unit 2 ( probability distributions )

PROBABILITY DISTRIBUTIONS C 5606/2/

UNIT 2

PROBABILITY DISTRIBUTIONS

OBJECTIVES

General Objective

To understand the concept of probability distributions

Specific Objectives

At the end of the unit you should be able to: Construct a probability distribution for a random variable. Find the mean, variance, and expected value for a discrete random

variable. Find the exact probability for X successes in n trials of a binomial

experiment. Find the mean, variance, and standard deviation for the variable of a

binomial distribution. Fit a theoretical distribution given a frequency list. Identify distributions as symmetrical or skewed. Identify the properties of the normal distribution. Find the area under the standard normal distribution, given various z

values Find probabilities for a normally distributed variable by transforming it into

a standard normal variable. Find specific data values for given percentages using the standard normal

distribution.

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2.0 INTRODUCTION

A variable can be defined as a characteristic or attribute that can assume different values. Various letters of the alphabet, such as X, Y, or Z, are used to represent variables. Since the variables in this unit are associated with probability, they are called random variables.

If a variable can assume only a specific number of values, such as the outcomes for the roll of a die or the outcomes for the toss of a coin, then the variable is called a discrete variable. Discrete variables have values that can be counted.

Variables that can assume all values in the interval between any given two values are called continuous variables. For example, if the temperature goes from 60o to 75o in a 24-hour period, it has passed through all possible number

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A random variable is variable whose values are determined by chance

INPUTINPUT


from 60 to 75. Continuous random variables are obtained from data that can be measured rather than counted.

2.1 PROBABILITY DISTRIBUTION

The procedure shown here for constructing a probability distribution for a discrete random variable uses the probability experiment of tossing three coins. Recall that when three coins are tossed, the sample space is represented as TTT, TTH, THT, HTT, HHT, HTH, THH, HHH, and if X is the random variable for the number of heads, then X assumes the value 0, 1,2, or 3.

Probabilities for the value of X can be determined as follows:

No heads

One head Two heads Three heads

TTT TTH THT HTT HHT HTH THH HHH

Hence, the probability of getting no heads is , one head is , two heads is ,

and three heads is . From these values, a probability distribution can be

constructed by listing the outcomes and assigning the probability of each outcome , as shown.

Number of heads, X 0 1 2 3Probability, P(X)

A discrete probability distribution consists of the values a random variable can assume and the corresponding probabilities of the values. The probabilities are determined theoretically or by observation.

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Example 2.1

1. Construct a probability distribution for rolling a single die.

2. Represent graphically the probability distribution for the sample space for tossing three coins.

3. In the rainy months, a store selling umbrella keeps track of the number of umbrellas it sells each day during a period of 90 days. The number of umbrellas sold per day is represented by the variable X. The results are shown here.

X Number of days0 451 302 15

Total 90

Compute the probability P(X) for each X, and construct a probability distribution and graph for the data.

Two requirements for a Probability Distribution

The sum of the probabilities of all events in the sample space must equal 1; that is The probability of each event in the sample space must be between or equal to 0 and 1. That is ≤P(X)≤1

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Solution to Example 2.1

1. Since the sample space is S = {1, 2, 3, 4, 5, 6},, and each outcome has a

probability of , the distribution is as follows:

Outcome X 1 2 3 4 5 6Probability, P(X)

2. The values X assumes are located on the x-axis, and the values for P(X) are located on the y axis.

Number of Heads, X

0 1 2 3

Probability, P(X) 1/8 3/8 3/8 1/8

3. The probability P(X) can be computed for each X by dividing the number of days X umbrellas sold by total days.

For 0 umbrellas:

For 1 umbrella:

For 2 umbrellas:

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The distribution is as follows:

Number of umbrellas sold 0 1 2Probability, P(X) 0.5 0.33 0.17

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Number of umbrellas


ACTIVITY 2A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

1. Define and give three examples of a random variable.

2. Give three examples of a discrete random variable.

3. What is a probability distribution?

Question 7 – 9: Determine whether the distribution represents a probability distribution. If it is not, state why.

4.. X 3 6 9 12 15P(X)

94

5. X 1 32 3 4 5P(X)

101

6. X 5 10 15P(X) 1.2 0.3 0.5

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For question 7 to 9, state whether the variable is discrete or continuous.

