midterm exam iii review - binghamton...

Midterm Exam III Review

Dr. Joseph Brennan

Math 148, BU

Dr. Joseph Brennan (Math 148, BU) Midterm Exam III Review 1 / 25

Permutations and Combinations

ORDER

In order to count the number of possible ways to choose, withoutreplacement, k objects from a collection of n distinct objects we must bespecific as to we acknowledge order.

A permutation is a choice where order matters.

A combination is a choice where order does not matter.

The only difference between a permutation and a combination is order.This leads to very similar counting formulas:

nPk =n!

(n − k)!

(n

k

)=

n!

k! · (n − k)!

Recall: An event E in the sample space S has probability

P(E ) =number of outcomes in E

number of outcomes in S


iClicker

A graduate class is comprised of 2 overachieving undergraduates, 7 firstyear graduates, 4 second year graduates, and 1 third year graduate. If classbegins with random presentations by three students, what is the probabilitythat an undergraduate and two first year graduates are chosen?

(A) 0%

(B) 6%

(C) 11%

(D) 19%

(E) 27% (21

)×(72

)(143

) ≈ 11%


Law of Averages

Law of Averages: If an experiment is independently repeated alarge number of times, the percentage of occurrences of a specific event Ewill be the theoretical probability of the event occurring, but of by someamount - the chance error.


iClicker

There are two hospitals: in the Hospital A 120 babies are born every day;in Hospital B 12 babies are born.

On average, the ratio of baby boys to baby girls born every day in eachhospital is 50/50. However, one day, in one of those hospitals, twice asmany baby girls were born as baby boys. In which hospital was it morelikely to happen?

(A) Hospital A

(B) Hospital B

(C) Hospital C

(D) Equally Likely

Hospital B. The probability of a random deviation of a particular size (fromthe population mean), decreases with the increase in the sample size.


Random Variable

Random Variable: An unknown subject to random change. Often arandom variable will be an unknown numerical result of study.

A random variable has a numerical sample space where each outcome hasan assigned probability. There is not necessarily equal assignedprobabilities.

Any random variable X , discrete or continuous, can be described with

A probability distribution.

A mean and standard deviation.


Random Variables

The distribution of a discrete random variable X is summarized in thedistribution table:

Value of X x1 x2 x3 ... xkProbability p1 p2 p3 ... pk

The symbols xi represent the distinct possible values of X and pi is theprobability associated to xi .

p1 + p2 + . . .+ pk = 1 (or 100%)


3 Coins Probability Histogram


Discrete Random Variable: µ

Mean: The mean µ of a discrete random variable is found bymultiplying each possible value by its probability and adding together allthe products:

µ = x1p1 + x2p2 + . . .+ xkpk =k∑

i=1

xipi


Discrete Random Variable: σ

Standard Deviation: The standard deviation σ of a discreterandom variable is found with the aid of µ:

σ =√

(x1 − µ)2p1 + (x2 − µ)2p2 + . . . (xk − µ)2pk

=

√√√√ k∑i=1

(xi − µ)2pi

When there are just two numbers, x1 and x2, in the distribution of X thedistribution’s standard deviation, σ, can be computed by using thefollowing short-cut formula:

σ = |x1 − x2|√p1p2

where pi is the probability of xi .


Box Models

Box Model: A model framing a statistical question as drawing tickets(with or without replacement) from a box. The tickets are to be labeledwith numerical values linked to a random variable.

The expected value of a random variable is the average of the ticketsoccupying the box model.

The standard deviation of a random variable is the standard deviationof the tickets.


iClicker

Did you know there are 4, 8, 10, 12, and 20 sided die? Create a box modelfor rolling a 4-sided die. What is the standard deviation of the box?

(A) 1.11

(B) 1.65

(C) 2

(D) 2.21

(E) 2.5

Solution: First find the mean:

µ = 1× 1

4+ 2× 1

4+ 3× 1

4+ 4× 1

4= 2.5 σ = 1.11

σ =

√(1− 2.5)2 × 1

4+ (2− 2.5)2 × 1

4+ (3− 2.5)2 × 1

4+ (4− 2.5)2 × 1

4


The Sum of n Independent Outcomes

When the same experiment is repeated independently n times, thefollowing is true for the sum of outcomes:

The expected value of the sum of n independent outcomes of anexperiment:

nµ

The standard error of the sum of n independent outcomes of anexperiment: √

nσ

The second part of the above rule is called the the Square Root Law.

Note that the above rule is true for any sequence of independent randomvariables, discrete or continuous!


The Binomial Setting

1 There are a fixed number of n of repeated trials.

2 The trials are independent. In other words, the outcome of anyparticular trial is not influenced by previous outcomes.

3 The outcome of every trial falls into one of just two categories, whichfor convenience we call success and failure.

4 The probability of a success, call it p, is the same for each trial.

5 It is the total number of successes that is of interest, not their orderof occurrence.

NOTE: The Binomial Setting can be framed as a box model with only 1’sand 0’s where draws are performed with replacement.


