﻿ ﻿Binomial Distributions, Geometric Distributions, and Sampling Distribution - Review the Knowledge You Need to Score High - 5 Steps to a 5 AP Statistics 2017 (2016)

## 5 Steps to a 5 AP Statistics 2017 (2016)

### Binomial Distributions, Geometric Distributions, and Sampling Distributions

IN THIS CHAPTER

Summary: In this chapter we finish laying the mathematical (probability) basis for inference by considering the binomial and geometric situations that occur often enough to warrant our study. In the last part of this chapter, we begin our study of inference by introducing the idea of a sampling distribution, one of the most important concepts in statistics. Once we”ve mastered this material, we will be ready to plunge into a study of formal inference (Chapters 1114 ).

Key Ideas

Binomial Distributions

Normal Approximation to the Binomial

Geometric Distributions

Sampling Distributions

Central Limit Theorem

Binomial Distributions

A binomial experiment has the following properties:

• The experiment consists of a fixed number,n , of identical trials.
• There are only two possible outcomes (that”s the “bi” in “binomial”): success (S) or failure (F ).
• The probability of success,p , is the same for each trial.
• The trials are independent (that is, knowledge of the outcomes of earlier trials does not affect the probability of success of the next trial).
• Our interest is in abinomial random variable X , which is the count of successes in n trials. The probability distribution of X is the binomial distribution .

(Taken together, the second, third, and fourth bullets above are called Bernoulli trials . One way to think of a binomial setting is as a fixed number n of Bernoulli trials in which our random variable of interest is the count of successes X in the n trials. You do not need to know the term Bernoulli trials for the AP exam.)

The short version of this is to say that a binomial experiment consists of n independent trials of an experiment that has two possible outcomes (success or failure), each trial having the same probability of success (p ). The binomial random variable X is the count of successes.

In practice, we may consider a situation to be binomial when, in fact, the independence condition is not quite satisfied. This occurs when the probability of occurrence of a given trial is affected only slightly by prior trials. For example, suppose that the probability of a defect in a manufacturing process is 0.0005. That is, there is, on average, only 1 defect in 2000 items. Suppose we check a sample of 10,000 items for defects. When we check the first item, the proportion of defects remaining changes slightly for the remaining 9,999 items in the sample. We would expect 5 out of 10,000 (0.0005) to be defective. But if the first one we look at is not defective, the probability of the next one being defective has changed to 5/9999 or 0.0005005. It”s a small change, but it means that the trials are not, strictly speaking, independent. A common rule of thumb is that we will consider a situation to be binomial if the population size is at least 10 times the sample size.

Symbolically, for the binomial random variable X , we say X has B (n, p ).

example: Suppose Dolores is a 65% free throw shooter. If we assume that that repeated shots are independent, we could ask, “What is the probability that Dolores makes exactly 7 of her next 10 free throws?” If X is the binomial random variable that gives us the count of successes for this experiment, then we say that X has B (10, 0.65). Our question is then: P (X = 7) = ?

We can think of B (n, p, x ) as a particular binomial probability. In this example, then, B (10, 0.65, 7) is the probability that there are exactly 7 successes in 10 repetitions of a binomial experiment where p = 0.65. This is handy because it is the same syntax used by the TI-83/84 calculator (binompdf(n,p,x)) when doing binomial problems.

If X has B (n, p ), then X can take on the values 0, 1, 2, …, n . Then,

gives the binomial probability of exactly x successes for a binomial random variable X that has B(n, p ).

Now,

On the TI-83/84,

and this is found in the MATH PRB menu. n! (“n factorial”) means n (n – 1)(n – 2) … (2)(1), and the factorial symbol can also be found in the MATH PRB menu.

example: Find B (15, .3, 5). That is, find P (X = 5) for a 15 trials of a binomial random variable X that succeeds with probability 0.3.

solution:

(On the TI-83/84, can be found

in the MATH PRB menu. To get , enter 15n Cr 5. )

Calculator Tip: On the TI-83/84, the solution to the previous example is given by binompdf(15,0.3,5) . The binompdf function is found in the DISTR menu of the calculator. The syntax for this function is binompdf(n, p, x) . The function binomcdf(n, p, x) = P(X = 0) + P(X = 1) + … P(X = x ). That is, it adds up the bino-mial probabilities from n = 0 through n = x . You must remember the “npx” order—it”s not optional. Try a mnemonic like “never play xylophone.”

example: Consider once again our free-throw shooter (Dolores) from the earlier example. Dolores is a 65% free-throw shooter and each shot is independent. If X is the count of free throws made by Dolores, then X has B (10, 0.65) if she shoots 10 free throws. What is P (X = 7)?

solution:

example: What is the probability that Dolores makes no more than 5 free throws? That is, what is P (X ≤ 5)?

solution:

There is about a 25% chance that she will make 5 or fewer free throws. The solution to this problem using the calculator is given by binomcdf (10,0.65,5).

example: What is the probability that Dolores makes at least 6 free throws?

solution: P (X6 ) = P (X = 6 ) + P (X = 7) + … + P (X = 10) = 1-binomcdf(10,0.65,5)=0.751.

