�, we can use another method called 𝑝-values, if specifically asked to:

Rule:

𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 2(1 − 𝑃(𝑧 ≤ |𝑇|))

Where 𝑇 is called the test statistic and

𝑇 =�̅� − 𝜇

𝜎

√𝑛

Now this all sounds difficult. Let’s look at a basic example:

Example 8

A confectionary company claims that the mean weight of its packets of crisps is 45g, with a standard deviation of 4g. A random sample of 80 packets is taken and the mean of these is found to be 44g. (i) Calculate the test statistic for this sample. (ii) Determine the 𝑝-value for this sample statistic. (i)

𝑇 =�̅� − 𝜇

𝜎

√𝑛

=44 − 45

4

√80

= −2.24 (ii)

𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 2(1 − 𝑃(𝑧 ≤ |𝑇|))

= 2(1 − 𝑃(𝑧 ≤ |−2.24|))

= 2(1 − 𝑃(𝑧 ≤ 2.24))

= 2(1 − 0.9875) = 0.025

So you might be wondering what these 𝑝 – values are for? Well, all you have to know is the following: If 𝑝-value ≥ 0.05: do not reject 𝐻0.

If 𝑝-value < 0.05: reject 𝐻0.

Memory aid: “If 𝑝 is low, 𝐻0 must go.”

Example 9

An internet provider claims that its service speed is 50Mb/s, with a standard deviation of 1.2Mb/s. To test this claim a random selection of 72 households are tested and found to have a mean speed of 49.7Mb/s. At the 5% level of significance, is there evidence to show that the mean speed is not 50Mb/s? Justify your answer using p-values. 𝐻0: 𝜇 = 50 𝐻1: 𝜇 ≠ 50

𝑇 =�̅� − 𝜇

𝜎

√𝑛

=49.7 − 50

1.2

√72

= −2.12

𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 2(1 − 𝑃(𝑧 ≤ |𝑇|))

= 2(1 − 𝑃(𝑧 ≤ |−2.12|))

= 2(1 − 𝑃(𝑧 ≤ 2.12))

= 2(1 − 0.9830) = 0.034

0.034 < 0.05 => reject 𝐻0


Question 8.6

It is known that the results of a certain exam have a mean score of 70 and a standard deviation of 6. A class of 36

students are given this exam and they receive the mean result of 68.5.

(i) Calculate the sample statistic for this sample.

(ii) Calculate the 𝑝-value for this sample statistic.

(iii) Use the 𝑝-value to determine if the mean score of the sample differs from that of the population, at 5%

significance.


Question 8.7

(a) The mean lifetime of light bulbs produced by a company has, in the past, been 1500 hours. A sample of 100 bulbs,

recently produced by the company, had a mean lifetime of 1475 hours with a standard deviation of 110 hours. Test the hypothesis that the mean lifetime of the bulbs has not changed, using a 0.05 level of significance.

(b) Find the 𝑝-value of the test you performed in a part (a) above and explain what this value represents in the context

of this question.

Summary Points

Although you should know everything in this section, the following few lines are all you need in most exams.

1. �̂� − 1.96√𝑝(1−𝑝)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

𝑝(1−𝑝)

𝑛

2. �̅� − 1.96𝑠

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝑠

√𝑛

3. If inside the confidence interval => do not reject 𝐻0.

If outside the confidence interval => reject 𝐻0.

4. Central Limit Theorem:

(i) Sample means will be normal.

(ii) Mean of sample means = mean of the population.

(iii) Standard deviation of the sample means 𝜎

√𝑛

5. 𝑇 =�̅�−𝜇

𝜎

√𝑛

6. 𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 2(1 − 𝑃(𝑧 ≤ |𝑇|))

7. If 𝑝-value ≥ 0.05: do not reject 𝐻0.

If 𝑝-value < 0.05: reject 𝐻0.

𝑝-value

meaning


9) Solutions to Statistics

Question 1.1

Use the median because it excludes outliers: 12, 14, 15, 15, 15, 15, 16, 16, 16, 16, 18, 18, 18, 18, 18, 20, 20, 20, 21, 21, 22, 22, 24, 70, 73

Question 2.1

Range = maximum value – minimum value = 50.1 – 20.3 = 29.8 seconds Interquartile range = Upper quartile value – Lower quartile value Order the data: 20.3, 21.7, 37.6, 42.9, 44.2, 50.1

Lower quartile value: 1

4× 6

= 1.5 => 2nd value = 21.7

Upper quartile value: 3

4× 6

= 4.5 =>5th value = 44.2 => Interquartile range = 44.2 – 21.7 = 22.5 seconds

Question 2.2

Standard deviation = 18.23915203

Question 2.3

Standard deviation = 1.492201952

Question 2.4

Mean = 3 Standard deviation = 1.104536102

Question 3.1

a)

b)

↑ 𝑚𝑒𝑑𝑖𝑎𝑛 = 18°


Similarity: Mode (both = 3) Difference: All girls do at least 1 sport.

c) Answer: No

Justification: See table above in part (b).

d) Improvement 1: Don’t ask people from a GAA club, as these are more likely to play sports. Improvement 2: Ask a more specific question, as the question asked doesn’t define what ‘playing’ means.

Question 4.1

i)

ii) Use diagram above: => 16% iii)

iv) Use diagram above: => 97.5%

Question 4.2

𝑥 = 172.5 + 2(7)

= 186.5𝑐𝑚

Question 4.3

a)

b)


c)

𝑥 = 490 + 2(100)

= 690

Question 4.4

Case 1 (S.D. = 8%)

Case 2 (S.D. = 16%)

= > We prefer Case 1 because only 16% of the students are better than as opposed to more than 16%

Question 4.5

Mean = c Mode = a Median = b

Question 4.6

a) The data are skewed to the left: C The data are skewed to the right: A The mean is equal to the median: B and D The mean is greater than the median: A There is a single mode: A and B and C

b) Answer: D

Justification: The data is most spread from the mean in this diagram.

Question 4.7

a) (i) 25 (ii) 15 (iii) 50 b) Area: Remains unchanged

Width: As standard deviation increases, width increases. Height: As standard deviation increases, height decreases.

Question 4.8

New standard deviation: d Mean: m + 5


Question 4.9

New standard deviation: 𝑑

8

New mean: 𝑚

8

Question 4.10

i.e. × 6 New mean: 20 × 6 = 120 hours New standard deviation: 5 × 6 =30 hours

Question 4.11

i.e. +2 New minimum: +2 New maximum: +2 New mean: + 2 New standard deviation: - New median: +2 New IQR: -

Question 5.1

a) Top left: Correlation ≈ 0 Top right: Correlation ≈ −0.8 Bottom left: Correlation ≈ 0.4 Bottom right: Correlation ≈ −0.4

b) Answer: 0.76

Question 5.2

Description Correlation coefficient No correlation

Very weak negative correlation Perfect negative correlation Strong negative correlation Strong positive correlation Very weak positive correlation

0 −0.2 −1

−0.8 0.8 0.2

Question 5.3

i)

ii) There is a strong negative correlation, which indicates that the older you are, the lower your time.


Question 5.4

i) Answer: 0 ii) Quadratic

Question 5.5

Just because two variables are correlated does not necessarily mean they cause each other. Example: Number of shark attacks –v- Number of ice-creams sold. Lurking variable: Temperature

Question 6.1

i)

ii) Answer: 0.6232364936

iii) There is a moderately strong positive correlation, which indicates than the longer spent in education; the

higher your annual income.

iv) [see diagram above]

v) €39,000

vi) Pick two points off the line: (13, 35) (18, 55) 𝑥1 𝑦1 𝑥2 𝑦2

𝑚 =𝑦2 − 𝑦1

𝑥2 − 𝑥1

=55 − 35

18 − 13

= 4 vii) The slope indicates that for every additional year spent in education you can expect to make an additional

€4,000.

Question 6.2

(a) Answer: −0.6 (b) Age = 46 Max. heart rate = 137 bpm (c) Answer: 176 bpm (d) Pick two points off the line: (12, 198) (88, 145)

𝑥1 𝑦1 𝑥2 𝑦2

𝑚 =𝑦2 − 𝑦1

𝑥2 − 𝑥1

=145 − 198

88 − 12

= −0.697 (e)

𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1) 𝑦 − 198 = −0.697(𝑥 − 12)

𝑦 − 198 = −0.697𝑥 + 8.364 𝑦 = 206.364 − 0.697𝑥

𝑀𝐻𝑅 = 206.364 − 0.697(𝑎𝑔𝑒)

(f) Pick a young person (age = 20):


New rule: 𝑀𝐻𝑅 = 206.364 − 0.697(20) = 192.424bpm Old rule: 𝑀𝐻𝑅 = 220 − 20 = 200bpm => new rule gives lower MHR for a young person. Now, pick an old person (age = 80): New rule: 𝑀𝐻𝑅 = 206.364 − 0.697(80) = 150.604 𝑏𝑝𝑚 Old rule: 𝑀𝐻𝑅 = 220 − 80 = 140𝑏𝑝𝑚 => new rule gives higher MHR for an old person.

(g) Old rule: 𝑀𝐻𝑅 = 220 − 65 New Rule: 𝑀𝐻𝑅 = 206.364 − 0.697(65) = 155bpm = 161.059bpm × 75% × 75% = 116.25bpm = 120.79bpm

=>speed up a bit ya boyo!

Question 7.1

𝑃(𝑧 > 1.49) = 1 − 𝑃(𝑧 ≤ 1.49) = 1 − 0.9319 = 0.0681

Question 7.2

𝑃(𝑧 ≤ −0.46) = 1 − 𝑃(𝑧 ≤ 0.46) = 1 − 0.6772 = 0.3228

Question 7.3

𝑃(𝑧 > −0.06) = 1 − 𝑃(𝑧 ≤ −0.06) = 1 − [1 − 𝑃(𝑧 ≤ 0.06)] = 1 − [1 − 0.5239] = 0.5239

Question 7.4

𝑃(1.7 ≤ 𝑧 ≤ 2.4) = 𝑃(𝑧 ≤ 2.4) − 𝑃(𝑧 ≤ 1.7) = 0.9918 − 0.9554 = 0.0364

Question 7.5

𝑧 =𝑥 − 𝜇

𝜎

=190 − 150

20

= 2 => 𝑃(𝑧 ≤ 2)

= 0.9772


Question 7.6

𝑧 =𝑥 − 𝜇

𝜎

𝑧 =168 − 180

8

= −1.5

𝑧 =184.5 − 180

8

= 0.5625 => 𝑃(−1.5 ≤ 𝑧 ≤ 0.5625)

= 𝑃(𝑧 ≤ 0.5625) − 𝑃(𝑧 ≤ −1.5) = 𝑃(𝑧 ≤ 0.5625) − [1 − 𝑃(𝑧 ≤ 1.5)]

= 0.7123 − [1 − 0.9332] = 0.6455

Question 7.7

a)

𝑧 =𝑥 − 𝜇

𝜎

=160 − 151

15

= 0.6 => 𝑃(𝑧 < 0.6)

= 0.7257 => 72.57%

b)

𝑧 =𝑥 − 𝜇

𝜎

=180 − 151

15

= 1.93 => 𝑃(𝑧 > 1.93)

= 1 − 𝑃(𝑧 ≤ 1.93) = 1 − 0.9732

= 0.0268 => 2.68%


Question 7.8

a)

b)

𝑧 =𝑥 − 𝜇

𝜎

=295 − 300

6

= −0.83̇

𝑧 =𝑥 − 𝜇

𝜎

=296 − 300

6

= −0. 6̇

𝑧 =𝑥 − 𝜇

𝜎

=295 − 300

6

= 1.16̇

=> 𝑃(𝑧 < −0.83̇)

= 1 − 𝑃(< 0.83̇)

= 1 − 0.7967 = 0.2033

× 1000 𝑝𝑎𝑐𝑘𝑒𝑡𝑠 = 203 𝑝𝑎𝑐𝑘𝑒𝑡𝑠

=> 𝑃(−0. 6̇ ≤ 𝑧 ≤ −1.16̇)

= 𝑃(𝑧 ≤ 1.16̇) − 𝑃(𝑧 ≤ −0. 6̇)

= 𝑃(𝑧 ≤ 1.16̇) − [1 − 𝑃(𝑧 ≤ 0. 6̇)]

= 0.8790 − [1 − 0.7486] = 0.6276

× 1000 𝑝𝑎𝑐𝑘𝑒𝑡𝑠 = 627 𝑝𝑎𝑐𝑘𝑒𝑡𝑠

Question 7.9

Probability of being less than 𝑥 = 60% = 0.6 => z – score = 0.25 but

𝑧 =𝑥 − 𝜇

𝜎

=> 𝑥 = 𝑧𝜎 + 𝜇 = (0.25)(15) + 180

= 183.75cm

Question 7.10

a)

𝑧 =𝑥 − 𝜇

𝜎

=160 − 170

13

= −0.77

𝑧 =𝑥 − 𝜇

𝜎

=185 − 170

13

= 1.15 => 𝑃(−0.77 ≤ 𝑧 ≤ 1.15)

= 𝑃(𝑧 ≤ 1.15) − 𝑃(𝑧 ≤ −0.77) = 𝑃(𝑧 ≤ 1.15) − [1 − 𝑃(𝑧 ≤ 0.77)]