7. The number of cups of coffee a fast-food restaurant serves each day.

8. The weight of a rhinoceros.

9. The number of lecturers in your polytechnic

For question 10 to 12, construct a probability distribution for the data and draw a graph for the distribution.

10. The probabilities that a patient will have 0, 1, 2, or 3 medical tests performed on entering a hospital are respectively.

11. The probabilities of a machine manufacturing 0, 1, 2, 3, 4, or 5 defective parts in one day are 0.75, 0.17, 0.04, 0.025, 0.01, and 0.005, respectively.

12.. A die is loaded in such a way that the probabilities of getting 1, 2, 3, 4, 5, and 6 are , respectively.

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FEEDBACK TO ACTIVITY 2A

1. A random variable is a variable whose values are determined by chance. Examples will vary.

2. The number of commercials a radio station plays during each hour. The number of times a student uses his or her calculator during a mathematics exam. The number of leaves on a specific type of tree.

3. A probability distribution is a distribution that consists of the values a random variable can assume along with the corresponding probability of these values.

4. Yes5. Yes6. No, the probability values cannot be greater than one.7. Discrete8. Continuous9. Discrete

10 to 12. Refer to the examples given in the input.

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2.2 RANDOM VARIABLES OF THE CONTINUOUS TYPE

If X is a continuous variable and f(x) is a function assigned to it, then

with the following requirements satisfied..

i) For every possible x values

ii)

Therefore f(x) is a probability density function for X (pdf)

Example 2.2

Prove Y is a continuous random variable given the following pdf:

f(y) =

i) Find P(Y>1)ii) P(Y<1)iii) P(0.5<Y 1)iv) Draw the graph for f(y) and double check your answer (i)

through (iii) by finding the area under the graph.

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INPUTINPUT



It is proven that Y is a continuous random variable.

i)

ii)

iii)

iv) P(Y>1) = the area of triangle = ½ X ½ =1/4

1

1/2

1 2 y

f(y)

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(ii) P(Y<1) = Area of trapezium =

iii) P( area of trapezium =

2 y

f(y)

1

1/2

1/2 1 2 y

f(y) 1 3/4

1/2

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ACTIVITY 2B

TEST YOUR UNDERSTANDING BEFORE PROCEEDING TO THE NEXT INPUT…!

1. The continuous random variable has the following pdf:

0.2

f(y) = 0.2 + cy

0 others

Find the constant c and a) b) c) P(Y>0.5) d) P(Y)>0.5 / P(Y)>0.1

2. The continuous random variable x has the following pdf :f(x) = x 0≤x≤2

c(2x-3) 2≤x≤30 others

a) Find the constant cb) Sketch y = f(x)c) Find P (X≤1)d) Find P (X>2.5)e) Find P (1≤X≤2.3)

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3. The continuous random variable x has the following pdf : 1/x2 for 1 <x <∞f(x) =

0 others

If A = { x : 1<x<2} and B = { x: 4 <x<5 }, find a) P(A∩B)b) P(AUB)

4. Find mode for :-

a) P (X) = x = ( )2 x = 1,2,3

b) f(x) = 12x2 (1-x) 0<x<1

c) f(x) = ( )x2ex 0<x<∞

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FEEDBACK TO ACTIVITY 2B

1. c = 1.2 (a) 0.25, (b) 0.35, (c) 0.55, (d) P(Y>0.5) = 0.55, P(Y)>0.1 = 0.774 P(Y)>0.5 / P(Y)>0.1 = 0.71

2. a) c = ¼

b) The sketch of y = f(x)]

c) P = ¼ d) P = 5/16 e) P = 0.3475

1 2 3

1/2

1/4

y = 1/4

15

1

3/4

y = ¼(2x-3)


3. a) P(A∩B) = 0

b) P(AUB) =

4. a) X = 1

b) X =

c) X = 2

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2.3 THE BINOMIAL DISTRIBUTION

Many types of probability problems have only two outcomes, or they can be reduced to two outcomes. For example when a coin is tossed, it can land heads or tails. When a baby is born, it will be either male or female. A multiple-choice, even though there are four or five answer choices, can be classified as correct or incorrect. Situation like these are called binomial experiments.