(Note that P (X > 6) = 1 – binomcdf(10,0.65,6)) .

The mean and standard deviation of a binomial random variable X are given by μ x = np ; . A binomial distribution for a given n and p (meaning you have all possible values of x along with their corresponding probabilities) is an example of a probability distribution as defined in Chapter 7 . The mean and standard deviation of a binomial random variable X could be found by using the formulas from Chapter 9 ,

but clearly the formulas for the binomial are easier to use. Be careful that you don”t try to use the formulas for the mean and standard deviation of a binomial random variable for a discrete random variable that is not binomial.

example: Find the mean and standard deviation of a binomial random variable X that has B (85, 0.6).

solution: μ x = (85)(0.6) = 51; .

Normal Approximation to the Binomial

Under the proper conditions, the shape of a binomial distribution is approximately normal, and binomial probabilities can be estimated using normal probabilities. Generally, this is true when np ≥ 10 and n (1 – p ) ≥ 10 (some books use np5 and n (1 – p ) ≥ 5; that”s OK). These conditions are not satisfied in Graph A (X has B (20, 0.1)) below, but they are satisfied in Graph B (X has B (20, 0.5))

It should be clear that Graph A is noticeably skewed to the right, and Graph B is approximately normal in shape, so it is reasonable that a normal curve would approximate Graph B better than Graph A. The approximating normal curve clearly fits the binomial histogram better in Graph B than in Graph A.

When np and n (1 – p ) are sufficiently large (that is, they are both greater than or equal to 10 or 5), the binomial random variable X has approximately a normal distribution with

Another way to say this is: If X has B (n, p ), then X has approximately , provided that np ≥ 10 and n (1 – p ) ≥ 10 (or np ≥ 5 and n( 1 – p ) ≥ 5).

example: Nationally, 15% of community college students live more than 6 miles from campus. Data from a simple random sample of 400 students at one community college are analyzed.

(a) What are the mean and standard deviation for the number of students in the sample who live more than 6 miles from campus?

(b) Use a normal approximation to calculate the probability that at least 65 of the students in the sample live more than 6 miles from campus.

solution: If X is the number of students who live more than 6 miles from campus, then X has B (400, 0.15).

(a) μ = 400(0.15) = 60; .

(b) Because 400(0.15) = 60 and 400(0.85) = 340, we can use the normal approximation to the binomial with mean 60 and standard deviation 7.14. The situation is pictured below:

Using Table A, we have .

By calculator, this can be found as normalcdf(65,1000,60,7.14) = 0.242.

The exact binomial solution to this problem is given by

In reality, you will need to use a normal approximation to the binomial only in limited circumstances. In the example above, the answer can be arrived at quite easily using the exact binomial capabilities of your calculator. The only time you might want to use a normal approximation is if the size of the binomial exceeds the capacity of your calculator (for example, enter binomcdf(50000000,0.7,3250000) . You”ll most likely see ERR:DOMAIN, which means you have exceeded the capacity of your calculator, and you didn”t have access to a computer. The real concept that you need to understand the normal approximation to a binomial is that another way of looking at binomial data is in terms of the proportion of successes rather than the count of successes. We will approximate a distribution of sample proportions with a normal distribution and the concepts and conditions for it are the same.

Geometric Distributions

In the Binomial Distributions section of this chapter, we defined a binomial setting as an experiment in which the following conditions are present:

• The experiment consists of a fixed number,n , of identical trials.
• There are only two possible outcomes: success (S) or failure (F ).
• The probability of success,p , is the same for each trial.
• The trials are independent (that is, knowledge of the outcomes of earlier trials does not affect the probability of success of the next trial).
• Our interest is in abinomial random variable X , which is the count of successes in n trials. The probability distribution of X is the binomial distribution .

There are times we are interested not in the count of successes out of n fixed trials, but in the probability that the first success occurs on a given trial, or in the average number of trials until the first success. A geometric setting is defined as follows.

• There are only two possible outcomes: success (S) or failure (F ).
• The probability of success,p , is the same for each trial.
• The trials are independent (that is, knowledge of the outcomes of earlier trials does not affect the probability of success of the next trial).
• Our interest is in ageometric random variable X , which is the number of trials necessary to obtain the first success.

Note that if X is a binomial , then X can take on the values 0, 1, 2, …, n . If X is geometric , then it takes on the values 1, 2, 3, …. There can be zero successes in a binomial, but the earliest a first success can come in a geometric setting is on the first trial.