= 0.8749 − [1 − 0.7794] = 0.6543


b)

Probability of being less than 𝑥 = 96% = 0.96 => z – score = 1.75 but

𝑧 =𝑥 − 𝜇

𝜎

=> 𝑥 = 𝑧𝜎 + 𝜇 = (1.75)(13) + 170

= 192.75cm

Question 7.11

a)

𝑧 =𝑥 − 𝜇

𝜎

=19.75 − 20

0.1

= −2.5

𝑧 =𝑥 − 𝜇

𝜎

=20.25 − 20

0.1

= 2.5 => 𝑃(−2.5 ≤ 𝑧 ≤ 2.5)

= 𝑃(𝑧 ≤ 2.5) − 𝑃(𝑧 ≤ −2.5) = 𝑃(𝑧 ≤ 2.5) − [1 − 𝑃(𝑧 ≤ 2.5)]

= 0.9938 − [1 − 0.9938] = 0.9876

=> 0.0124 rejected × 10,000 = 124 batteries rejected

b)

𝑧 =𝑥 − 𝜇

𝜎

=19.75 − 20.05

0.1

= −3

𝑧 =𝑥 − 𝜇

𝜎

=20.25 − 20.05

0.1

= 2 => 𝑃(−3 ≤ 𝑧 ≤ 2)

= 𝑃(𝑧 ≤ 2) − 𝑃(𝑧 ≤ −3) = 𝑃(𝑧 ≤ 2) − [1 − 𝑃(𝑧 ≤ 3)]

= 0.9772 − [1 − 0.9987] = 0.9759

=> 0.0241 rejected

=> % increase =0.0241 − 0.0124

0.0124× 100%

= 94.35%


Question 7.12

(i)

Probability of being less than 𝑥 = 60% = 0.6 => 𝑧 −score = 0.25 but

𝑥 =𝑥 − 𝜇

𝜎

=> 𝑥 = 𝑧𝜎 + 𝜇 = (0.25)(20) + 120

= 125 (ii)


𝑥 =𝑥 − 𝜇

𝜎

=> 𝑥 = 𝑧𝜎 + 𝜇 = (0.84)(20) + 120

= 136.8 Due to the symmetry: y is the same distance from the mean as x is , => y = 103.2

Question 7.13

(i)

𝑧 =𝑥 − 𝜇

𝜎

=390 − 490

100

= −1

𝑧 =𝑥 − 𝜇

𝜎

=590 − 490

100

= 1 => 𝑃(−1 ≤ 𝑧 ≤ 1)

= 𝑃(𝑧 ≤ 1) − 𝑃(𝑧 ≤ −1) = 𝑃(𝑧 ≤ 1) − [1 − 𝑃(𝑧 ≤ 1)]

= 0.8413 − [1 − 0.8413] = 0.6826

= 68.26%


(ii)

𝑧 =𝑥 − 𝜇

𝜎

=795 − 490

100

= 3.05 => 𝑃(𝑧 ≤ 3.05)

= 0.9989 = 99.89%

=> This students is better than 99.89% of students. (iii)


𝑧 =𝑥 − 𝜇

𝜎

=> 𝑥 = 𝑧𝜎 + 𝜇 = (0.99)(100) + 490

= 589 (iv)

𝑧 =𝑥 − 𝜇

𝜎

=650 − 490

100

= 1.6 => 𝑃(𝑧 > 1.6)

= 1 − 𝑃(𝑧 ≤ 1.6) = 1 − 0.9452

= 0.0548 × 20 people = 1 person

490


Question 7.14

(i)

(ii) Roughly right skewed. (iii) Step 1: Put the data in order

32, 50, 76, 79, 121, 122, 128, 139, 140, 177, 190, 231, 231, 251, 266, 266, 273, 274, 278, 289, 304, 317, 417, 467, 523, 579, 591, 614, 639, 691, 703, 741, 763, 958, 973, 1253.

Step 2: 20×36

100= 7.2

Step 3: i.e. a decimal => use 8th value =139 kg

(iv) Step 1: see above

Step 2: 65×36

100= 23.4

Step 3: i.e. a decimal => use 24th value =467 kg

(v) 20% of the weights are less than 139kg and 65% of the weights are less than 467 kg

(vi) 30

36 are lighter than 700kg,, i.e. 83.3%

The manager is not fully correct.

Question 7.15

10

20 of the scores are less than 85, i.e: 50th percentile.

Question 7.16

4

14 of the heights are less than 61 ie: 28.5714285th percentile.

Questions 7.17

90 × 18

100= 16.2

=> the 17th result


Question 7.18

Maths:

𝑧 =𝑥 − 𝜇

𝜎

=84 − 90

10

= −0.6

Irish:

𝑧 =𝑥 − 𝜇

𝜎

=48 − 55

5

= −1.4

i.e. Irish is more standard deviations below the mean . => Maths is better

Question 8.1

(i)

�̂� =136

400

= 0.34 = 34%

(ii)

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

0.34 − 1.96√0.34(1 − 0.34)

400≤ 𝑝 ≤ 0.34 + 1.96√

0.34(1 − 0.34)

400

0.294 ≤ 𝑝 ≤ 0.386 95% of these confidence intervals would contain the true proportion.

(iii) “To within ±2%" means that the margin of error should be less than 2% (or 0.02).”

⇒ 𝐸 < 0.02

1.96√�̂�(1 − �̂�)

𝑛< 0.02

1.96√0.34(1 − 0.34)

𝑛< 0.02

÷ 1.96: √0.2244

𝑛<

1

98

𝑆𝑞𝑢𝑎𝑟𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠: 0.2244

𝑛<

1

9,604

2,155.1376 < 𝑛 ⇒ 𝑛 = 2,156

(iv) Same centre but wider, since 99% would give a bigger z-score than 95% on pages 36/37 of the log tables,

meaning E would be bigger.

(v) Since we haven’t got �̂�, we will use 𝐸 =1

√𝑛

0.01 =1

√𝑛

0.01√𝑛 = 1

√𝑛 =1

0.01

√𝑛 = 100 𝑛 = 10,000


Question 8.2

(i)

=3

12

= 25% (ii)

𝐸 = 1.96√𝑝(1 − 𝑝)

𝑛

= 1.96√0.25(1 − 0.25)

400

=0.042 (iii)

𝐻0: 𝑝 = 0.25 𝐻1: 𝑝 ≠ 0.25

�̂� − 1.96√𝑝(1 − 𝑝)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

𝑝(1 − 𝑝)

𝑛

0.35 − 0.042 ≤ 𝑝 ≤ 0.35 + 0.042 0.308 ≤ 𝑝 ≤ 0.392

𝐻0 is outside the confidence interval so we reject 𝐻0. Therefore, fail to reject the manager’s claim.

Question 8.3

(i) Linear (ii)

Shape: normal Mean = 50

S. D. = 28.868

√80

= 3.227540519 (iii) No the Central Limit Theorem only works for large samples (30 or more). (iv)

𝑧 =𝑥 − 𝜇

𝜎

√𝑛

𝑧 =52 − 50

3.227540519

= 0.6196689 = 0.062

𝑧 =58 − 50

3.227540519

= 2.478667575 = 2.48

𝑃(0.62 ≤ 𝑧 ≤ 2.48) = 𝑃(𝑧 ≤ 2.48) − 𝑃(𝑧 ≤ 0.62)

= 0.9934 − 0.7342 = 0.26


Question 8.4

Formula: �̅� − 1.96𝜎

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝜎

√𝑛

�̅� = 53 𝑠 = 5

𝑛 = 100 We use 𝑠 = 5 as the standard deviation, 𝜎, of the population is not available:

=> �̅� − 1.96𝑠

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝑠

√𝑛

53 − 1.965

√100≤ 𝜇 ≤ 53 + 1.96

5

√100

52.02 ≤ 𝜇 ≤ 53.98

Question 8.5

𝐻0: 𝜇 = 366 𝐻1: 𝜇 ≠ 366

�̅� − 1.96𝑠

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝑠

√𝑛

360 − 1.9624

√150≤ 𝜇 ≤ 360 + 1.96

24

√150

356.159 ≤ 𝜇 ≤ 363.841 𝐻0 is outside confidence interval so we reject 𝐻0.

Question 8.6

(i)

𝑇 =�̅� − 𝜇

𝜎

√𝑛

=68.5 − 70

6

√36

= −1.5 (ii)

𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 2(1 − 𝑃(𝑧 ≤ |𝑇|))

= 2(1 − 𝑃(𝑧 ≤ |−1.5|))

= 2(1 − 𝑃(𝑧 ≤ 1.5))

= 2(1 − 0.9332) = 0.1336

(iii) 0.1336 ≥ 0.05 => do not reject 𝐻0


Question 8.7

(a) 𝐻0: 𝜇 = 1500 𝐻1: 𝜇 ≠ 1500

�̅� − 1.96𝑠

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝑠

√𝑛

1475 − 1.96110

√100≤ 𝜇 ≤ 1475 + 1.96

110

√100

1453.44 ≤ 𝜇 ≤ 1496.56 1500 is outside the confidence interval.

=> reject 𝐻0 (b) 𝑝-value:

𝑇 =�̅� − 𝜇

𝜎

√𝑛

=1475 − 1500

110

√100

= −2.27

𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 2(1 − 𝑃(𝑧 ≤ |𝑇|))

= 2(1 − 𝑃(𝑧 ≤ |−2.27|))

= 2(1 − 𝑃(𝑧 ≤ 2.27))

= 2(1 − 0.9884) = 0.0232

0.0232 < 0.05 => reject 𝐻0

meaning: The mean lifetime of the bulbs has changed.


Probability Theory

The fundamental principle of counting:

If one event has 𝑚 possible outcomes and a second event has 𝑛 possible outcomes, then the total number of possible

outcomes is 𝑚 × 𝑛.

A sample space is the set of all possible outcomes of an experiment.

Problems: There are 6 types of problems in Probability:

1) And/ or

Rule 1: In probability “and” means mulitply.

Rule 2: In probability “or” means add.

Example 1 Two die are rolled. Find the probability that you get a 5 on the first dice and a number greater than 2 on the second dice.

𝑃(𝑛𝑢𝑚𝑏𝑒𝑟 = 5) =1

6

𝑃(𝑛𝑢𝑚𝑏𝑒𝑟 > 2) =4

6

𝑃(𝑛𝑢𝑚𝑏𝑒𝑟 = 5 𝑎𝑛𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 > 2) =1

6×

4

6

=4

36

=1

9

Example 2 There are 20 cars in a carpark. 7 are black, 4 are red, 3 are white and 6 are blue. Find the probability that a car chosen at random is white or red.

𝑃(𝑤ℎ𝑖𝑡𝑒) =3

20

𝑃(𝑟𝑒𝑑) =4

20

𝑃(𝑤ℎ𝑖𝑡𝑒 𝑜𝑟 𝑟𝑒𝑑) =3

20+

4

20

=7

20

Question 1.1


Question 1.2

Each night Kim goes to the gym or the pool. If she goes to the gym one night, the probability she goes to the pool the

next night is 0.4, and if she goes to the pool one night, the probability she goes to the gym the next night is 0.7. Suppose

she goes to the gym one Monday night

(i)What is the probability that she goes to the pool on each of the next three nights?

(ii) What is the probability that she goes to the pool on exactly two of the next three nights after Monday?


Note: You are required to know what is in a deck of cards:

Standard Deck of 52 Playing Cards:

Diamonds (Red): 2♦ 3♦ 4♦ 5♦ 6♦ 7♦ 8♦ 9♦ 10 ♦ J♦ Q♦ K♦ A♦

Hearts (Red): 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ Q♥ K♥ A♥

Clubs (Black): 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣ A♣

Spades(Black): 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♠

What is the probability of choosing, at random, one of the following cards from a normal pack of 52 playing cards?

1. A red card 26

52 =

1

2

2. A black card or ‘not a red card’ 26

52 =

1

2

3. A spade 13

52 =

1

4

4. Not a spade 39

52 =

3

4

5. An ace 4

52 =

1

13

6. Not an ace 48

52 =

12

13

7. The ace of spades 1

52

8. A picture card 12

52 =

3

13

9. A number card or ‘not a picture card’ 40

52 =

10

13

10. A card that is either a heart or a club 26

52 = ½

11. A card that is neither a heart or a club 26

52 = ½

12. A 4 or 5 8

52 =

2

13

13. A 4 or 5 but not a spade 6

52 =

3

26

14. An even numbered card 20

52 =

5

13

Picture cards


Question 1.3


2) Relative frequency

This is all about using the past to predict the future.

The more trials we have from the past the more accurate our prediction of the future will be.

Example 1 Last season Lionel Messi took 25 penalties and scored 20. a) What is the relative frequency of Messi scoring a penalty?

𝐴𝑛𝑠𝑤𝑒𝑟: 20

25 (𝑜𝑟 80%)

b) If Messi takes 15 penalties next season, how many would you expect him to score? 𝐴𝑛𝑠𝑤𝑒𝑟: 80% 𝑜𝑓 15 = 12

Question 2.1

Three students have four coins, which they flip repeatedly. They keep a record if all four coins show the same outcome

or not (that is, whether or not they get HHHH or TTTT). Here are the results:

Name Number of Trials All four coins show the

same outcome All four coins do NOT

show the same outcome Peter 40 7 33 Mary 30 3 27 Tim 90 11 79 Total 160 21 139

i) Which students’s data is likely to give the best estimate of the probability of getting all four coins to show the same outcome? Give a reason for your answer.

ii) The actual probability that all four coins will show the same outcome is 1

8 . How many times would you

theoretically expect this to happen in 160 trials?

iii) Give a reason why the total result of the students experiment is different from what is theoretically expected.