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A binomial experiment is a probability experiment that satisfies the following four requirements:

1. Each trial can have only two outcomes or outcomes that can be reduced to two outcomes. The outcomes can be considered as either success or failure.

2. There must be a fixed number of trials.3. The outcomes of each trial must be independent of each other.4. The probability of a success must remain the same for each trial.

INPUTINPUT


A binomial experiment and its results give rise to a special probability distribution called the binomial distribution. The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called a binomial distribution.

The probability of a success in a binomial experiment can be computed with the following formula.

Binomial Probability Formula

In a binomial experiment, the probability of exactly X successes in n trials is

Notation for the Binomial Distribution

P(S) The symbol for the probability of successP(F) The symbol for the probability of failurep The numerical probability of a successq The numerical probability of a failure

P(S) = p and P(F) =1 – p = qn The number of trialsX The number of success

Note that 0≤X≥n

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Example 2.3

1. A coin is tossed three times. Find the probability of getting exactly two heads.

2. If a student randomly guesses at five multiple-choice questions, find the probability that the student gets exactly three correct. Each question has five possible choices.

3. Solve the problem in (a) using the binomial distribution table.

4. A fake survey reported that 5% of Malaysians are afraid of being alone at night. If a random sample of 20 Malaysians is selected, find the probabilities using the binomial table:

0 There are exactly five people in the sample who are afraid being alone at night.

1 There are at most three people in the sample who are being afraid of being alone at night.

2 There are at least three people in the sample who are afraid of being alone at night.


1. This problem can be solved by looking at the sample space. There are three ways to get two heads. S = {HHH, HHT,HTH, THH, TTH, THT, HTT, TTT}

The answer is , or 0.375

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2. From the standpoint of a binomial experiment, you can show that it meet the four requirements.

a.There are only two outcomes for each trial, heads or tails. b.There is a fixed number of trials (three) c. The outcomes are independent of each other (the outcome of one toss in

no way affects the outcome of another toss). d.The probability of a success (heads) is in each case.

In this case, n = 3, X = 2, p = , and q = . Hence substituting in the formula gives P(2 heads) =

3. In this case n = 5, X = 3, and P = , since there is one chance in five of guessing a correct answer. Then,

P(3) =

4. You need the binomial distribution table. Fit in the proper values and the answer is 0.375.

i) n = 20, p = 0.05, and X = 5. From the table, one gets 0.02.

ii) n = 20, and p = 0.05. “At most three people” means 0, or 1, or 2, or 3. Hence the solution is P(1) + P(2) + P(3) = 0.358 + 0.377 + 0.189 + 0.060 = 0.984

iii) n = 20 and p = 0.05. “At least three people” means 3, 4, 5,…, 20. This problem can best be solved by finding P(0) + p(1) + P(2) and subtracting from 1.

P(0) + P(1) + P(2) = 0.358 + 0.377 + 0.189 = 0.924

1 – 0.924 = 0.076

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ACTIVITY 2C


1. Which of the following are binomial experiments or can be reduced to binomial experiments?

i) Surveying 100 people to determine if they like Siti Nurhaliza.ii) Tossing a coin 100 times to see how many heads occur.iii) Testing four different brands of aspirin to see which brands are effective.iv) Testing one brand of aspirin using 10 people to see which brands are effective.

2. Compute the probability of X successes, using the binomial formula. I) n = 6, X = 3, p = 0.03 ii) n = 4, x = 2, p = 0.18 iii) n = 5, x = 3, p = 0.63 iv) n = 9, X = 0, p = 0.42

3. A student takes a 10-question, true-false exam and guesses on each question. Find the probability of passing if the lowest passing grade is 6 correct out of 10. Based on your answer, would it be a good idea not to study and depend on guessing?

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Feedback to Activity 2C

1. Yes, ii) Yes, iii) No, iv) Yes

2. Refer to the table

3. 0.377; no, your score will be about 40%

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2.4 MEAN, VARIANCE AND STANDARD DEVIATION FOR THE BINOMIAL DISTRIBUTION

In the use of the binomial distribution, the outcomes must be independent. For example, in a selection of components from a batch to be tested, each component must be replaced before the next one is selected. Otherwise, the outcomes are not independent. However, a dilemma arises because there is a chance that the same component could be selected again. This situation can be avoided by not replacing the component and using the hyper geometric distribution to calculate the probabilities.