If X is geometric, the probability that the first success occurs on the nth trial is given by P (X = n ) = p (1 – p )n1 . The value of P (X = n ) in a geometric setting can be found on the TI-83/84 calculator, in the DISTR menu, as geometpdf(p,n) (note that the order of p and n are, for reasons known only to the good folks at TI, reversed from the binomial). Given the relative simplicity of the formula for P (X = n ) for a geometric setting, it”s probably just as easy to calculate the expression directly. There is also a geometcdf function that behaves analogously to the binomcdf function, but is not much needed in this course.

example: Remember Dolores, the basketball player whose free-throw shooting percentage was 0.65? What is the probability that the first free throw she manages to hit is on her fourth attempt?

solution: P (X = 4) = (0.65) (1 – 0.65)4–1 = (0.65) (0.35)3 = 0.028. This can be done on the TI-83/84 as follows: geometpdf(p,n) = geometpdf(0.65,4) = 0.028.

example: In a standard deck of 52 cards, there are 12 face cards. So the probability of drawing a face card from a full deck is 12/52 = 0.231.

(a) If you draw cards with replacement (that is, you replace the card in the deck before drawing the next card), what is the probability that the first face card you draw is the 10th card?

(b) If you draw cards without replacement, what is the probability that the first face card you draw is the 10th card?

solution:

(a) P (X = 10) = (0.231) (1 – 0.231)9 = 0.022. On the TI-83/84: geometpdf(0.231,10) =0.0217).

(b) If you don”t replace the card each time, the probability of drawing a face card on each trial is different because the proportion of face cards in the deck changes each time a card is removed. Hence, this is not a geometric setting and cannot be answered by the techniques of this section. It can be answered, but not easily, by the techniques of the previous chapter.

Rather than the probability that the first success occurs on a specified trial, we may be interested in the average wait until the first success. The average wait until the first success of a geometric random variable is 1/p . (This can be derived by summing (1) · P (X = 1) + (2) · P (X = 2) + (3) · P (X = 3) + ... = 1p + 2p (1 – p ) + 3p (1 – p )2 + …, which can be done using algebraic techniques for summing an infinite series with a common ratio less than 1.)

example: On average, how many free throws will Dolores have to take before she makes one (remember, p = 0.65)?

solution: .

Since, in a geometric distribution P (X = n ) = p (1 – p ) n –1 , the probabilities become less likely as n increases since we are multiplying by 1 – p , a number less than one. The geometric distribution has a step-ladder graph that looks like this:

Sampling Distributions

Suppose we drew a sample of size 10 from an approximately normal population with unknown mean and standard deviation and got = 18.87. Two questions arise: (1) what does this sample tell us about the population from which the sample was drawn, and (2) what would happen if we drew more samples?

Suppose we drew 5 more samples of size 10 from this population and got and . In answer to question (1), we might believe that the population from which these samples was drawn had a mean around 20 because these averages tend to group there (in fact, the six samples were drawn from a normal population whose mean is 20 and whose standard deviation is 4). The mean of the 6 samples is 19.64, which supports our feeling that the mean of the original population might have been 20.

The standard deviation of the 6 samples is 0.68, and you might not have any intuitive sense about how that relates to the population standard deviation, although you might suspect that the standard deviation of the samples should be less than the standard deviation of the population because the chance of an extreme value for an average should be less than that for an individual term (it just doesn”t seem very likely that we would draw a lot of extreme values in a single sample).

Suppose we continued to draw samples of size 10 from this population until we were exhausted or until we had drawn all possible samples of size 10 . If we did succeed in drawing all possible samples of size 10, and computed the mean of each sample, the distribution of these sample means would be the sampling distribution of .

Remembering that a “statistic” is a value that describes a sample, the sampling distribution of a statistic is the distribution of that statistic for all possible samples of a given size. It”s important to understand that a dotplot of a few samples drawn from a population is not a distribution (it”s a simulation of a distribution)—it becomes a distribution only when all possible samples of a given size are drawn.

Sampling Distribution of a Sample Mean

Suppose we have the sampling distribution of . That is, we have formed a distribution of the means of all possible samples of size n from an unknown population (thus, we know little about its shape, center, or spread). Let μ and σ represent the mean and standard deviation of the sampling distribution of , respectively.

Then

for any population with mean μ and standard deviation σ .

(Note: the value given for σ above is correct only if the sample there is true independence between trials, such as when sampling with replacement, or when the population is infinite, such as when tossing a coin. The formula still works well enough if the sample size n is small compared to the population size N . A general rule is that n should be no more than 10% of N to use the value given for σ (that is, N > 10n ). If n is more than 10% of N , the exact value for the standard deviation of the sampling distribution is

In practice this usually isn”t a major issue because

is close to one whenever N is large in comparison to n . (You don”t have to know this formula for the AP exam. However, you do need to understand that when our sample size is larger than 10% of the population, an adjustment must be made to the standard deviation formula.)

example: A large population is known to have a mean of 23 and a standard deviation of 2.5. What are the mean and standard deviation of the sampling distribution of means of samples of size 20 drawn from this population?

solution:

Central Limit Theorem

The discussion above gives us measures of center and spread for the sampling distribution of but tells us nothing about the shape of the sampling distribution. It turns out that the shape of the sampling distribution is determined by (a) the shape of the original population and (b) n , the sample size. If the original population is normal, then it”s easy: the shape of the sampling distribution will be normal if the population is normal.