3) Probability rules

Some text books completely over-do this topic, making it confusing. All you need to know is the following 4 rules:

1) Union probability: 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)

2) Conditional Probability (ie: the probability of 𝐴 given 𝐵):

𝑃(𝐴 | 𝐵) =𝑃(𝐴 ∩ 𝐵)

𝑃(𝐵)

3) Mutually exclusive events: 𝐴 ∩ 𝐵 = ∅

Two events are mutually exclusive if they cannot occur at the same time.

eg. Rolling a dice and getting an even number and the number 5.

4) Independent events: 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴). 𝑃(𝐵)

Two events are independent if the outcome of one does not affect the other .

eg. Rolling a dice and tossing a coin.

Question 3.1


Question 3.2

Question 3.3


Question 3.4

Question 3.5

In a certain school, 31% of students got an A in Junior Cert Maths, 16% got an A in Junior Cert Science and 10% got an A

in both. A student is selected at random. Find the probability that:

i) This student got an A in Maths, given that she got an A in Science.

ii) This student got an A in Science, given that she got an A in Maths.


iii) This student got an A in Maths or Science.

iv) This student got an A in neither Maths nor Science.


Question 3.6

The Venn diagram shows the probability of events A, B & C.

Find the following:

i) 𝑃(𝐴) ii) 𝑃(𝐴 ∩ 𝐵) iii) 𝑃(𝐴|𝐵) iv) 𝑃(𝐵|𝐴) v) 𝑃(𝐴|(𝐵 ∪ 𝐶)) vi) 𝑃(𝐶|(𝐴 ∩ 𝐵))


Question 3.7

On a randomly chosen day, the probability that Bill travels to school by car, by bicycle or on foot is 1

2,

1

6 and

1

3

respectively. The probability of being late when using these methods of travel is 1

5,

2

5 and

1

10 respectively.

Find the probability that on a randomly chosen day: (i) Bill travels by foot and is late.

(ii) Bill is not late.

(iii) Given that Bill is late, find the probability that he did not travel on foot.


4) Binomial distribution (also known as Bernoulli trials)

Bernoulli is a buzzer. This question is to be expected on this year’s Leaving Cert.

𝑛 stands for the number of trials

𝑟 stands for the desired number of successes

𝑝 stands for the probability of success

𝑞 stands for the probability of failure

Note: 𝑝 + 𝑞 = 1

Note: 𝑝 & 𝑞 must be written as a decimal or fraction and not as a percentage.

=> If you know 𝑝 you can find 𝑞.

You need to know the following theory:

4 requirements for Bernoulli:

There is a finite number of trials.

There are only two outcomes - success and failure.

Trials are independent of each other.

The probability of success is the same for each trial.

Note: The ( ) is a button on your calculator called the ‘C’ button. Casio: above the ÷ button. Sharp: above the number 5 button. Question 4.1

The probability that David answers any one phone call is 70%.

His phone rings 10 times

a) Find, correct to 3 decimal places, the probability that he answer 8 times?

b) Find, correct to 4 decimal places, the probability he answers half of the time?

c) Find, correct to 3 decimal places, the probability that he answers all the calls?


d) Find, correct to 3 significant figures, the probability that he answers none of the calls.

e) Find, correct to 3 decimal places, the probability that he answers fewer than nine times.

f) Find, correct to 3 decimal places, the probability that he answers at least 3 phone calls.

g) Find, the probability that he answers his first call on the 4th call.


h) Find, correct to 3 decimal places, the probability that he answers his 3rd call on the 8th call.

Question 4.2


(ii)


Question 4.3


Question 4.4


Question 4.5


Question 4.6

A certain basketball player scores 60% of the free-throw shots she attempts. During a particular game, she gets six free-

throws.

a) What assumption(s) must be made in order to regard this as a sequence of Bernoulli trials?

b) Based on such assumptions, find, correct to three decimal places, the probability that:

i) she scores on exactly four of the six shots

ii) she scores for the second time on the fifth shot.


5) Expected value

DSOG rule:

1) List the possible outcomes 2) Multiply the values by the probabilities for each outcome 3) Sum up these answers

Example 1 If you spin the following spinner, how much do you expect to win, if it is free to play?

Note that the expected value isn’t always one of the actual real life values. It is the mean value you would get if you

played the game ‘many’ times.

Also note that values can have a minus value (ie: you may lose money, although that isn’t possible in the game above).

Question 5.1 The spinner on a wheel of fortune can land with an equal chance on any one of ten regions. Three regions are red, four

are blue, two are yellow and one is green. A player wins €4 if the spinner stops on red and €2 if it stops on green. The

player loses €2 if it stops on blue and €3 if it stops on yellow. Find the expected value for this game?

Answer:

€5: €5 ×1

2 = €2.50

€10: €10 × 1

4= €2.50

€15: €15 × 1

4= €3.75

€8.75 ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅ =>Expected to win €8.75


Question 5.2 A carnival game consists of tossing a dart, which lands at a random spot

within the square target seen. The lined circle wins €5, the non-lined circle

wins €7 and either black triangle wins €10. If it costs €5 to play this game,

do you expect to make money or lose money?


Question 5.3


6) Factorial/ arranging/ choosing/ hangman

There is often confusion over when and when not to use the following:

Factorials…..denoted by !

Arranging…..denoted by 𝑛 𝑃 𝑟

Choosing….. denoted by by 𝑛 𝐶 𝑟

Hangman method…..denoted by _ × _ × _ × _

Students often ask when does order matter?! When does order not matter?!

Lets clear up this confusion…

Note: There are many correct ways to do each question in this section. I always stick to my rules. I recommend you do

too. Trust me.

1) Factorials…..denoted by !

Casio: above the 𝑥−1button near the top right

Sharp: above the number 4 button

We use ‘factorials’ under the following 4 conditions:

i) We are arranging things

ii) There are no restrictions

iii) Order matters

iv) No repititions are allowed

Example 1 How many ways can you arrange Tom, Nick, Mary and Sarah for a photograph? Now, looking at the 4 conditions i) We are arranging because the question uses the word ‘arrange’.

ii) We have been given no restrictions. iii) Unless a question tells us that order doesn’t matter, then we assume that it does matter. So, in this arrangment: ‘Tom, Nick, Mary, Sarah’ is different that than, lets say: ‘Mary, Nick, Sarah, Tom’. (Note: There is an exception to this rule as seen in ‘choosing’ on the next page). iv) In this example we can’t physically have repititions, because a person can’t be in two or more places at one time. All 4 requirements are satisfied, so we can use factorials. => Answer = 4! = 24


2) Arranging…..denoted by 𝑛 𝑃 𝑟

Casio: above the × button


We use ‘arranging’ under the following 4 requirements:

i) We are arranging things

ii) There are restrictions

iii) Order matters

iv) No repititions are allowed

So you can see that these are the same as factorials with one exception: There are restrictions

Example 2 How many ways can you arrange Tom, Nick, Mary and Sarah for a photograph, taking 2 at a time. This is very similar to the last question except that now there is a clear restriction, in that only 2 people can pose for the photo, so we use 𝑛 𝑃 𝑟 Answer: 4 𝑃 2 = 12

3) Choosing, denoted by 𝑛 𝐶 𝑟

Casio: above the ÷ button


We use ‘choosing’ under the following 3 requirements:

i) We are choosing a number of things less than the total

ii) Order doesn’t matter

iii) No repetitions are allowed

Example 3 Tom, Nick, Mary and Sarah are sitting at the end of Dun Laoghaire Pier. Their friend Johnny comes along and wants to bring two of them to Teddys. How many choices does Johnny have? Now, looking at the 3 requirements. i) We are choosing 2 from 4

ii) Nick and Mary would be the same as Mary and Nick so order doesn’t matter. iii) There can’t be repititions because one person cannot be two people!!

All 3 requirements are satisfied so we can use 𝑛 𝐶 𝑟 Answer = 4 𝐶 2 = 6

Note: Very often the wording of the question gives you a hint. If you see the words ‘choose’ or ‘choices’ it is hinting to

use 𝑛 𝐶 𝑟. If you see ‘arranging’ or ‘arrangments’ it is hinting to use ! or 𝑛 𝑃 𝑟.

Before we look at the “hangman method” lets look at some examples which don’t involve people. These are different

because you may think repetitions are allowed. Unless they say otherwise, we assume repetitions are not allowed.

Example 4 How many ways can the word FOX be arranged? Answer: 3! = 6


Example 5 In how many ways can the word BRIAN be arranged, taking three letters at a time? Answer: = 5 𝑃 3 = 60

Example 6 In how many ways can the word FACEBOOK be arranged, taking five letters at a time? Be careful!! The letter O is repeated twice, so for example FBKOO would be the same as FBKOO even though the O’s have been swapped around. Therefore after we do 8 𝑃 5 we must divide our answer by the amount of ways that the two O’s can be arranged, which is 2!

=> Answer =8 𝑃 5

2!

= 3360

Example 7 How many ways can you arrange the following numbers: 57, 53, 53, 59, 60, 53, 64, There are 7 numbers to arrange but one of them is repeated 3 times

=>Answer =7!

3!

= 840

Example 8 How many ways can you arrange the word HARVART. Be careful! The letters A and R are repeated twice, therefore after we do 7! we must divide by 2! 2!

Answer =7!

2!2!

= 1260

Although key words can often be a hint, sometimes there are no key words:

Example 9

Solution: i) Now from geometry we would know a quadrilateral is a four sided figure, which means it has four corners or ‘vertices’. So essentially what we are being asked here is how many ways we can choose 4 vertices from the 6 on the circle. The word ‘choose’ is essentially hidden in the question! => Answer = 6 𝐶 4 = 15 ii) Here they are telling us the ‘a’ and ‘b’ are already chosen, so to make up the quadrilateral we need another 2 vertices from the remaining 4. => Answer = 4 𝐶 2


= 6 Example 10 A table tennis club has 10 players: 7 men and 3 women. 5 players must be chosen to represent the club in a national competition. How many choices are there if i) There are no restrictions. ii) There must be more men than women. Solution: i) This part is quite easy, we just have to choose 5 from 10. All requirements for ‘choosing’ are satisfied so we use 𝑛 𝐶 𝑟 Answer: = 10 𝐶 5 = 252 ii) There is a restriction in place here, so we can’t just use 𝑛 𝐶 𝑟. However, we can use it a few times. Think about it, if there must be more men than woman the options are as follows: 5 men and 0 women or 4 men and 1 woman or 3 men and 2 women.

Now the way to choose 5 men from 7 is 7 𝐶 5 The way to choose 0 women from the 3 is 3 𝐶 0 The way to choose 4 men from the 7 is 7 𝐶 4 The way to choose 1 woman from the 3 is 3 𝐶 1 The way to choose 3 men from the 7 is 7 𝐶 3 The way to choose 2 women from the 3 is 3 𝐶 2

But remember in probability ‘and’ means multiply and ‘or’ means add. Therefore we have: (7 𝐶 5 × 3 𝐶 0) + (7 𝐶 4 × 3 𝐶 1) + (7 𝐶 3 × 3 𝐶 2) = 231

4) Hangmen method…. denoted by _ × _ × _ × _

This method does not have a button on your calculator.

We use it when none of the other methods work, or are hard to work. This is usually when there are restrictions of

some sort in the question.

Example 11

Solution: i) This part is a simple arrangement question with no restrictions so we use factorials. Answer = 4! = 24 ii) Here there is a restriction. For the number to be greater than 6,000 the first number must be 6 or 8. ie: We have 2 options for the first digit so we write:

2 × _ × _ × _ Note: We haven’t used the 2 of our options, we have only used 1 option. So, now we have 3 options left for the second space, 2 options left for the third space and 1 option left for the last space.

2 × 3 × 2 × 1 = 12

iii) Here we only have 1 option for the last space: _ × _ × _ × 1

Then we have 3 options left for the first space, 2 options left for the second space and one option left for the third space.

= 3 × 2 × 1 × 1 = 6


Example 12 How many ways can the word LEAVING be arranged if: i)There are no restrictions ii) It must end with V iii) It must begin with a vowel iv) The 3 vowels must be together v) G and V must be together vi) G and V must be apart Solution: i) This part is a simple arrangement question with no restrictions so we use factorials. Answer = 7! = 5040 ii) Here the V is stuck at the end so I will use the hangman method:

= _ × _ × _ × _ × _ × _ × 1 = 6 × 5 × 4 × 3 × 2 × 1 × 1

= 720 iii) Here I will use the hangman method. On each space I list out the number of options for that space. I begin by filling in the 3 options for the first space:

= 3 × _ × _ × _ × _ × _ × _ Now 1 letter has been used (Note: only 1 letter has be used, not 3. We had 3 options but we only used 1, don’t get this confused). Therefore I have 6 letters as options for the second space, then 5 for the next space and so on:

= 3 × 6 × 5 × 4 × 3 × 2 × 1 = 2,160

iv) If the three vowels must be together we treat them as ‘one thing’. So our ‘things’ are:

So we have 5 things to arrange so we use 5! But the 3 vowels can be arranged in 3! ways themselves. => Answer = 5! × 3! = 720 v) This is similar to the previous part of the question. Treat G and V as ‘one thing’ so we have 6 ‘things’ to arrange, so we use 6! But the G and V can be arranged 2! ways themselves => Answer = 6! × 2! = 1,440 vi) We know from part (i) that there are 5,040 ways of arranging the letters. We know from part (v) that 1,440 of these have G and V together. Therefore 5,040 - 1,440 will have G and V apart. Answer = 5,040 − 1,440 = 3,600


Question 6.1 A football squad has 3 goalkeepers and 17 outfielders. How many ways can the manager choose a team if it must have 1 goalkeeper and 10 outfielders?