Mean, Variance, and Standard Deviation for the binomial Distribution

mean variance

standard deviation =

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INPUTINPUT


Example 2.4

1. A coin is tossed is tossed four times. Find the mean, variance, and standard deviation of the number of heads that it will be obtained.

2. A die is rolled 480 times. Find the mean, variance, and standard deviation of the number of 2 s that will be rolled.


1. With the formulas for the binomial distribution and n = 4, p = , and q = , the result are

2. n = 480, p = . This is a binomial situation, where getting a 2 is a success and not getting 2 is a failure; hence,

On average, there will be 80 twos. The standard deviation is 8.

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ACTIVITY 2C


1.A statistical bulletin reported that 2% of all births result in twins. If a random sample of 8000 births is taken, find the mean, variance, and standard deviation of the number of births that would result in twins.

2.If 2% of automobile carburetors are defective, find the mean, variance, and standard deviation of 500 carburetors.

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FEEDBACK TO ACTIVITY 2C

1. 160, 156.8, 12.5

2.

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2.5 THE NORMAL DISTRIBUTION

What is Normal?

It is known that the normal range of systolic blood pressure 110 to 140. The normal interval for a person’s triglycerides is from 30 to 200 mg/dl. By measuring these variables, a doctor can determine if a patient’s vital statistics are within the normal interval, or if some type of treatment is needed to correct the condition and avoid future illness. The question then is ‘How does one determine the so-called normal interval?’

Recall that a continuous variable can assume all values between any two given values of the variables. Many continuous variables have distributions that are bell-shaped and are called approximately normally distributed variables. For example, if we select a random sample of 100 adult women, measure their heights, and construct a histogram, we get a graph similar to the one in fig (a). Now if we increase the sample size and decrease the width of the classes, the histogram will look like the ones shown in fig. (d) and (c). Finally, if it were possible to measure exactly the heights of all adult females in Malaysia and plot them, the histogram would approach what is called the normal distribution, shown in (d).

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INPUTINPUT


When the data values are evenly distributed about the mean, the distribution is said to be symmetrical. See fig (a) below. When the majority of the data values fall to the left or right of the mean, the distribution is said to be skewed. See fig (b) and (c)

Normal and Skewed Distributions

For the sake of simplicity, a theoretical curve, called the normal distribution curve can be used to study many variables that are not perfectly normally distributed but are nevertheless approximately normal. The mathematical equation for the

normal distribution curve is , where e 2.718, =3.14, =

population mean, = population standard deviation.

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Another important aspect in statistics is that the area under the normal distribution curve is more important than frequencies. Therefore, when the normal distribution is pictured, the y axis, which indicates the frequencies, is sometimes omitted.

The shape and position of the normal distribution curve depend on two parameters, the mean and standard deviation. Each normally distributed variable has its own normal distribution curve, which depends on the values of the variable’s mean and standard deviation.

Fig (a) shows two normal distributions with the same mean values but different standard deviations. The larger the standard deviation, the more dispersed, or spread out the distribution is. Fig (b) shows two normal distributions with the same standard deviation with the same standard deviation but with different means. These curves have the same shapes but are located at different positions on the x axis. Fig (c) shows two normal distributions with different means and different standard deviation.

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The normal distribution is a continuous, symmetric, bell-shaped distribution of a variable.

Summary of the Properties of the Theoretical Normal Distribution

1. The normal distribution curve is bell-shaped.2. The mean, median, and mode are equal and

located at the center of the distribution.3. The normal distribution curve is unimodal (i.e., it

has only one mode).4. The curve is symmetrical about the mean, which is

equivalent to saying that its shape is the same on both sides of a vertical line passing through the center.

5. The curve is continuous – i.e., there are no gaps or holes. For each value of X, there is a corresponding value of Y.

6. The curve never touches the x axis. Theoretically, no matter how far in either direction the curve extends, it never meets the x axis – but it gets increasingly closer

7. The total area under the normal distribution curve is equal to 1.00 or 100%.

8. The area under the normal curve that lies within one standard deviation of the mean is approximately 0.68, or 68%; within two standard deviations, about 0.95, or 95%; and within three standard deviations, about 0.997, or 99.7%. See the figure below that shows the area in each region.