If the shape of the original population is not normal, or unknown, and the sample size is small, then the shape of the sampling distribution will be similar to that of the original population. For example, if a population is skewed to the right, we would expect the sampling distribution of the mean for small samples also to be somewhat skewed to the right, although not as much as the original population.

When the sample size is large, we have the following result, known as the central limit theorem : For large n , the sampling distribution of will be approximately normal. The larger is n , the more normal will be the shape of the sampling distribution.

A rough rule of thumb for using the central limit theorem is that n should be at least 30, although the sampling distribution may be approximately normal for much smaller values of n if the population doesn”t depart much from normal. The central limit theorem allows us to use normal calculations to do problems involving sampling distributions without having to have knowledge of the original population. Note that calculations involving z -procedures require that you know the value of s, the population standard deviation. Since you will rarely know s, the large sample size essentially says that the sampling distribution is approximately , but not exactly, normal. That is, technically you should not be using z -procedures unless you know s but, as a practical matter, z -procedures are numerically close to correct for large n . Given that the population size (N ) is large in relation to the sample size (n ), the information presented in this section can be summarized in the following table:

example: Describe the sampling distribution of for samples of size 15 drawn from a normal population with mean 65 and standard deviation 9.

solution: Because the original population is normal, is normal with mean 65 and standard deviation . That is, has .

example: Describe the sampling distribution of for samples of size 15 drawn from a population that is strongly skewed to the left (like the scores on a very easy test) with mean 65 and standard deviation 9.

solution: μ = 65 and σ = 2.32 as in the above example. However this time the population is skewed to the left. The sample size is reasonably large, but not large enough to argue, based on our rule of thumb (n ≥ 30), that the sampling distribution is normal. The best we can say is that the sampling distribution is probably more mound shaped than the original but might still be somewhat skewed to the left.

example: The average adult has completed an average of 11.25 years of education with a standard deviation of 1.75 years. A random sample of 90 adults is obtained. What is the probability that the sample will have a mean

(a) greater than 11.5 years?

(b) between 11 and 11.5 years?

solution: The sampling distribution of has μ = 11.25 and

Because the sample size is large (n = 90), the central limit theorem tells us that large sample techniques are appropriate. Accordingly,

(a) The graph of the sampling distribution is shown below:

(b) From part (a), the area to the left of 11.5 is 1 – 0.0869 = 0.9131. Since the sampling distribution is approximately normal, it is symmetric. Since 11 is the same distance to the left of the mean as 11.5 is to the right, we know that P ( < 11) = P ( > 11.5) = 0.0869. Hence, P (11 < < 11.5) = 0.9131 – 0.0869 = 0.8262. The calculator solution is normalcdf(11,11.5,11, 0.184)=0.8258.

example: Over the years, the scores on the final exam for AP Calculus have been normally distributed with a mean of 82 and a standard deviation of 6. The instructor thought that this year”s class was quite dull and, in fact, they only averaged 79 on their final. Assuming that this class is a random sample of 32 students from AP Calculus, what is the probability that the average score on the final for this class is no more than 79? Do you think the instructor was right?

solution:

If this group really were typical, there is less than a 1% probability of getting an average this low by chance alone. That seems unlikely, so we have good evidence that the instructor was correct.

(The calculator solution for this problem is normalcdf(-1000,79, 82,1.06).)

Sampling Distributions of a Sample Proportion

If X is the count of successes in a sample of n trials of a binomial random variable, then the proportion of success is given by = X /n . is what we use for the sample proportion (a statistic). The true population proportion would then be given by p .

Digression: Before introducing , we have used and s as statistics, and μ and σ as parameters. Often we represent statistics with English letters and parameters with Greek letters. However, we depart from that convention here by using as a statistic and p as a parameter. There are texts that are true to the English/Greek convention by using p for the sample proportion and II for the population proportion.

We learned in the first section of this chapter that, if X is a binomial random variable, the mean and standard deviation of the sampling distribution of X are given by

We know that if we divide each term in a dataset by the same value n , then the mean and standard deviation of the transformed dataset will be the mean and standard deviation of the original dataset divided by n . Doing the algebra, we find that the mean and standard deviation of the sampling distribution of are given by:

Like the binomial, the sampling distribution of will be approximately normally distributed if n and p are large enough. The test is exactly the same as it was for the binomial: If X has B (n, p ), and = X /n , then has approximately

provided that np ≥ 10 and n (1 – p ) ≥ 10 (or np ≥ 5 and n (1 – p) ≥ 5).

example: Harold fails to study for his statistics final. The final has 100 multiple- choice questions, each with 5 choices. Harold has no choice but to guess randomly at all 100 questions. What is the probability that Harold will get at least 30% on the test?

solution: Since 100(0.2) and 100(0.8) are both greater than 10, we can use the normal approximation to the sampling distribution of . Since

Therefore,

. The TI-83/84 solution is given by normalcdf(0.3,100,0.2,0.040)=0.0062.