Question 6.2 How many ways can the word DUBLIN be arranged if the vowels must be apart?


7) Solutions to probability

Question 1.1

= (𝐵𝑙𝑢𝑒) 𝑎𝑛𝑑 (𝑅𝑒𝑑) 𝑎𝑛𝑑 (𝑅𝑒𝑑 𝑜𝑟 𝐺𝑟𝑒𝑒𝑛)

= (6

13) × (

4

12) × (

3

11 +

3

11)

=12

143

Question 1.2

(i) = (𝑃𝑜𝑜𝑙) 𝑎𝑛𝑑 (𝑃𝑜𝑜𝑙) 𝑎𝑛𝑑 (𝑃𝑜𝑜𝑙) = (0.4) × (0.3) × (0.3) = 0.036

(ii) = (𝑃𝑜𝑜𝑙) 𝑎𝑛𝑑 (𝑃𝑜𝑜𝑙) 𝑎𝑛𝑑 (𝐺𝑦𝑚) or = (𝑃𝑜𝑜𝑙) 𝑎𝑛𝑑 (𝐺𝑦𝑚) 𝑎𝑛𝑑 (𝑃𝑜𝑜𝑙) or = (𝐺𝑦𝑚) 𝑎𝑛𝑑 (𝑃𝑜𝑜𝑙) 𝑎𝑛𝑑 (𝑃𝑜𝑜𝑙) = (0.4) × (0.3) × (0.7) + = (0.4) × (0.7) × (0.4) + = (0.6) × (0.4) × (0.3) = 0.268

Question 1.3

a)

b) (i) =

1

52

(ii) From the diagram above: 22

52

=11

26

(iii) =30

52×

29

51

= 0.33

Question 2.1

(i) Tim, because he had the most trials, meaning he had the least sampling error.

(ii) 160 ×1

8= 20

(iii) Sampling error


Question 3.1

(a)

(b) From diagram above: 0.2

(c) 𝑃(𝐴|𝐵) =𝑃(𝐴∩𝐵)

𝑃(𝐵)

=0.15

0.75

= 0.2 (d) If independent: 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴). 𝑃(𝐵)

0.15 = (0.2)(0.75) 0.15 = 0.15 => Independent

Question 3.2

a) 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) = 0.7 + 0.5 − 0.3 = 0.9

b) 𝑃(𝐴|𝐵) =𝑃(𝐴∩𝐵)

𝑃(𝐵)

=0.3

0.5

= 0.6 c) If independent: 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴). 𝑃(𝐵)

0.3 = (0.7)(0.5) 0.3 ≠ 0.35 => Not independent

Question 3.3

Independent => 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴). 𝑃(𝐵) 𝑃(𝐴 ∩ 𝐵) = 0.25𝑥

But 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) 0.55 = 0.25 + 𝑥 − 0.25𝑥 0.3 = 0.75𝑥

0.3

0.75= 𝑥

0.4 = 𝑥

Question 3.4

(i) Independent => 𝑃(𝐸1 ∩ 𝐸2) = 𝑃(𝐸1). 𝑃(𝐸2)

𝑃(𝐸1 ∩ 𝐸2) = (1

5) (

1

7)

=1

35

(ii) 𝑃(𝐸1 ∪ 𝐸2) = 𝑃(𝐸1) + 𝑃(𝐸2) − 𝑃(𝐸1 ∩ 𝐸2)

= (1

5) + (

1

7) − (

1

35)

=11

35


Question 3.5

(i) 𝑃(𝑀|𝑆) =𝑃(𝑀∩𝑆)

𝑃(𝑆)

=0.1

0.16

= 0.625

(ii) 𝑃(𝑆|𝑀) =𝑃(𝑆∩𝑀)

𝑃(𝑀)

=0.1

0.31

= 0.322580645 (iii) Maybe a Venn Diagram will help:

=> 0.37

(iv) From the diagram above: 0.63

Question 3.6

(i) 𝑃(𝐴) = 0.4 + 0.09 + 0.07 + 0.1 = 0.66

(ii) 𝑃(𝐴 ∩ 𝐵) = 0.09 + 0.07 = 0.16

(iii) 𝑃(𝐴|𝐵) =𝑃(𝐴∩𝐵)

𝑃(𝐵)

=0.16

0.09+0.07+0.1+0.04

= 0.53̇

(iv) 𝑃(𝐵|𝐴) =𝑃(𝐵∩𝐴)

𝑃(𝐴)

=0.16

0.66

= 0. 2̇4̇

(v) (𝐴|(𝐵 ∪ 𝐶)) =𝑃(𝐴∩(𝐵∪𝐶))

𝑃(𝐵∪𝐶)

=0.09+0.07+0.1

0.09+0.07+0.1+0.04+0.1+0.15

= 0.47̇2̇

(vi) (𝐶|(𝐴 ∩ 𝐵)) =𝑃(𝐶∩(𝐴∩𝐵))

𝑃(𝐴∩𝐵)

=0.07

0.16

= 0.4375

Question 3.7

(i) = (𝐹𝑜𝑜𝑡) 𝑎𝑛𝑑 (𝐿𝑎𝑡𝑒)

= (1

3) × (

1

10)

=1

30

(ii) = (𝐶𝑎𝑟) 𝑎𝑛𝑑 (𝑁𝑜𝑡 𝐿𝑎𝑡𝑒) 𝑜𝑟 (𝐵𝑖𝑘𝑒) 𝑎𝑛𝑑 (𝑁𝑜𝑡 𝐿𝑎𝑡𝑒) 𝑜𝑟 (𝐹𝑜𝑜𝑡) 𝑎𝑛𝑑 (𝑁𝑜𝑡 𝐿𝑎𝑡𝑒)

= (1

2) × (

4

5) + (

1

6) × (

3

5) + (

1

3) × (

9

10)

=4

5

(iii) 𝑃(𝑁𝑜𝑡 𝑓𝑜𝑜𝑡|𝐿𝑎𝑡𝑒) =𝑃(𝑁𝑜𝑡 𝑓𝑜𝑜𝑡 ∩ 𝐿𝑎𝑡𝑒)

𝑃(𝐿𝑎𝑡𝑒)

Ie: =(𝐶𝑎𝑟) 𝑎𝑛𝑑 (𝐿𝑎𝑡𝑒) 𝑜𝑟 (𝐵𝑖𝑘𝑒) 𝑎𝑛𝑑 (𝐿𝑎𝑡𝑒)

1

5

=(

12

) × (15

) + (16

) × (25

)

15

=5

6


Question 4.1

(a) (𝑛𝑟

) 𝑝𝑟𝑞𝑛−𝑟

= (108

) (0.7)8(0.3)2

= 0.233

(b) (𝑛𝑟


= (105

) (0.7)5(0.3)5

= 0.1029

(c) (𝑛𝑟


= (1010

) (0.7)10(0.3)0

= 0.028

(d) (𝑛𝑟


= (100

) (0.7)0(0.3)10

= 5.9049 × 10−6 = 0.0000059049 = 0.00000590 (see Number Systes chapter for explanation on rounding off to significant figures).

(e) Fewer than 9 means 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 Alternatively: 1 − (9 𝑡𝑖𝑚𝑒𝑠 𝑜𝑟 10 𝑡𝑖𝑚𝑒𝑠)

= 1 − [(109

) (0.7)9(0.3)1 + 0.028]

=243

250

(f) At least 3 means 3 or 4 or 5 or 6 or 7 or 8 or 9 or 10. Alternatively: 1 − (0 𝑡𝑖𝑚𝑒𝑠 𝑜𝑟 1 𝑡𝑖𝑚𝑒 𝑜𝑟 2 𝑡𝑖𝑚𝑒𝑠)

= 1 − [0.00000590 + (101

) (0.7)1(0.3)9 + (102

) (0.7)2(0.3)8]

= 0.998 (g) This means he: (𝑚𝑖𝑠𝑠𝑒𝑠) 𝑎𝑛𝑑 (𝑚𝑖𝑠𝑠𝑒𝑠) 𝑎𝑛𝑑 (𝑚𝑖𝑠𝑠𝑒𝑠) 𝑎𝑛𝑑 (𝑎𝑛𝑠𝑤𝑒𝑟𝑠)

= 0.3 × 0.3 × 0.3 × 0.7 = 0.189

(h) This means he answers (2 out of 7), and then the 8th

= (72

) (0.7)2(0.3)5 × 0.7

= 0.018

Question 4.2

(a) (i) p is the probability of success.

q is the probability of failure. n is the total number of trials. r is the desired number of successes.

(ii) There is a finite number of trials. There are only two outcomes - success and failure. Trials are independent of each other. The probability of success is the same for each trial.

(b)

(i) (𝑛𝑟


= (109

) (2

3)

9

(1

3)

1

= 0.087 (ii) Fewer than 9 means 0 or 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8

Alternatively: 1 − (9 𝑡𝑖𝑚𝑒𝑠 𝑜𝑟 10 𝑡𝑖𝑚𝑒𝑠)

= 1 − [0.087 + (1010

) (2

3)

10

(1

3)

0

]

= 0.896


Question 4.3

(i) (𝑛𝑟


= (40

) (4

5)

0

(1

5)

4

=1

625

(ii) (𝑛𝑟


= (43

) (4

5)

3

(1

5)

1

=256

625

(iii) At least 8 means 8 or 9 or 10

= (108

) (4

5)

8

(1

5)

2

+ (109

) (4

5)

9

(1

5)

1

+ (1010

) (4

5)

10

(1

5)

0

= 0.6777995264

Question 4.4

(i) (𝑛𝑟


= (60

) (3

4)

0

(1

4)

6

=1

4096

(ii) ie: 0 out of 2 and 4 out of 4

= (20

) (3

4)

0

(1

4)

2

× (44

) (3

4)

4

(1

4)

0

=81

4096

(iii) (𝑛𝑟


= (61

) (3

4)

1

(1

4)

5

=9

2048

(iv) At least 2 means 2 or 3 or 4 or 5 or 6:

= (62

) (3

4)

2

(1

4)

4

+ (63

) (3

4)

3

(1

4)

3

+ (64

) (3

4)

4

(1

4)

2

+ (65

) (3

4)

5

(1

4)

1

+ (66

) (3

4)

6

(1

4)

0

=4077

4096

Question 4.5

(i) (𝑛𝑟


= (53

) (0.5)3(0.5)2

=5

16

(ii) (𝑛𝑟


= (84

) (5

16)

4

(11

16)

4

= 0.149


Question 4.6 a) There is a finite number of trials.

There are only two outcomes – success and failure Trials are independent of each other. The probability of success is the same for each trial.

b) (i) (𝑛𝑟


= (64

) (0.6)4(0.4)2

= 0.311 (ii) This means she scores (1 out of 4), and then 5th

= (41

) (0.6)1(0.4)3 × 0.6

= 0.092

Question 5.1

Red: €4 ×3

10 = €1.20

Blue: −€2 ×4

10 = −€0.80

Yellow:−€3 ×2

10= −€0.60

Green: €2 ×1

10 = €0.20

€0

Question 5.2

Lined circle: €5 ×𝜋

16 = €

5𝜋

16

Non-lined circle €7 ×𝜋

16 = €

7𝜋

16

2 triangles: €10 ×1

4= €2.50

€4.86 => I expect to lose if it costs €5.

#devo

Area of circle = 𝜋𝑟2 = 𝜋𝑥2 Area of square = (4𝑥)2 = 16𝑥2 => probability of a circle

=𝜋𝑥2

16𝑥2

=𝜋

16

Question 5.3 (a)

Option 1: €10,000×1 = €10,000

Option 2: €50,000×0.5 = €25,000

Option 3: €75,000×0.3 = €22,500

Option 4: €100,000×0.2 = €20,000 (b) Take the €10,000 euro because it is certain.

OR Option 2 because it has the highest expected value.


Question 6.1

= 1 𝑔𝑜𝑎𝑙𝑘𝑒𝑒𝑝𝑒𝑟 𝑎𝑛𝑑 10 𝑜𝑢𝑡𝑓𝑖𝑒𝑙𝑑𝑒𝑟𝑠

= (31

) × (1710

)

= 58,344

Question 6.2

= 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑎𝑟𝑟𝑎𝑛𝑔𝑒 − 𝑁𝑜. 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑣𝑜𝑤𝑒𝑙𝑠 𝑎𝑟𝑒𝑎 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑑 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟 = 6! − (5! × 2!) = 480

D B L N UI ie: 5! × 2!