The Binomial Distribution

Let X denote the number of successes under the binomial setting. Then Xis a random variable which may take values 0, 1, 2, 3, ..., n. In particular,

X = 0 means no successes in n trails. Only failures were observed.

X = n means the outcomes of all n trails are successes.

X = 5 means 5 successes in n trials.

It turns out that X has a special discrete distribution which is called thebinomial distribution. The probabilities of values of X are computed as

P(X = k) =

(n

k

)pk(1− p)n−k , k = 0, 1, 2, . . . , n. (1)

So the binomial distribution is a probability distribution of a randomvariable X which has 2 parameters: p (probability of success) and n (thenumber of trials).


Binomial Mean and Standard Deviation

Let X be a binomial random variable with parameters n (number of trials)and p (probability of success in each trial). Then the mean and standarddeviation of X are

µ = np,

σ =√np(1− p).


Binomial Distribution and Normal Curves

NORMAL APPROXIMATION for BINOMIAL COUNTS

Let X be a random variable which has a binomial distribution withparameters n and p. When n is large, the distribution of X isapproximately normal.

X is approximately normal with mean np and standard deviation√np(1− p).

As a rule, we will use this approximation for values of n and p that satisfynp ≥ 10 and n(1− p) ≥ 10.


The Central Limit Theorem (CLT)

The Central Limit Theorem: When drawing at random withreplacement from a box, the probability histogram for the sum willapproximately follow the normal curve, even if the contents of the box donot. The larger the number of draws, the better the normal approximation.

The sample size n should be at least 30 (n ≥ 30) before the normalapproximation can be used.

For symmetric population distributions the distribution of x̄ is usuallynormal-like even at n = 10 or more.

For very skewed populations distributions larger values of n may beneeded to overcome the skewness.


Parameters & Statistics

Parameter: A numerical fact about a population.

Statistic: A numerical fact about a sample.

An investigator knows a statistic and wants to know a parameter.

Probability Methods: Sampling techniques which implements anobjective chance process to choose subjects from the population, leavingno discretion to the interviewer.

It is possible to compute the chance that any particular individual inthe population will get into the sample.

Simple Random Sampling: A sampling technique where selection ofindividuals is equally likely and drawing for the sample is performedwithout replacement.


Bias

Sampling Bias: A bias in which a sample is collected in such a way thatsome members of the intended population are less likely to be includedthan others.

The bias can lead to an over/underrepresentation of thecorresponding parameter in the population.

Almost every sample in practice is biased because it is practicallyimpossible to ensure a perfectly random sample.

Non-response Bias: A bias that results when respondents differ inmeaningful ways from nonrespondents.

Respondents and nonrespondents can differ in ways beyond theirwillingness to answer a questionnaire.

Quota Sampling: A sampling method in which interviewers areassigned a fixed quota of subjects to interview.


Variable Type

Given n draws with replacement from a box with

Mean µ (average for quantitative and percent for qualitative).

Standard Deviation σ.

Sum Average Number PercentExpected Value: n × µ µ n × µ µ

Standard Error:√n × σ σ/

√n

√n × σ σ/

√n


The Correction Factor

When drawing without replacement, to get the exact SE you mustmultiply by the correction factor:√

number of objects in box− number of draws

number of objects in box− 1

When the number of tickets in the box is large relative to the number ofdraws, the correction factor is nearly one.


Normal Curve for SE for Averages and Percentages

Suppose 1,000 draws are made with replacement from a box whoseaverage ticket value is 200. The standard error for averages is found tobe 10.

There is about a 68% chance for the average of the 1, 000 draws tobe in the range 190 to 210.

Suppose 1,000 draws are made with replacement from a 0− 1 box whosepercent of 1’s was 15%. The standard error for percent is found to be0.5%.

There is about a 68% chance for the percentage of successful drawsof the 1, 000 draws to be in the range 14.5% to 15.5%.


iClicker

Historically, Five Star Pizza has recorded Thursday night drivers as havingan average delivery time of 28 minutes with a standard deviation of 5minutes.

If the manager were to take a sample of 100 orders at random onThursday nights spread over a year, what is the chance that the averagedelivery time is greater than 30 minutes?

(A) 0%

(B) 2%

(C) 5%

(D) 10%

(E) 40%

Solution: The box has µ = 22 and σ = 5.

EV = 28 SE =5√100

= 0.5

We can answer this question using the normal curve:

P(X > 30) = 1− P

(Z <

30− 28

0.5

)= 1− P(Z < 4) ≈ 0


iClicker

Five Star Pizza has not been keeping track of their Monday night deliverytime and the manager has been receiving complaints. The manager takesa simple random sample of delivery 40 deliveries, and finds a sample meanof 36 minutes and a sample standard deviation of 12 minutes.

What is the 90% confidence interval for the average Monday night deliverytime?

(A) [35,37]

(B) [34,38]

(C) [32,40]

(D) [30,42](E) [28,44]

Solution: Begin by finding the z-score for confidence:

(90/2 + 50) = 95 with z-score: 1.66

The margin of error is found

m = zC × SE = 1.66× 12√40≈ 4

[x̄ −m, x̄ + m] = [32, 40]


midterm exam iii review - binghamton...

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