Harold should have studied.

Rapid Review

1. A coin is known to be unbalanced in such a way that heads only comes up 0.4 of the time.

(a) What is the probability the first head appears on the 4th toss?

(b) How many tosses would it take, on average, to flip two heads?

(a) P (first head appears on fourth toss) = 0.4 (1 – 0.4)4−1 = 0.4(0.6)3 = 0.0864

(b) Average wait to flip two heads = 2(average wait to flip one head) = .

1. The coin of problem #1 is flipped 50 times. LetX be the number of heads. What is

(a) the probability of exactly 20 heads?

(b) the probability of at least 20 heads?

(a) [on the TI-83/84: binompdf(50,0.4,20).]

(b) = 1-binomcdf(50,0.4,19)=0.554.

1. A binomial random variableX has B (300, 0.2). Describe the sampling distribution of .

Answer: Since 300(0.2) = 60 ≥ 10 and 300(0.8) = 240 ≥ 10, has approximately a normal distribution with μ = 0.2 and

1. A distribution is known to be highly skewed to the left with mean 25 and standard deviation 4. Samples of size 10 are drawn from this population, and the mean of each sample is calculated. Describe the sampling distribution of .

Since the samples are small, the shape of the sampling distribution would probably show some left-skewness but would be more mound-shaped than the original population.

1. What is the probability that a sample of size 35 drawn from a population with mean 65 and standard deviation 6 will have a mean less than 64?

Answer : The sample size is large enough that we can use large-sample procedures. Hence,

On the TI-83/84, the solution is given by normalcdf .

Practice Problems

Multiple-Choice

1. A binomial event has n = 60 trials. The probability of success on each trial is 0.4. Let X be the count of successes of the event during the 60 trials. What are μ x and σ x ?

(a) 24, 3.79

(b) 24, 14.4

(c) 4.90, 3.79

(d) 4.90, 14.4

(e) 2.4, 3.79

1. Consider repeated trials of a binomial random variable. Suppose the probability of the first success occurring on the second trial is 0.25. What is the probability of success on the first trial?

(a) ¼

(b) 1

(c) ½

(d) ⅛

(e)

1. To use a normal approximation to the binomial, which of the following does not have to be true?

(a) np ≥ 10, n (1 – p ) ≥ 10 (or: np ≥ 5, n (1 – p ) ≥ 5).

(b) The individual trials must be independent.

(c) The sample size in the problem must be too large to permit doing the problem on a calculator.

(d) For the binomial, the population size must be at least 10 times as large as the sample size.

(e) All of the above are true.

1. You form a distribution of the means of all samples of size 9 drawn from an infinite population that is skewed to the left (like the scores on an easy Stats quiz!). The population from which the samples are drawn has a mean of 50 and a standard deviation of 12. Which one of the following statements is true of this distribution?

(a) μ = 50, σ = 12, the sampling distribution is skewed somewhat to the left.

(b) μ = 50, σ = 4, the sampling distribution is skewed somewhat to the left.

(c) μ = 50, σ = 12, the sampling distribution is approximately normal.

(d) μ = 50, σ = 4, the sampling distribution is approximately normal.

(e) μ = 50, σ = 4, the sample size is too small to make any statements about the shape of the sampling distribution.

1. A 12-sided die has faces numbered from 1–12. Assuming the die is fair (that is, each face is equally likely to appear each time), which of the following would give the exact probability of getting at least 10 3s out of 50 rolls?

(a)

(b)

(c)

(d)

(e)

1. In a large population, 55% of the people get a physical examination at least once every two years. An SRS of 100 people are interviewed and the sample proportion is computed. The mean and standard deviation of the sampling distribution of the sample proportion are

(a) 55, 4.97

(b) 0.55, 0.002

(c) 55, 2

(d) 0.55, 0.0497

(e) The standard deviation cannot be determined from the given information.

1. Which of the following best describes the sampling distribution of a sample mean?

(a) It is the distribution of all possible sample means of a given size.

(b) It is the particular distribution in which μ = μ and σ = σ.

(c) It is a graphical representation of the means of all possible samples.

(d) It is the distribution of all possible sample means from a given population.

(e) It is the probability distribution for each possible sample size.

1. Which of the following is not a common characteristic of binomial and geometric experiments?

(a) There are exactly two possible outcomes: success or failure.

(b) There is a random variable X that counts the number of successes.

(c) Each trial is independent (knowledge about what has happened on previous trials gives you no information about the current trial).

(d) The probability of success stays the same from trial to trial.

(e) P (success) + P (failure) = 1.