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The Standard Normal Distribution

Since each normally distributed variable has its own mean and standard deviation, the shape and location of these curves will vary. In practical applications, one would have to have a table of areas under the curve for each variable. To simplify this situation, statisticians use what is called the standard normal distribution.

Areas under the Normal Distribution Curve

The standard normal distribution is a normal distribution with a mean 0 and a standard deviation of 1

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Standard Normal Distribution

The values under the curve indicate the proportion of area in each section. For example, the area between the mean and one standard deviation above or below the mean is about 0.3143, or 31.43%.

The formula for the standard normal distribution is

All normally distributed variables can be transformed into the standard normally distributed variable by using the formula for the standard score

z= or

This is the same formula used in Section 3-4. The use of this formula will be explained in the next session.

As stated earlier, the area the normal distribution curve is used to solve practical application problems, such as finding the percentage of adult women whose height is between 5 feet 4 inches and 5 feet 7 inches, or finding the probability that a new battery will last longer than four years. Hence, the major emphasis of this section will be to show the procedure for finding the area under the standard normal distribution curve for any z value. The application will be shown in the next section. Once the X values are transformed using the preceding formula, they are called z value. The z value is actually the number of standard deviations that a particular X value is away from the mean.

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For the solution of problems using the normal distribution, a four-step procedure is recommended with the use of the Procedure Table shown below.

STEP 1 Draw a picture

STEP 2 Shade the area desired.

STEP 3 Find the correct figure in the following Procedure Table.

STEP 4 Follow the directions given in the appropriate block of the Procedure Table to get the desired area.

Finding The Area under The Normal Distribution Curve

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Example 2.5

Find the probability (i.e. the area under the curve) for each question 1 through 17.

Question 1

a. b. c. d. e.

Question 2

a. If X~N(1, 4), find

b. If X~N(2, 4), find

c. If X~N(2, 9), find

d. If X~N(-5, 36), find

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Question 1

35


Question 2

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ACTIVITY 2D


1. Find the area to the right of Z = 2.43 and to the left of Z = -3.01.

2. Find the probability for each.

a) P(0<z<2.32) b) P(z<1.65) c) P(z>1.91)

3. Find the z value such that the areas under the normal distribution curve between 0 and the z value is 0.2123.

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FEEDBACK TO ACTIVITY 2D

1. 0.0088, or 0.88%

2. i) 0.4898, or 48.98% ii) 0.9505, or 95.05% iii) 0.0281, or 2.81%

3. 0.56

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2.6 APPLICATIONS OF THE NORMAL DISTRIBUTION

To solve problems by using the standard normal distribution, transform the original variable into a standard normal a standard normal distribution variable by using the formula

or

For example, suppose that the scores for a standardized test are normally distributed, have a mean of 100, and have a standard deviation of 15. When the scores are transformed into z values, the two distributions coincide, as shown in the figure below. (Recall that the z distribution has a mean of 0 and a standard deviation of 1)

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INPUTINPUT


Example 2.6

1. If the scores for the test have a mean of 100 and a standard deviation of 15, find the percentage of scores that will fall below 112.

2. Each month, a Malaysian household generates an average 28 kg of newspaper for garbage recycling. Assume the standard deviation is 2 kg. If a household is selected at random, find the probability of its generating

i) Between 27 and 31 kg per month. ii) More than 30.2 kg per month. Assume the variable is approximately normally distributed.

3. The normal distribution can also be used to answer questions of “ how many?” Automobile Association of Malaysia reports that the average time it takes to respond to an emergency call is 25 minutes. Assume the variable is approximately distributed and the standard deviation is 4.5 minutes. If 80 calls are randomly selected, approximately how many will be responded to in less than 15 minutes.

4. In order to qualify for ‘Askar Wataniah Politeknik’ training program, candidates must score in the top 10% on a general abilities test. The test has a mean of 200 and a standard deviation of 20. Find the lowest possible score to qualify. Assume the test scores are normally distributed.