Past and probable exam questions

Questions 1 → 48 are similar to those you are to expect in Section A of Paper 2.

Questions 49 → 63 are similar to those you are to expect in Section B of Paper 2.

1) Section A questions

Question 1

(a) State the fundamental principal of counting.

(b) How many different ways are there to arrange five distinct objects in a row?

(c) David is arranging DVD’s on a shelf. He has five comedy DVD’s and three horror DVD’s. He wants to keep the 5

comedy DVD’s togther and the 3 horror DVD’s together. In how many different ways can he arrange the DVD’s?

Question 2

A bag contains 5 red and 3 yellow discs only. When a disc is drawn from the bag, it is retruned before the next draw.

(i) What is the probability that the two draws will yield both discs yellow?

(ii) What is the probability that two draws will yield both discs the same colour.

Question 3

Question 4

Question 5


Question 6

(a) There are eight questions on an examination paper.

(i) In how many different ways can a candidate select six questions?

(ii) In how many different ways can a candidate select six questions if one particular question must always be selected

(b) How many ways are there to select a subcommittee of 7 members from a committee of 17?

(c) Determine the total number of five-card hands that can be drawn from a deck of 52 cards.

(d) There are five women and six men in a group. From this group a committee of 4 is to be chosen. How many different ways can a committee be formed that contains three women and one man?

Question 7

(a) An “exacta” in horse racing is when you correctly guess which horse will finish first and second. If there are eight horses in the race, how many different possible outcomes for the “exacta” are there? (b) How many different ways can the letters in the word “store” be arranged?

(c) How many different ways are there to arrange the word SCHOOL?

Question 8

Question 9


Question 10

Question 11

Question 12

Question 13

Find the value of k.


Question 14

Question 15

Two events, A and B, from a given event space, are such that 𝑃(𝐴) =1

5 and 𝑃(𝐵) =

1

3.

(a) Calculate 𝑃(𝐴′ ∩ 𝐵) when 𝑃(𝑃 ∩ 𝐵) =1

8.

(b) Calculate 𝑃(𝐴′ ∩ 𝐵) when A and B are mutually exclusive events.

Question 16

In the above figure, 𝑥 and 𝑦 represent the number of elements in (𝐴 ∩ 𝐵) and (𝐵 ∩ 𝐴′) respectively. Given that the total number of elements in the above universe is 35 and ⋕ (𝐴 ∪ 𝐵) = 27 and ⋕ (𝐵′) = 17, find the number of elements in: (a) (𝐴 ∪ 𝐵)′

(b) (𝐴 ∩ 𝐵′).

Given also that 𝑥 = 2𝑦.

(c) Calculate the value of 𝑦 and the value of 𝑥.


Question 17

Question 18

Question 19


Question 20

Question 21

Question 22

Question 23


Question 24

(a)

(b)

Question 25

Question 26


Question 27


Question 28


Question 29

(i)

(ii)


Question 30

Question 31


Question 32

Question 33

Question 34

The number of points scores in 15 basketball games are:

53, 50, 47, 56, 63, 48, 55, 57, 49, 50, 61, 60, 62, 48, 54

a) Find 𝑃30 (the 30th percetile score)

b) The final was the game with 49 points. What is the relative standing in the competition for this game?


Question 35 This question is provided with permission from www.educate.ie ©





Question 39

Question 40


Question 41


Question 42 A car rental company has been using Evertread tyres on their fleet of economy cars. All cars in this fleet are identical. The company manages the tyres on each car in such a way that the four tyres all wear out at the same time. The company keeps a record of the lifespan of each set of tyres. The records show that the lifespan of these sets of tyres is normally distributed with a mean 45 000 km and standard deviation 8000km. i) A car from the economy fleet is chosen at random. Find the probability that the tyres on this car will last for at

least 40 000km. ii) Twenty cars from the economy fleet are chosen at random. Find the probability that the tyres on at least

eighteen of these cars will last for more than 40 000km. Question 43

Question 44


Question 45 The probability of a person having an allergy to a particular nut is 0.1.

A test is available to see if a person has this allergy.

In 80% of the cases where the person has this allergy the test gives a positive result.

If the patient does not have this allergy there is a 0.005 probability of getting a positive result.

(a) Complete the tree diagram for the two events: A person has the allergy. The test gives a positive result.

Mular takes the test and gets a positive result. (b) Work out the probability that Mular has this allergy.

(c) Find the probability that Tome selected a red ball.


Question 46

Question 47


Question 48


2) Section B questions These questions include a mixture of topics from Statistics and Probability, and are similar to those which appear in

Section B of Paper 2.

Question 49


Question 50


Question 51


Question 52

(c) Describe the distribution for Class B. Your description should refer to the shape of the distributions and estimate the median.

(d) Which do you think is the better class? Give reasons for your answer.

(e) A student takes two tests, A and B. The probability of passing in Test A is 1

2 and the probability of passing in test

B is 2

3. The probability of passing in both tests is

1

3. Find the probability of passing in at least one of these two

tests.

Question 53


Question 54


Question 60


Question 61


Question 62


Question 63

Note: This question contains a mixture of topics.


3) Solutions to past and probable exam questions

Question 1 a) If one event has m possible outcomes and a second event has n possible outcomes, then the total number of

outcomes is m × n b) 5!

= 120 c) 5! × 3! × 2!

= 1440 Comment: Handy question.

Question 2 (i) (Yellow) and (Yellow)

3

8 ×

3

8

=9

64

(ii) (Yellow) and (Yellow) or (Red) and (Red)

3

8 ×

3

8 +

5

8 ×

5

8

=17

32

Comment: Nothing major here.

Question 3

= (72

)

= 21 Comment: Meh. This is fine

Question 4 (i) 7!

= 5040

(ii) 1 × 6 × 5 × 4 × 3 × 2 × 1 = 720

(iii) 1 × 5 × 4 × 3 × 2 × 1 × 1

= 120

Comment: This question is okay. Question 5

(i) 6! = 720

(ii) 1 × 4 × 3 × 2 × 1 × 2 = 48

(iii) So we have F R M T EA but EA can be arranged 2! ways => 5! × 2! = 240

Comment: Boring but fine.


Question 6

a) (i) (86

)

= 28

(ii) (75

)

= 21

b) (177

)

= 19,448

c) (525

)

= 2,598,960 d) (3 women) and (1 man)

= (53

) × (61

)

= 60

Comment: An old-school type of question, but make sure you know it.

Question 7 a) 8P2 (ie: order matters)

= 56 b) 5!

= 120

c) 6!

2! (the 2! Comes from the two O’s)

= 360 Comment: Again, an old-school type of question. Nothing too major.

Question 8

𝐻0: 𝑝 = 0.2 𝐻1: 𝑝 ≠ 0.2

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

2

30− 1.96

√2

30(1 −

230

)

30≤ 𝑝 ≤

2

30+ 1.96

√2

30(1 −

230

)

30

−0.023 ≤ 𝑝 ≤ 0.156 𝐻0 is outside the confidence interval so we reject 𝐻0. Comment: Standard question.

Question 9

a)

𝑧 =𝑥 − 𝜇

𝜎

=15 − 16.12

1.6

= −0.7 => 𝑃(𝑧 < −0.7) = 1 − 𝑃(𝑧 < 0.7)

= 1 − 0.7580 = 0.242

b)


Probability of being less than ‘a’ is 30% = 0.3

=> 𝑧 −score = NOT IN LOG TABLES! => Find b (because then we can find a due to symmetry) Probability of being less than b is 70% = 0.7

=> 𝑧 −score = 0.52

but 𝑧 =𝑥−𝜇

𝜎

=> 𝑥 = 𝑧𝜎 + 𝜇 = (0.52)(1.6) + 16.12

= 16.952 => 𝑎 = 15.288 𝑠𝑒𝑐𝑜𝑛𝑑𝑠

Comment: Be careful of the fastest runners, because they will actually have the lowest time. Think about it! Lovely question.

Question 10 a)

𝑧 =𝑥 − 𝜇

𝜎

=830 − 850

50

= −0.4 => 𝑃(𝑧 < −0.4) = 1 − 𝑃(𝑧 < 0.4)

= 1 − 0.6554 = 0.3446

b) 0.3446 × 500 = 172.2 = 172 bulbs

c)

Probability of being less 818 is 20% = 0.2

=> 𝑧 −score = NOT IN LOG TABLES Now, we know that the probability of being less than ‘a’ is 80% = 0.8

𝑧 − score = 0.84

but 𝑧 =𝑥 − 𝜇

𝜎

=> 𝜎 =𝑥 − 𝜇

𝑧

=902 − 860

0.84

= 50 d)

I would prefer company Y, because their bulbs have a higher mean, and the standard deviations are equal.

Comment: Beautiful question on z-scores. Know it.

Question 11

[we know 𝑎 = 902 due to symmetry]


a) Normal distributed => median = mean = 200gb)

Probability of being less than 210 is 80% =0.8

=> 𝑧 − score = 0.84

but z =𝑥 − 𝜇

𝜎

=> 𝜎 =𝑥 − 𝜇

𝑧

=210 − 200

0.84= 11.9g

c)

𝑧 =𝑥 − 𝜇

𝜎

=180 − 200

11.9= −1.68

=> 𝑃(𝑧 ≤ −1.68) = 1 − 𝑃(𝑧 ≤ 1.68)

= 1 − 0.9535 = 0.0465

Comment: Nice question. Nothing we can’t handle here. I would love some popcorn now.

Question 12 a)

𝑧 =𝑥 − 𝜇

𝜎

=91 − 100

15= −0.6

=> 𝑃(𝑧 ≤ −0.6) = 1 − 𝑃(𝑧 ≤ 0.6)

= 1 − 0.7257 = 0.2743

b)

60%


The probability of it being less than (100 + 𝑘) is 79.1%

=0.791 => 𝑧 − score = 0.81


𝜎

=> 𝑥 = 𝑧𝜎 + 𝜇 = (0.81)(15) + 100

= 112.15 ie: 100 + 𝑘 = 112.15

=> 𝑘 = 12.15 Comment: A nice question, perhaps suitable for those students with a high IQ!!

Question 13

Due to symmetry, the mean equals 50. The probability of being less than 55 is 70%

= 0.7 => 𝑧 − score = 0.52


𝜎

=> 𝜎 =𝑥 − 𝜇

𝑧

=55 − 50

0.52

= 9.6 Comment: Good little question.


Question 14 a)

b) 𝑃(𝐴) = 0.54

𝑃(𝐵) = 0.33

c) 𝑃(𝐴|𝐵′) =𝑃(𝐴∩𝐵′)

𝑃(𝐵′)

=0.32

0.67

= 0.47761194 d) If independent: 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴). 𝑃(𝐵)

0.22 = (0.54)(0.33) 0.22 ≠ 0.1782

=> not independent

Comment: This is straight-forward once you’ve learned your rules.

Question 15 a)

=>5

24

b) Mutually exclusive => 𝑃(𝐴 ∩ 𝐵) = 0

=>1

3

Comment: Again, it’s all about knowing your rules here. Question 16

a) (A ∪ B)′ = 35 − 27

= 8 b)

Here, we’re trying to find w: We’re told #(𝐵′) = 17 ie: 𝑤 + 8 = 17 => 𝑤 = 9


c)

9 + 2𝑦 + 𝑦 + 8 = 35

=> 𝑦 = 6 => 𝑥 = 12

Comment: Students won’t fancy this. A question to separate the top students from the rest.

Question 17 a) 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)

i) When mutually exclusive: 𝑃(𝐴 ∩ 𝐵) = 0 => 𝑃(𝐴 ∪ 𝐵) = 𝑎 + 𝑏

ii) When independent: 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴). 𝑃(𝐵) = 𝑎𝑏 => 𝑃(𝐴 ∪ 𝐵) = 𝑎 + 𝑏 − 𝑎𝑏

b) 1st Draft

2nd Draft

=> 𝑃(𝑅 ∪ 𝑄) = 0.5

c) 𝑃(𝑅 ∩ 𝑄) = 0.035

d) 𝑃(𝑅) = 0.185

Comment: Know your rules!!

Told 𝑃(𝑅|𝑄) = 0.1

𝑃(𝑅|𝑄) =𝑃(𝑅 ∩ 𝑄)

𝑃(𝑄)

0.1 =𝑥

0.35

=> 𝑥 = 0.035

Question 18

a) = 0.1 + 0.15 + 0.05

= 0.3 b) = 0.15 + 0.05

= 0.2

c) 𝑃(𝐾| 𝐽 ) =𝑃(𝐾∩𝐽)

𝑃(𝐽)

=0.15

0.25

= 0.6 d) If independent: 𝑃(𝐽 ∩ 𝐾) = 𝑃(𝐽). 𝑃(𝐾)

0.15 = (0.25)(0.2) 0.15 ≠ 0.05 => Not independent

e) Not independent => one does effect the other. ie: teacher is correct.

Comment: This is a nice interesting question. Not too hard either.


Question 19

a) (𝑛𝑟


= (202

) (0.3)2(0.7)18

= 0.027845872 (Note: We are considering a fault as being a success) b) More than 3 faults means 4 faults or 5 faults or 6 faults or …. 0r 20 faults.