1. A school survey of students concerning which band to hire for the next school dance shows 70% of students in favor of hiring The Greasy Slugs. What is the approximate probability that, in a random sample of 200 students, at least 150 will favor hiring The Greasy Slugs?

(a)

(b)

(c)

(d)

(e)

Free-Response

1. A factory manufacturing tennis balls determines that the probability that a single can of three balls will contain at least one defective ball is 0.025. What is the probability that a case of 48 cans will contain at least two cans with a defective ball?
2. A population is highly skewed to the left. Describe the shape of the sampling distribution of drawn from this population if the sample size is (a) 3 or (b) 30.
3. Suppose you had lots of time on your hands and decided to flip a fair coin 1,000,000 times and note whether each flip was a head or a tail. Let X be the count of heads. What is the probability that there are at least 1000 more heads than tails? (Note : This is a binomial, but your calculator may not be able to do the binomial computation because the numbers are too large for it.)
4. In Chapter 9 , we had an example in which we asked if it would change the proportion of girls in the population (assumed to be 0.5) if families continued to have children until they had a girl and then they stopped. That problem was to be done by simulation. How could you use what you know about the geometric distribution to answer this same question?
5. At a school better known for football than academics (a school its football team can be proud of), it is known that only 20% of the scholarship athletes graduate within 5 years. The school is able to give 55 scholarships for football. What are the expected mean and standard deviation of the number of graduates for a group of 55 scholarship athletes?
6. Consider a population consisting of the numbers 2, 4, 5, and 7. List all possible samples of size two from this population and compute the mean and standard deviation of the sampling distribution of . Compare this with the values obtained by relevant formulas for the sampling distribution of . Note that the sample size is large relative to the population—this may affect how you compute σ by formula.
7. Approximately 10% of the population of the United States is known to have blood type B. What is the probability that between 11% and 15%, inclusive, of a random sample of 500 adults will have type B blood?
8. Which of the following is/are true of the central limit theorem? (More than one answer might be true.)
9. μ = μ .

III. The sampling distribution of a sample mean will be approximately normally distributed for sufficiently large samples, regardless of the shape of the original population.

1. The sampling distribution of a sample mean will be normally distributed if the population from which the samples are drawn is brakes.
2. A brake inspection station reports that 15% of all cars tested have brakes in need of replacement pads. For a sample of 20 cars that come to the inspection station,

(a) what is the probability that exactly 3 have defective brakes?

(b) what is the mean and standard deviation of the number of cars that need replacement pads?

1. A tire manufacturer claims that his tires will last 40,000 miles with a standard deviation of 5000 miles.

(a) Assuming that the claim is true, describe the sampling distribution of the mean lifetime of a random sample of 160 tires. Remember that “describe” means discuss center, spread, and shape.

(b) What is the probability that the mean life time of the sample of 160 tires will be less than 39,000 miles? Interpret the probability in terms of the truth of the manufacturer”s claim.

1. The probability of winning a bet on red in roulette is 0.474. The binomial probability of winning money if you play 10 games is 0.31, and drops to 0.27 if you play 100 games. Use a normal approximation to the binomial to estimate your probability of coming out ahead (that is, winning more than 1 /2 of your bets) if you play 1000 times. Justify being able to use a normal approximation for this situation.
2. Crabs off the coast of Northern California have a mean weight of 2 lbs with a standard deviation of 5 oz. A large trap captures 35 crabs.

(a) Describe the sampling distribution for the average weight of a random sample of 35 crabs taken from this population.

(b) What would the mean weight of a sample of 35 crabs have to be in order to be in the top 10% of all such samples?

1. The probability that a person recovers from a particular type of cancer operation is 0.7. Suppose 8 people have the operation. What is the probability that

(a) exactly 5 recover?

(b) they all recover?

(c) at least one of them recovers?

1. A certain type of lightbulb is advertised to have an average life of 1200 hours. If, in fact, lightbulbs of this type only average 1185 hours with a standard deviation of 80 hours, what is the probability that a sample of 100 bulbs will have an average life of at least 1200 hours?
2. Your task is to explain to your friend Gretchen, who knows virtually nothing (and cares even less) about statistics, just what the sampling distribution of the mean is. Explain the idea of a sampling distribution in such a way that even Gretchen, if she pays attention, will understand.
3. Consider the distribution shown at the right.

Describe the shape of the sampling distribution of for samples of size n if

(a) n = 3.

(b) n = 40.

1. After the Challenger disaster of 1986, it was discovered that the explosion was caused by defective O-rings. The probability that a single O-ring was defective and would fail (with catastrophic consequences) was 0.003 and there were 12 of them (6 outer and 6 inner). What was the probability that at least one of the O-rings would fail (as it actually did)?
2. Your favorite cereal has a little prize in each box. There are 5 such prizes. Each box is equally likely to contain any one of the prizes. So far, you have been able to collect 2 of the prizes. What is:

(a) the probability that you will get the third different prize on the next box you buy?

(b) the probability that it will take three more boxes to get the next prize?