5. For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure. If the mean systolic blood

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pressure is 120 and the standard deviation is 8, find the upper and lower readings that would qualify people to participate in the study.


1.

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42

Hence, the probability that a randomly selected household generates between 27 and 31 kg of newspaper per month is 62.47%.


3.

STEP 2 Find z values for 15.

STEP 3 Find the appropriate area. The area obtained from the table is 0.4868, which corresponds to the area between z = 0 and z = -2.2

STEP 4 Subtract 0.4868 from 0.5000 to get 0.0132.

STEP 5 To find how many calls will be made in less than 15 minutes, multiply the sample size (80) by the area (0.0132) to get 1.056. Hence, 1.056, or approximately one, call will be responded to in under 15 minutes.

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4.

STEP 1 Subtract 0.1000 from 0.5000 to get area under the normal distribution between 200 and X: 0.5000 – 0.1000 = 0.4000

STEP 2 Find the z value that corresponds to an area of 0.4000 by looking up 0.4000 in the table. If the specific value cannot be found, use the closest value, for example 0.3997. The corresponding z value is 1.28.

STEP 3 Substitute in the formula and solve for X.

of which the answer is X = 226.

A score of 226 should be used as cutoff. Anybody scoring 226 or higher qualifies.

5. Assume that blood pressure readings are normally distributed; then cutoff points are as shown in the figure below.

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Note that two values are needed, one above the mean and one below the mean. Find the value to the right of the value first. The closest z value for an area of 0.3000 is 0.84. Substituting in the formula , one gets

, On the other side, z = -0.84; hence

Therefore , the middle 60% will have blood pressure readings of 113.28<X<126.72

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ACTIVITY 2E


1. The mean income of daily labors is RM60. Assume the standard deviation RM10. If a labor is selected at random, what is the probability that his salary is -

a) more than RM70 b) below RM62 c) more than RM64 d) between RM62 and RM72 e) between RM55 and RM80

2. Absorption tests on 500 bricks gave a mean absorbtion and standard deviation of 10.3% and 2.1% respectively. Determine the number of bricks with absorbtion

a) between 8.9% and 11.1%b) greater than 15%.

3. Chip-board partition units are produced with a mean length of 75mm and standard deviation 6 mm. Determine the rejection rate if the permissible length deviations are 15 mm (assume the lengths to be normally distributed).

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FEEDBACK TO ACTIVITY 2E

1. a) P(X>70) = 0.1567b) P(X<62) = 0.5793c) P(X>54) = 0.7257d)e)

2. a) P(8.9<X<11.1) = 0.3966 b) P(X>15) = 2.24

3. P(735>X>765) = 0.0124 that is the rejection rate is 1.24%

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SELF-ASSESMENT 2

You are approaching success. Try all the questions in this self-assessment section and check your answer given on the next page. If you face any problems, consult your instructor. Good luck.

1. What is a probability distribution? Give an example.2. Determine whether the distribution represents a probability distribution. If it

is not, state why

X 3 6 9 12 15P(X) 4/9 2/9 1/9 1/9 1/9

Construct a probability distribution for the data and draw a graph for the distribution.

State whether The Variable is Discrete or Continuous (3-4)

3. The number of cups of coffee a fast food restaurant serves each day.4. The weight of a rhinoceros5. The probability that a patient will have 0, 1,2 or 3 medical tests performed

on entering a hospital are , , and ,

respectively.

6. The probabilities that a customer will purchase 0,1,2, or 3 books are 0.45, 0.30 ,0.15 and 0.10, respectively.

7. The probabilities that a customer selects 1, 2, 3, 4 or 5 items at a convenience store are 0.32, 0.12, 0.23, 0.18 and 0.15, respectively.

8. Construct a probability distribution for drawing a card from a deck of 40 cards consisting of 10 cards numbered 1, 10 cards numbered 2 and 15 cards numbered 3 and 5 cards numbered 4

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FEEDBACK TO SELF-ASSESSMENT 2

Have you tried the questions??? If “YES”, check your answers now.

1. A probability distribution is a distribution that consists of the values a random variable can assume along with the corresponding probabilities of these values.

2. Yes

3. Discrete

4. Continuous

5 – 8: Refer to example 2.1

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unit 2 ( probability distributions )

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