Alternatively: 1 – [0 faults or 1 fault or 2 faults or 3 faults]

= 1 − [(200

) (0.3)0(0.7)20 + (201

) (0.3)1(0.7)19 + 0.027845872 + (202

) (0.3)3(0.7)17]

= 0.892913196

c) (𝑛𝑟


= (106

) (0.892913196)6(0.107086804)4

= 0.01399647172

Comment: Watch out for the final part here. A famous style of question from the UK.

Question 20

i) (𝑛𝑟


= (88

) (0.8)8(0.2)0

= 0.16777216

ii) (𝑛𝑟


= (86

) (0.8)6(0.2)2

= 0.29360128 Comment: Nothing difficult here.

Question 21

i) (𝑛𝑟


= (108

) (0.8)8(0.2)2

= 0.3020 ii) 80% of 10

=8 Comment: Nice little question.

Question 22 a) Alternative Method: i) (heads) and (heads) and (heads)

= 𝑝 × 𝑝 × 𝑝 = 𝑝3

ii) (heads) and (heads) and (tails)

= (𝑝) × (𝑝) × (1 − 𝑝) or + (heads) and (tails) and (heads)

= (𝑝) × (1 − 𝑝) × (𝑝) or + (tails) and (heads) and (heads) = (1 − 𝑝) × (𝑝) × (𝑝) = 3(𝑝)(𝑝)(1 − 𝑝) = 3𝑝2(1 − 𝑝) = 3𝑝2 − 3𝑝3

i) (𝑛𝑟


= (33

) (𝑝)3(1 − 𝑝)0

= 𝑝3

ii) (𝑛𝑟


= (32

) (𝑝)2(1 − 𝑝)1

= 3𝑝2(1 − 𝑝) = 3𝑝2 − 3𝑝3

b)


𝑝3 = 3𝑝2 − 3𝑝3 ÷ 𝑝2: 𝑝 = 3 − 3𝑝

4𝑝 = 3

𝑝 =3

4

Comment: A short, clever question.

Question 23 (a) 2 heads and 2 tails

(𝑛𝑟


= (42

) (0.5)2(0.5)2

= 0.375 (b) 4 heads or 0 heads

= (44

) (0.5)4(0.5)0 + (40

) (0.5)0(0.5)4

= 0.125 (c) (heads) and (heads) and (tails)

= 0.5 × 0.5 × 0.5 = 0.125

Comment: No surprises here. This has got exam question written all over it.

Question 24 (a) Total Probability = 1

=> 0.1 + 𝑝 + 0.2 + 𝑞 + 0.3 = 1 => 𝑝 + 𝑞 = 0.4

Expected value = 3.5 => (1)(0.1) + (2)(𝑝) + (3)(0.2) + (4)(𝑞) + (5)(0.3) = 3.5

=> 2𝑝 + 4𝑞 = 1.3 (b) Total probability = 1

=> 𝑝2 + 𝑝2 +𝑝

4+

4𝑝 + 1

8= 1

× 8: 8𝑝2 + 8𝑝2 + 2𝑝 + 4𝑝 + 1 = 8 16𝑝2 + 6𝑝 − 7 = 0

𝑎 = 16 𝑏 = 6 𝑐 = −7

𝑝 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

𝑝 =−(6) ± √(6)2 − 4(16)(−7)

2(16)

𝑝 =1

2 𝑜𝑟 𝑝 = −

7

8

The value of −7

8 would result in a negative value when subbed back in.

=> 𝑝 =1

2

Comment: UK style question, which is making it’s way over the Irish Sea.


Question 25

(a) =18

37

= 0.486 (b) (not red) and (not red) and (not red) and (not red) and (red)

= 19

37 ×

19

37 ×

19

37 ×

19

37 ×

18

37

= 0.034 (c) (Once) or (twice) or (Three times) or (four times)

= (41

) (18

37)

1

(19

37)

3

+ (42

) (18

37)

2

(19

37)

2

+ (43

) (18

37)

3

(19

37)

1

+ (44

) (18

37)

4

(19

37)

0

= 0.930 Alternative method: Total probability – [doesn’t fall into a red] = 1 − [doesn’t fall into a red]

= 1 − [ 19

37 ×

19

37 ×

19

37 ×

19

37 ]

= 0.930 Comment: A lovely question, although should they be promoting gambling?!

Question 26

(a) =1

4 ×

1

3

=1

12

(b) (1 + 4) 𝑜𝑟 (4 + 1) 𝑜𝑟 (2 + 3) 𝑜𝑟 (3 + 2)

= (1

4 ×

1

3) + (

1

4 ×

1

3) + (

1

4 ×

1

3) + (

1

4 ×

1

3)

=1

3

(c) From list above: 1

4

Comment: The final part can cause confusion. Make sure you know this!

Question 27 (a) (b)

(c) 4 pence (d) The longer the distance travelled, the lower the cost per mile. Comment: Very basic.


Question 28 (a)

(b) There is a very strong positive correlation, which indicates that the higher your mark on Paper 1 mark, the higher your mark on Paper 2. (c) From line of best fit: 39 Comment: Could easily be on the Ordinary Level exam.

Question 29 (i) B (ii) The equations given are written in the form: 𝑦 = 𝑐 + 𝑚𝑥 m is the slope:

𝑚 =𝑦2 − 𝑦1

𝑥2 − 𝑥1

(20, 167) (34, 185) 𝑥1 𝑦1 𝑥2 𝑦2

𝑚 =185 − 167

34 − 20

= 1.285714286 … which is close to 1.3, meaning the answer is B, C or E. The y measurement is ‘height’, so now the answer must be B or C. In the equation 𝑦 = 𝑐 + 𝑚𝑥, ′𝑐′ stands for where the line cuts the y-axis where x = 0. Be careful, because the x – axis starts at 20, so at 0 it would be lower than 167. => Answer = B Comment: The second part is a bit messy, requiring you to understand the “rise over run method“ from Co-ordinate Geometry of the Line.


Question 30 (a) The equation is written in the form 𝑦 = 𝑐 + 𝑚𝑥 ‘c’ is where the line cutes the y-axis at x = 0 ie:the y intercept is (0, 131) ‘m’ is the slope which is given as −2.68

Slope =rise

run

Let’s pick a run of 30

=> −2.68 =rise

run

=> rise = −80.4 => a point on the line would be (30, 49.6)

(b) rainfall = 131 − 2.68(35) = 37.2mm Comment: Again, a bit messy, requiring you to understand the “rise over run method“ from Co-ordinate Geometry of the Line.

Question 31 15 classes × 30 students = 450 6th form = 150 Total: 600 1

4 of students are in 6th form

=>1

4 of the 40 need to be taken from 6th form

=> 10 students from the 6th form (5 boys, 5 girls) The other 30 students need to be taken from 15 other classes … => 2 from each of the classes (1 boy, 1 girl) Summary:

Within each of the above groups, students names can be randomly selected, perhaps by pulling name(s) out of a hat. Comment: Wow, that would cause lot’s of problems.


Question 32 (a) The population is all the subscribers. (b) A list of names from those who bought the magazine last year. (c) Those who have paid. (d) Advantage of census: No sampling error Disadvantage of census: Takes more time

(e) (𝑛𝑟


= (2510

) (0.4)10(0.6)15

= 0.161157938 (f) 𝐻0: 𝑝 = 0.4 (40% of magazine subscribers wish to change the name of the magazine). 𝐻1: 𝑝 ≠ 0.4

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

6

25− 1.96

√6

25(1 −

625

)

25≤ 𝑝 ≤

6

25+ 1.96

√6

25(1 −

625

)

25

0.073 ≤ 𝑝 ≤ 0.407 𝐻0 is inside the confidence interval so we do not reject 𝐻0. Comment: A mixture of theory and numerical problems. Lots going on here, making it a well constructed exam question.

Question33

(a) 50,000

100

= 500 Start at a random point between 1 and 100 and then select every 500th name from there on. When selecting the random starting point, I could pick a number between 1 and 100 out of a hat, or alternatively use a random number generator. (b) (i) Quota sampling: Advantage: Quick and cheap to organise. Disadvantage: Because the sample is non-random, there could be bias. (ii) Systematic sampling: Advantage: Easier to conduct than a simple random sample. Disadvantage: The system may interact with some hidden pattern in the population. Comment: Theory, theory, theory! Learn it!

Question 34 (a) Step 1: Put the data in order: 47, 48, 48, 49, 50, 50, 53, 54, 55, 56, 57, 60, 61, 62, 63

Step 2: 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒 ×𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠

100

=30×15

100

= 4.5 Step 3: Decimal number => round up to next value, ie: the 5th value

= 50 𝑝𝑜𝑖𝑛𝑡𝑠 (b) Rule: Put the number of values less than your number over the total number of values and then multiply by 100:

=3

15× 100

= 20𝑡ℎ percentile Comment: A rare question but we should know this if we’ve learned the material.


Question 35

(a) (i) (𝑛𝑟


= (44

) (0.6)4(0.4)0

= 0.1296

(ii) (𝑛𝑟


= (41

) (0.6)1(0.4)3

= 0.1536

(iii) (𝑛𝑟


= (40

) (0.6)0(0.4)4

= 0.0256 (b) I would say no, because if one seed germinates, then this may push the others out of the way and take all the soils nutrients. (c) Expected value= 4 × 0.6 = 2.4 Comment: Ah howiye Brocolli. Good question.

Question 36

(a)

i) Median, M = 175 cm ii) Lower quartile, LQ = 170 cm iii) Interquartile range = Upper quartile value – Lower quartile value

= 180 − 170 = 10

(b) From diagram: 75% (c) Probability of being less than 180 is 75% = 0.75

=> 𝑧 − 𝑠𝑐𝑜𝑟𝑒 = 0.67

𝑏𝑢𝑡 𝑧 =𝑥 − 𝜇

𝜎

=> 𝜎 =𝑥 − 𝜇

𝑧

=180 − 175

0.67

= 7.462686567 (d) 𝑃(170 ≤ 𝑥 ≤ 180)

𝑧 =𝑥 − 𝜇

𝜎

𝑧 =170 − 175

7.462686567 𝑧 =

180 − 175

7.46268567

= −0.67 = 0.67 => 𝑃(−0.67 ≤ 𝑧 ≤ 0.67)

= 𝑃(𝑧 ≤ 0.67) − (𝑃(𝑧 ≤ −0.67) = 𝑃(𝑧 ≤ 0.67) − [1 − (𝑃(≤ 0.67)]

= 0.7486 − [1 − 0.7486] = 0.4972 (≈ 0.5)

Comment: Students will find this okay.


Question 37 (a) (i) At least 3 means 3 or 4 or 5

= 0.3 + 0.3 + 0.1 = 0.7

(ii) Mean = (0.1)(1)+(0.2)(2)+(0.3)(3)+(0.3)(4)+(0.1)(5)

0.1+0.2+0.3+0.3+0.1

= 3.1 (b)

(i) =

85

300

= 0.283̇

(ii) 84+73+58

300

= 0.716

(iii) 𝑃(𝑀|𝐹) =𝑃(𝑀∩𝐹)

𝑃(𝐹)

=(

58300

)

(142300

)

= 0.408450704 Comment: A bit of a head-melter, so you have to be careful here.

Question 38 (a)

(b)

Due to symmetry: Mean = 30.6

30.6 + 2𝜎 = 36.4 => 𝜎 = 2.9

(c)

𝑧 =𝑥 − 𝜇

𝜎


𝑧 =30 − 30.6

2.9

= −0.21 => 𝑃(𝑧 > −0.21)

= 1 − 𝑃(𝑧 ≤ −0.21) = 1 − [1 − 𝑃(𝑧 ≤ 0.21)]

= 1 − [1 − 0.5832] = 0.5832

Comment: No problems here.

Question 39 The population I am interested in is all small and medium sized Irish companies. As my sampling frame I would get a list of small and medium sized Irish companies, perhaps off a government database. I would then take a sample of 100 of these using a random sampling method.

I chose 100 because this will result in a ‘small’ sampling error of 10% (𝐸 =1

√𝑛=

1

√100= 0.1)

I would then gather the data by ringing each business. Potential bias could come from the fact that some smaller businesses may not be on the database and these may be the ones without e-mail and internet access. Comment: A nice theory question, but too many students tend to waffle.

Question 40 (a) (i) A sample space is the set of all possible outcomes of an experiment. (ii) Two events are mutually exclusive if they cannot occur at the same time. (iii) Two events are independent if the outcome of one does not affect the other. (b) (i)

(ii) If independent: 𝑃(𝐸 ∩ 𝐹) = 𝑃(𝐸). 𝑃(𝐹)

4

30= (

20

30) (

6

30)

2

15=

2

15

=> Independent‼ Comment: Straight out of the notes. No surprises here.


Question 41 (a) (i)

𝑧 =𝑥 − 𝜇

𝜎

=68 − 60

5

= 1.6 => 𝑃(𝑧 ≤ 1.6)

= 0.9452 (ii)

Due to symmetry and using our answer from part (i), answer is 0.8904 Hormone A The effect of hormone A was to increase the height of all the plants.

Hormone B The effect of hormone B was to reduce the number of really small plants and the number of really tall plants. The mean was unchanged.

Hormone C The effect of harmonic C was to increase the number of small plants and the number of tall plants. The mean was unchanged.

Comment: The first part is straight forward, but the second part is a clever question.