(c) the average number of boxes you will have to buy before getting the third prize?

1. We wish to approximate the binomial distribution B (40, 0.8) with a normal curve N (μ, σ ). Is this an appropriate approximation and, if so, what are μ and σ for approximating the normal curve?
2. Opinion polls in 2002 showed that about 70% of the population had a favorable opinion of President Bush. That same year, a simple random sample of 600 adults living in the San Francisco Bay Area showed only 65% had a favorable opinion of President Bush. What is the probability of getting a rating of 65% or less in a random sample of this size if the true proportion in the population was 0.70?

Cumulative Review Problems

1. An unbalanced coin has p = 0.6 of turning up heads. Toss the coin three times and let X be the count of heads among the three coins. Construct the probability distribution for this experiment.
2. You are doing a survey for your school newspaper and want to select a sample of 25 seniors. You decide to do this by randomly selecting 5 students from each of the 5 senior-level classes, each of which contains 28 students. The school data clerk assures you that students have been randomly assigned, by computer, to each of the 5 classes. Is this sample

(a) a random sample?

(b) a simple random sample?

1. Data are collected in an experiment to measure a person”s reaction time (in seconds) as a function of the number of milligrams of a new drug. The least squares regression line (LSRL) for the data is = 0.2 + 0.8(mg ). Interpret the slope of the regression line in the context of the situation.
2. If P(A) = 0.5, P(B) = 0.3, and P(A or B) = 0.65, are events A and B independent?
3. Which of the following is (are) examples of quantitative data and which is (are) examples of qualitative data ?

(a) The height of an individual, measured in inches.

(b) The color of the shirts in my closet.

(c) The outcome of a flip of a coin described as “heads” or “tails.”

(d) The value of the change in your pocket.

(e) Individuals, after they are weighed, are identified as thin, normal, or heavy.

Solutions to Practice Problems

Multiple-Choice

1. The correct answer is (a).
2. The correct answer is (c). If it is a binomial random variable, the probability of success, p , is the same on each trial. The probability of not succeeding on the first trial and then succeeding on the second trial is (1 – p )(p ). Thus, (1 – p )p = 0.25. Solving algebraically, p = ½.
3. The correct answer is (c). Although you probably wouldn”t need to use a normal approximation to the binomial for small sample sizes, there is no reason (except perhaps accuracy) that you couldn”t.

For small samples, the shape of the sampling distribution of will resemble the shape of the sampling distribution of the original population. The shape of the sampling distribution of is approximately normal for nsufficiently large.

1. The correct answer is (d). Because the problem stated “at least 10,” we must include the term where x = 10. If the problem had said “more than 10,” the correct answer would have been (b) or (c) (they are equivalent). The answer could also have been given as
2. The correct answer is (d).
3. The correct answer is (a).
4. The correct answer is (b). This is a characteristic of a binomial experiment. The analogous characteristic for a geometric experiment is that there is a random variable X that is the number of trials needed to achieve the first success.
5. The correct answer is (c). This is actually a binomial situation. If X is the count of students “in favor,” then X has B (200, 0.70). Thus, P (X ≥ 150) = P (X = 150) + P (X = 151) + … + P (X = 200). Using the TI-83/84, the exact binomial answer equals 1–binomcdf (200,0.7.0,149)=0.0695 . None of the listed choices shows a sum of several binomial expressions, so we assume this is to be done as a normal approximation. We note that B (200, 0.7) can be approximated by N = N(140, 6.4807). A normal approximation is OK since 200(0.7) and 200(0.3) are both much greater than 10. Since 75% of 200 is150, we have P(X ≥ 150) =

Free-Response

1. If X is the count of cans with at least one defective ball, then X has B (48, 0.025).

On the TI-83/84, the solution is given by 1–binomcdf(48,0.025,1).

1. We know that the sampling distribution of will be similar to the shape of the original population for small n and approximately normal for large n (that”s the central limit theorem). Hence,

(a) if n = 3, the sampling distribution would probably be somewhat skewed to the left.

(b) if n = 30, the sampling distribution should be approximately normal.

Remember that using n ≥ 30 as a rule of thumb for deciding whether to assume normality is for a sampling distribution just that: a rule of thumb. This is probably a bit conservative. Unless the original population differs markedly from mound shaped and symmetric, we would expect to see the sampling distribution of be approximately normal for considerably smaller values of n .