Question 42 (i)

𝑧 =𝑥 − 𝜇

𝜎

=40,000 − 45,000

8,000

= −0.63 => 𝑃(𝑧 > −0.63)

= 1 − 𝑃(𝑧 ≤ −0.63) = 1 − [1 − (𝑃 ≤ 0.63)] = 1 − [1 − 0.7357] = 0.7357

(ii) At least 18 means 18 or 19 or 20

(𝑛𝑟


= (2018

) (0.7357)18(0.2643)2 + (2019

) (0.7357)19(0.2643)1 + (2020

) (0.7357)20(0.2643)0

= 0.07057462622 Comment: Fine

Question 43 (a)

𝑧 =𝑥 − 𝜇

𝜎

=39.7 − 40

0.2

= −1.5 => 𝑃(𝑧 < −1.5) = 1 − 𝑃(𝑧 < 1.5) = 1 − 0.9332 = 0.0668

(b) At least 2 means 2 or 3 or 4 or 5

(𝑛𝑟


= (52

) (0.0668)2(0.9332)3 + (53

) (0.0668)3(0.9332)2 + (54

) (0.0668)4(0.9332)1 + (55

) (0.0668)5(0.9332)0

= 0.03895420077 Comment: Very straight-forward. This type of questions keeps appearing over and over again.


Question 44 (a)

(b) Heads =

5

18+

7

24

=41

72

(c) 𝑃(𝑅|𝐻) =𝑃(𝑅∩𝐻)

𝑃(𝐻)

=(

5

18)

(41

72)

=20

41

(d) (Red) and (Red) or (Blue) and (Blue)

(5

12) × (

5

12) + (

7

12) × (

7

12)

=37

72

Comment: These tree diagrams can be hard to get your head around but once they click, they click!

Question 45 (a)

(b) 𝑃(𝐴𝑙𝑙𝑒𝑟𝑔𝑦|𝑃𝑜𝑠𝑖𝑡𝑣𝑒) =𝑃(𝐴𝑙𝑙𝑒𝑟𝑔𝑦 ∩ 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒)

𝑃(𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒)

=0.08

0.08+0.045

= 0.64 Comment: Another tree diagram. Make sure you understand this!


Question 46 a) John would need to get €0, €0, €3, €5 but in any order.

The aount of arrangements of 4 different objects would be 4!, but since 2 of these are the same we must divide by 2!.

=>4!

2!

= 12 b) Game A: Expected value for player:

€0 ×2

5 = €0

€3 ×1

5 = €0.60

€5 ×1

5 = €1

€6 ×1

5 = €1.20

€2.80

=> Profit for charity = €0.20 Game B: Expected value for player:

€0 ×1

6 = €0

€1 ×1

6 = €

1

6

€2 ×1

6 = €

1

3

€3 ×1

6 = €

1

2

€4 ×1

6 = €

2

3

€5 ×1

6 = €

5

6

€2.50 => Profit for charity = €0.50

=> Answer: Game B c)

(𝑛𝑟


= (62

) (1

6)

2

(5

6)

4

=3,125

15,552

Comment: Have to think about this one carefully, but once you get your head around what’s going on, it’s plane sailing.5/18

Question 47 (a)


(b) (i)

=10

36

=5

18

(ii)

=26

36×

26

36×

26

36

= 0.3767 (c)

(2 out of 9) and (10th)

= (92

) (5

18)

2

(13

18)

7

×5

18

= 0.0791 Comment: Very Handy!

Question 48

(a)

�̅� − 1.96𝑠

√𝑛≤ 𝜇 ≤ �̅� + 1.96

𝑠

√𝑛

90.45 − 1.9620.73

√100≤ 𝜇 ≤ 90.45 + 1.96

20.73

√100

86.39 ≤ 𝜇 ≤ 94.51 (b) 𝐻0: 𝜇 = 94 The mean amount spent by shoppers on a Saturday is €94. 𝐻1: 𝜇 ≠ 94 The mean amount spent by shoppers on a Saturday is different than €94.

86.39 ≤ 𝜇 ≤ 94.51 𝐻0 is inside the confidence interval so we do not reject 𝐻0. i.e.: The mean amount spent by shoppers on a Saturday is €94 (c)

𝑇 =𝑥 − 𝜇

𝜎

√𝑛

=90.45 − 95

20.73

√100

= −1.71

p – value:

= 2(1 − 𝑃(𝑧 ≤ |𝑇|))

= 2(1 − 𝑃(𝑧 ≤ |−1.71|))

= 2(1 − 𝑃(𝑧 ≤ 1.71))

= 2(1 − 0.9564) = 0.0872

𝑖. 𝑒. : ≥ 0.05 => do not reject 𝐻𝑜

Comment: No surprises here. Nice question.


Question 49 (a) (i)

1st Histogram: Run 2nd Histogram: Cycle by looking at the minimums 3rd Histogram: Swim

(ii) ≈ 25 minutes (iii) Well, all data lies within mean ± 3 standard deviations (roughly) => The range must be ≈ 6 standard deviations. ie: 32 − 10 = 6𝜎 => 𝜎 = 3.67 minutes (iv) There may not be any times which occurred more than once. (b) Cycle –v- Swim: Moderately strong positive correlation, which indicates that the longer your swim time, the longer your cycle time. Run –v- Swim: Same as above Run –v- Cycle: Same as above but slightly stronger. (d)

Run/Swim: 𝑦 = 0.53𝑥 + 15.2 Brian: 𝑥 = 17.6 mins 𝑦 = 0.53(17.6) + 15.2 = 24.528 mins

Run/Cycle: 𝑦 = 0.58𝑥 + 0.71 Brian: 𝑥 = 35.7 mins 𝑦 = 0.58(35.7) + 0.71 = 21.416 mins

Take the average of the 2 run times:24.528 + 21.416

2

= 22.972 mins (e) Based on the empirical rule:

𝜇 = 88.1 mins, 𝜎 = 10.3mins 𝜇 − 2𝜎 = 88.1 − 2(10.3) = 67.5 mins 𝜇 + 2𝜎 = 88.1 + 2(10.3) = 108.7 mins 95% of athletes took between 67.5 and 108.7 minutes to complete the race.

(f)

𝑧 =𝑥 − 𝜇

𝜎

=100 − 88.1

10.3

= 1.155 => 𝑃(𝑧 < 1.155)

= 0.877 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡ℎ𝑙𝑒𝑡𝑒𝑠 = 224 × 0.877

≈ 196 (g) 2nd on 6th means: (1 out of 5) and then the 6th

= (51

) (0.123)1(0.877)4 × 0.123

= 0.0447485122 ≈ 4.5%

Comment: A question that needs close attention.


Question 50 (i) & (iv)

(ii) 0.6232364836 (iii) There is a moderately strong positive correlation, which indicates that the longer you spend in education, the higher your annual income. (iv) see above (v) €40,000 (vi)

𝑚 =𝑦2 − 𝑦1

𝑥2 − 𝑥1

=64 − 40

19 − 14

= 4.8

𝑥1 𝑦1 (14, 40) 𝑥2 𝑦2 (19, 64)

(vii) For every additional year in education, you can expect your annual income to increase by €4,800 (viii) Problem 1: Just calling people in Dublin because perhaps salaries are higher in the capital. Problem 2: Just calling in the evening, because maybe a certain portion of people only work at these hours. Problem 3: Calling over the phone only, because people can easily lie over the phone. Comment: This was a nice question. More of this please.


Question 51 (a) −0.6 (b) Age = 46 Max. heart rate = 137 bpm (c) 176 bpm (d)

𝑚 =𝑦2 − 𝑦1

𝑥2 − 𝑥1

=145 − 198

88 − 12

= −0.697

𝑥1 𝑦1 (12, 198) 𝑥2 𝑦2 (88, 145)

(e) 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1)

𝑦 − 198 = −0.697(𝑥 − 12) 𝑦 − 198 = −0.697𝑥 + 8.364

𝑦 = 206.364 − 0.697𝑥 𝑀𝐻𝑅 = 206.364 − 0.697𝑎𝑔𝑒

(f) Young person (age = 20) Old rule:

𝑀𝐻𝑅 = 220 − 20 = 200

New rule:

𝑀𝐻𝑅 = 206.364 − 0.697(20) = 192.424

Old person (age = 80) Old rule:

𝑀𝐻𝑅 = 220 − 80 = 140

New rule:

𝑀𝐻𝑅 = 206.364 − 0.697(80) = 150.604

=> new rule gives lower MHR for young people, but a higher MHR for old people. (g) Old rule:

𝑀𝐻𝑅 = 220 − 65 = 155

× 75% = 116.25 New rule

𝑀𝐻𝑅 = 206.364 − 0.697(65) = 161.059

× 75% = 120.79 => speed up a bit there ya auld bollox

Comment: This question was fine until the last couple of parts, where it wasn’t too obvious what the Examiner was looking for!

Question 52 (a)

(b) Shape: Bi-modal Median: 65% (c) Shape: Left-skewed Median: 87%


(d) Class B, because it has a higher median and mean. (e)

=>1

6+

1

3+

1

3

=5

6

(f) (i) (𝑛𝑟


= (60

) (2

3)

0

(1

3)

6

=1

729

(ii) (𝑛𝑟


= (63

) (2

3)

3

(1

3)

3

=160

729

Comment: Boring but fine.

Question 53 (a) (i)

(ii) Shape: Right skewed Median: Well there are 115 earthquakes, so the middle one is the 58th There are 55 from 0 -200

=> we want 3

12 into 200 – 300

= 225 days (iii) The shape of the distribution isn’t normal. (iv) Amount between 100 – 200 is 24

=>24

115

Reason: Because there is no other way! (v) Don’t just look at earthquakes that caused 1000 deaths. (b) (i)The statement is not completely true. A more accurate statement would be:


Strong earthquakes (> 7 magnitude) are more likely to cause tsunamis. (ii) Tsunamis: 102 No Tsunami: 156

=>102

102 + 156

≈ 0.4 (iii) Tsunamis: 210 No tsunamis: 182

=>210

210 + 182

=1528

Now, at least 4 means 4 or 5 or 6

(𝑛𝑟


= (64

) (15

28)

4

(13

28)

2

+ (65

) (15

28)

5

(13

28)

1

+ (66

) (15

28)

6

(13

28)

0

= 0.413 Comment: Some parts are nice here, but some are awkward. Project Maths at its finest.

Question 54

(a) (i) They’ve gone up: (in 2009 they were mostly between 1% and 5%) (in 2011 they were mostly between 5% and 15%) (ii) They’ve gone up: (in 2009 they were mostly between 2.3% and 4.1%) (in 2011 they were mostly between 4% and 6%) (iii) there appears to be a stronger relationship. (b) How many people are represented by each dot. (c) It relates to which variable causes the other ie: Does higher arrears rates cause higher interest rates or vice-versa.

(d) (i) =145,414

475,136

= 0.31

(ii) =11,644

145,414

= 0.08

(iii) 𝑃(𝑁|𝐴) =𝑃(𝑁∩𝐴)

𝑃(𝐴)

=(

11,644

475,136)

(24,011

475,136)

= 0.48 Comment: Ridiculous question. (e) 𝐻0: 𝑝 = 0.31 (31% of the properties are in negative equity). 𝐻1: 𝑝 ≠ 0.31

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

552

2000− 1.96

√552

2000(1 −

5522000

)

2000≤ 𝑝 ≤

552

2000+ 1.96

√552

2000(1 −

5522000

)

2000

0.256 ≤ 𝑝 ≤ 0.296 𝐻0 is outside the confidence interval so we reject 𝐻0. i.e.: The percentage of properties in negative equity has changed. Comment: Scandalous question. Argued by many to be the worst question to appear on a Leaving Cert Maths exam for many, many years.

475,136 − 317,355 − 145,414 + 11,644


Question 55 (a) (c) (f) (iii)

(b) 𝑟 = 0.6378 (c) �̅� = 267 �̅� = 518 As domestic yearly income increases the foreign yearly income increases. (d) (600, 1200) Perhaps a boom in the economy. (e) (i) �̅� = 230 𝜎𝑥 = 54 (ii) �̅� = 442 𝜎𝑦 = 160

(iii) 𝑟 = −0.5085 (f) (i)

𝑦 = �̅� +𝑟𝜎𝑦

𝜎𝑥

(𝑥 − �̅�)

𝑦 = 442 +−0.5085(160)

54(𝑥 − 230)

𝑦 = −1.51𝑥 + 789.3 (ii) slope = −1.5 (iii) y – intercept = 789.3

slope =rise

run

Choose a run of 500

=> −1.5 =rise

500

=> rise = −750 (See diagram)

(iv) It now results in a negative correlation, which indicates that as domestic income increases, foreign income decreases. Comment: Students won’t like this.


Question 56 (a) (i) A numerical characteristic of interest in each element of the sample or population. (ii) Discrete: The number of people at a match Continuous: The heights of people. (iii) Expected value = sum of values times probabilities. (b) (i) A number with only two factors (itself and 1) (ii)

(iii) Prime numbers: 2, 3, 5, 7, 11

=>15

36

=5

12

(iv) = 1 −5

12

=7

12

(c)

(d) (i) = −1.75 + 1.25 = −0.50 This means Bob can expect to lose an average of €0.50 each time he plays. (ii) = 30 × −0.50 = −€15 (e) We need the 𝑥𝑃(𝑥) for a Prime sum to be 2.25.

5

12× 𝑥 = 2.25

=> 𝑥 = €5.40 Comment: Not the easiest question we’ve ever seen. Students won’t love it anyway!


Question 57 (a) Expected value = sum of values times probabilities = (0)(0.8) + (1,000)(0.15) + (5,000)(0.04) + (10,000)(0.01) = €450 (b) They can expect the average of each claim to be €450. (c) = 0.15 + 0.04 + 0.01 = 0.2 (d) (𝑀)(0.8) + (𝑀 − 500)(0.15) + (𝑀 − 4,500)(0.04) + (𝑀 − 9,500)(0.01) = 200

𝑀 − 350 = 200 𝑀 = €550

(e) (i)

There is a strong negative correlation, which indicates that the heavier the car, the less the number of claims. (ii)

𝑚 =𝑦2 − 𝑦1

𝑥2 − 𝑥1

=17 − 66

5000 − 1000

= −0.01

𝑥1 𝑦1 (1000, 66) 𝑥2 𝑦2 (5000, 17)

(iii) Maybe the heavier cars don’t bother making a claim because damage may be minimal, so my answer is no. (f) (i) Outlier is at (18, 20) Learners usually go very slow and have an experienced driver with them. (ii) There is a strong negative correlation, which indicates that the older you are, the less claims you make. (g) According to this data the insurance company could claim that its lowest risk is a person aged 69 driving a car of weight 5,800kg and the highest risk is a person aged 21 driving a car of weight 1,300kg. Comment: A question for the stronger student. Requires you to keep the head!


Question 58 (a)

𝑀𝑒𝑎𝑛 =(6)(72) + (18)(76) + (38)(80) + (18)(84) + (18)(88) + (2)(92)

100

= 81.2%

(b)

Median: Well there are 100 players, so the middle one is between the 50th and 51st. There are 24 from 70-78.

=> we want 26

38 into 78 – 82 for the 50th

and

we want 27

38 into 78 − 82 for the 51st.

= 80.736 and = 80.842 Median is between these two, so = 80.8 (c) (i) Mean = 81.2 Median = 80.8

=> % difference =81.2−80.8

81.2× 100%

= 0.49% (ii) Because: mean ≈ median (iii) = 4.75 (d) (i) Can’t use normal percentile rules because we haven’t got the 100 pieces of data to put them in order.

The probability of being less than x is 90% =0.9

=> 𝑧 − score = 1.28


𝜎

=> 𝑥 = 𝑧𝜎 + 𝜇

= (1.28)(4.75) + 81.2 = 87.28 = 87.3%


(ii)

𝑧 =𝑥 − 𝜇

𝜎

=85 − 81.2

4.75

= 0.8

=> 𝑃(𝑧 < 0.8)

= 0.7881

= 78.8% (e) Condition 1: There is a finite number of trials. Condition 2: There are only two outcomes – success and failure Condition 3: Trials are independent of each other.

(f) (i) (𝑛𝑟


= (55

) (0.927)5(0.073)0

= 0.685 = 68.5%

(ii) (𝑛𝑟


(53

) (0.927)3(0.073)2

= 0.042 = 4.2% (iii) Third on fifth means: 2 out of 4 , and then the fifth:

= (42

) (0.927)2(0.073)2 × 0.927

= 0.025 = 2.5%

(g) (i) 4 outliers: 79°, 4°, 13°, 68° Perhaps the instrument used was faulty or perhaps the student didn’t understand ‘angle of elevation’. (ii) mean = 28.4 (iii)

𝐻0: 𝑝 = 0.75 𝐻1: 𝑝 ≠ 0.75

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

18

30− 1.96

√1830

(1 −1830

)

30≤ 𝑝 ≤

18

30+ 1.96

√1830

(1 −1830

)

30

0.425 ≤ 𝑝 ≤ 0.775 𝐻0 is inside the confidence interval so we do not reject 𝐻0.

Comment: A bit of everything in one big question. Strong exam appearance potential.


Question 59 (a) (i) Normal (ii) Mean is symmetrically between 97.4g and 108.6g =103 g 95% of all data lies within mean ±2 standard deviations

=> 103 + 2𝜎 = 108.6 𝜎 = 2.8 𝑔

(b)

𝑧 =𝑥 − 𝜇

𝜎

=100 − 103

2.8

= −1.07

=> 𝑃(𝑧 < −1.07)

= 1 − 𝑃(𝑧 < 1.07)

= 1 − 0.8577

= 0.1423

(c) 1 out of 1,000 = 0.001

Probability of being less than 100g is 0.001

=> 𝑧 − score = NOT IN LOG TABLES Using symmetry, the probability of being less than b is 0.999

=> 𝑧 − score = 3.08 => 𝑧 − score for 100 is − 3.08


𝜎

=> 𝜇 = 𝑥 − 𝑧𝜎

=> 𝜇 = 100 − (−3.08)(2.8) = 108.624 g

(d) At least 1 means: 1 or 2 or 3 or 4 or 5 Alternatively: 1 − [0 bars]

= 1 − (𝑛𝑟


= 1 − (50

) (0.1423)0(0.8577)5

= 0.5358300187


(e) 𝐻0: 𝑝 = 0.7 𝐻1: 𝑝 ≠ 0.7

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

62

100− 1.96

√62

100(1 −

62100

)

100≤ 𝑝 ≤

62

100+ 1.96

√62

100(1 −

62100

)

100

0.525 ≤ 𝑝 ≤ 0.715 𝐻0 is inside the confidence interval so we do not reject 𝐻0.

(f) (i) Central Limit theorem: Mean of samples equals mean of population

=> 𝜇 = 103𝑔

Standard deviation of the sample means equals the standard deviation of the population divided by the square root of the sample size

=> 0.885 =𝜎

√10

=> 𝜎 = 2.8 (ii) It agrees with the manufacturers details. Comment: This is a great question. Some fine parts and some hard parts, exactly as an exam question should be.

Question 60 (a) (i) The population is divided into subgroups, so that subjects within each subgroup have a common characteristic. Then a simple random sample is drawn from each subgroup to make the full sample. (ii) Short haul passengers. Long hall economy class passengers. Long hall executive class passengers. Long hall business class passengers. (b) (i) Flight delayed

=231

1000 𝑜𝑟 =

231

979

= 0.231 𝑜𝑟 = 0.236 Not satisfied

=238

1000 𝑜𝑟 =

238

902

= 0.238 𝑜𝑟 = 0.264 (ii) No The employee is suggesting that being ‘delayed’ and ‘not satisfied’ are independent events. However, clearly one would affect the other. (c) Graph (ii) Because most passengers will keep their bag under 20kg to avoid extra charges. (d) (i)

(ii) Told that the mean age is greater than the median age => right skewed.

[either answer accepted]

[either answer accepted]


(e) (i) 𝐻0: 𝑝 = 0.7 (70% of customers are satisfied with the overall service of the company) 𝐻1: 𝑝 ≠ 0.7

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

664

1000− 1.96

√664

1000(1 −

6641000

)

1000≤ 𝑝 ≤

664

1000+ 1.96

√664

1000(1 −

6641000

)

1000

0.635 ≤ 𝑝 ≤ 0.693 𝐻0 is outside the confidence interval so we reject 𝐻0.

i.e.: the company’s claim that 70% of customers are satisfied with the overall service of the company is false.

(ii) No Because: With 1000:

𝐸 =1

√𝑛

=1

√1000

= 0.032 With 2000:

𝐸 =1

√𝑛

=1

√2000

= 0.022 (f) (i) (iv)

(ii) = 0.88 (iii) There is a strong positive correlation, which indicates that the older you are the more you will spend. (iv) (see above). Comment: Michael O’Leary will be loving this.


Question 61

a) Students Pensioners

b) List the data in order: 1773.4/1784.8/1803.4/1813.4/1828.6/1867/1903.3/1954.9/2049.6/2060.4

Median is between 5th and 6th value:

𝑚𝑒𝑑𝑖𝑎𝑛 =1828.6 + 1867

2

= 1847.8 = 1,847,800

Interquartile range = Upper quartile – Lower quartile Lower quartile:

1

4× 10 = 2.5

=> use 3rd value = 1803.4

Upper quartile: 3

4× 10 = 7.5

=> use 8th value = 1954.9

=> Interquartile range = 1954.9 − 1803.4 = 151.5 = 151,500

c) (i)

(ii)

(iii) I’m not a fecking economist but here are my thoughts:

I agree. Reason 1: There are more people on the dole which increases the social welfare expenditure. Reason 2: As less people are in the labour force, there is less income tax available to the Government.

d) (i) Disagree

From Liam’s chart it can be seen that decline from 2008-2012 was steeper for males than females. From Niamh’s chart it can be seen that the number of females at work as a percentage of the total persons at work increased from 2008-2012, where as it decreased for the males.

(ii) Liam’s because it is visually clearer. (iii)

Comment: Loads of graphs, which make it more appealing to certain students. Students will find this okay.


Question 62

a) (i)

(ii) 𝑃(𝑇𝑒𝑠𝑡𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒) = 0.00297 + 0.03988 = 0.04285 (iii)

𝑃(ℎ𝑎𝑠 𝑑𝑖𝑠𝑒𝑎𝑠𝑒|𝑡𝑒𝑠𝑡𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒) =𝑃(ℎ𝑎𝑠 𝑑𝑖𝑠𝑒𝑎𝑠𝑒 ∩ 𝑡𝑒𝑠𝑡 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒)

𝑃(𝑡𝑒𝑠𝑡𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒)

=0.00297

0.04285

= 0.0693 (iv)

b) (i) 𝐻0: 𝑝 = 0.51 𝐻1: 𝑝 ≠ 0.51

�̂� − 1.96√�̂�(1 − �̂�)

𝑛≤ 𝑝 ≤ �̂� + 1.96√

�̂�(1 − �̂�)

𝑛

296

500− 1.96

√296500

(1 −296500

)

500≤ 𝑝 ≤

296

500+ 1.96

√296500

(1 −296500

)

500

0.549 ≤ 𝑝 ≤ 0.635 𝐻0 is outside the confidence interval so we reject 𝐻0.

i.e.: Drug A is more successful than the generic drug.

(ii) For 𝐻0 to be accepted, Drug B must be less successful than the generic drug:

�̂� − 𝐸 < 0.51 𝑥

500−

1

√500< 0.51

=> 𝑥 < 277.36 => 277 patients

Comment: Very similar to The Dublin School of Grinds mock exam question from a previous year.


Question 63

(a) 𝑃(𝑆, 𝑆, 𝑆) = (0.7)(0.8)(0.8)

= 0.448 (b)

𝑃(𝑈, 𝑈, 𝑆) = (0.3)(0.4)(0.6) = 0.072

(c) (𝑆, 𝑆, 𝑆) or (𝑈, 𝑈, 𝑆) or (𝑆, 𝑈, 𝑆) or (𝑈, 𝑆, 𝑆)

= 0.448 + 0.072 + (0.7)(0.2)(0.4) + (0.3)(0.6)(0.8) = 0.72

(d) (i)

If 𝑝𝑛 was successful, then 𝑝𝑛+1 = (𝑝𝑛)(0.8) If 𝑝𝑛 was unsuccessful, then 𝑝𝑛+1 = (1 − 𝑝𝑛)(0.6)

=> total 𝑝𝑛+1 = (𝑝𝑛)(0.8) + (1 − 𝑝𝑛)(0.6) = 0.8𝑝𝑛 + 0.6 − 0.6𝑝𝑛

= 0.6 + 0.2𝑝𝑛 (ii)

Given: 𝑝𝑛+1 = 0.6 + 0.2𝑝𝑛 Told 𝑝𝑛+1 ≈ 𝑝𝑛 ≈ 𝑝 => 𝑝 = 0.6 + 0.2𝑝

0.8𝑝 = 0.6 𝑝 = 0.75

(e) (i)

𝑎𝑛+1

𝑎𝑛

=𝑝 − 𝑝𝑛+1

𝑝 − 𝑝𝑛

=0.75 − (0.6 + 0.2𝑝𝑛)

0.75 − 𝑝𝑛

=0.15 − 0.2𝑝𝑛

0.75 − 𝑝𝑛

=0.2(0.75 − 𝑝𝑛)

0.75 − 𝑝𝑛

= 0.2

i. e. :1

5 (this is a constant => geometric)

(ii)

Well 𝑝 − 𝑝𝑛 is 𝑎𝑛 according to part (i).

It also tells us that 𝑎𝑛 is a geometric sequence with 𝑟 =1

5 or 0.2.

The general term of a geometric sequence series is 𝑎𝑟𝑛−1 We know 𝑎𝑛 = 𝑝 − 𝑝𝑛

=> 𝑎1 = 𝑝 − 𝑝1 = 0.75 − 0.7

= 0.05 => (0.05)(0.2)𝑛−1 < 0.00001

0.2𝑛−1 < 0.00002 log 0.2𝑛−1 < log 0.00002

(𝑛 − 1) log 0.2 < log 0.0002

𝑛 − 1 >log 0.00002

log 0.2

𝑛 − 1 > 5.292 𝑛 > 6.292 => 𝑛 = 7

Comment: Parts (a), (b) and (c) were fine, but then it turned into an absolutely ridiculous question.

(inequality changes sign since we

divided across by log0.2 which is

negative)

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