1. Since the binomcdf function can”t be used due to calculator overflow, we will use a normal approximation to the binomial. Let X = the count of heads. Then μ X =(1,000,000)(0.5) = 500,000 (assuming a fair coin) and . Certainly both np and n (1 – p ) are greater than 10, so the conditions needed to use a normal approximation are present. If we are to have at least 1000 more heads than tails, then there must be at least 500,500 heads (and, of course no more than 499,500 tails). Thus, P (there are at least 1000 more heads than tails) = P (X ) ≥ 500500 = .
2. The average wait for the first success to occur in a geometric setting is 1/p , where p is the probability of success on any one trial. In this case, the probability of a girl for any one birth is p = 0.5. Hence, the average wait for the first girl is . So, we have one boy and one girl, on average, for each two children. The proportion of girls in the population would not change.
3. If X is the count of scholarship athletes that graduate from any sample of 55 players, then X has B (55, 0.20). μ X = 55(0.20) = 11 and .
4. Putting the numbers 2, 4, 5, and 7 into a list in a calculator and doing 1-Var Stats, we find μ = 4.5 and σ =1.802775638. The set of all samples of size 2 is {(2,4), (2,5), (2,7), (4,5), (4,7), (5,7)} and the means of these samples are {3, 3.5, 4.5, 4.5, 5.5, 6}. Putting the means into a list and doing 1-Var Stats to find μ and σ , we get μ = 4.5 (which agrees with the formula) and σ = 1.040833 (which does not agree with . Since the sample is large compared with the population (that is, the population isn”t at least 10 times as large as the sample), we use , which does agree with the computed value. (Note, students have not been asked to use this formula on the AP exam.)
5. There are three different ways to do this problem: exact binomial, using proportions, or using a normal approximation to the binomial. The last two are essentially the same.

(i) Exact binomial . Let X be the count of persons in the sample that have blood type B. Then X has B (500, 0.10). Also, 11% of 500 is 55 and 15% of 500 is 75. Hence, P (55 ≤ X ≤ 75) = P (X ≤ 75) – P (X ≤ 54) = binomcdf(500,0.10,75)– binomcdf(500,0.10,54)=0.2475.

(ii) Proportions. We note that μ X = np = 500(0.1) = 50 and n (1 – p ) = 500(0.9) = 90, so we are OK to use a normal approximation. Also, μ = p = 0.10 and . P (0.11) < ≤ 0.15) = P (0.7463 ≤ z ≤ 3.731) = 0.2276. On the TI 83/84: normalcdf(0.7463,3.731) .

(iii) Normal approximation to the binomial. The conditions for doing a normal approximation were established in part (ii). Also, μ X = 500(0.1) = 50 and . P (55 ≤ X ≤ 75) = P (0.7454 ≤ z ≤ 3.7268) = 0.2279.

1. All four of these statement are true. However, only III is a statement of the central limit theorem. The others are true of sampling distributions in general.
2. If X is the count of cars with defective pads, then X has B (20, 0.15).

(a) . On the TI-83/84, the solution is given by binompdf(20,0.15,3).

(b) μ X = np = 20(0.15) = 3, .

1. μ = 40,000 miles and .

(a) With n = 160, the sampling distribution of will be approximately normally distributed with mean equal to 40,000 miles and standard deviation 395.28 miles.

(b) .

If the manufacturer is correct, there is only about a 0.6% chance of getting an average this low or lower. That makes it unlikely to be just a chance occurrence, and we should have some doubts about the manufacturer”s claim.

1. If X is the number of times you win, then X has B (1000, 0.474). To come out ahead, you must win more than half your bets. That is, you are being asked for P (X > 500). Because (1000)(0.474) = 474 and 1000(1 – 0.474) = 526 are both greater than 10, we are justified in using a normal approximation to the binomial. Furthermore, we find that

Now,

That is, you have slightly less than a 5% chance of making money if you play 1000 games of roulette.

Using the TI-83/84, the normal approximation is given by normalcdf(500, 10000,474,15.79) = 0.0498. The exact binomial solution using the calculator is 1-binomcdf(1000,0.474,500)=0.0467.

1. μ = 2 lbs = 32 oz and .

(a) With samples of size 35, the central limit theorem tells us that the sampling distribution of is approximately normal. The mean is 32 oz and standard deviation is 0.845 oz.

(b) In order for to be in the top 10% of samples, it would have to be at the 90th percentile, which tells us that its z -score is 1.28 [that”s InvNorm(0.9) on your calculator]. Hence,

Solving, we have = 33.08 oz. The mean weight of a sample of 35 crabs has to be at least 33.08 oz, or about 2 lb 1 oz, to be in the top 10% of samples of this size.

1. If X is the number that recover, then X has B (8, 0.7).

(a) . On the TI-83/84, the solution is given by binompdf(8,0.7,5).

(b) . On the TI-83/84, the solution is given by binompdf(8,0.7,8).

(c) . On the TI-83/84, the solution is given by 1–binompdf(8,0.7,0).

1. The first thing Gretchen needs to understand is that a distribution is just the set of all possible values of some variable. For example the distribution of SAT scores for the current senior class is just the values of all the SAT scores. We can draw samples from that population if, say, we want to estimate the average SAT score for the senior class but don”t have the time or money to get all the data. Suppose we draw samples of size n and compute for each sample. Imagine drawing ALL possible samples of size n from the original distribution (that was the set of SAT scores for everybody in the senior class). Now consider the distribution (all the values) of means for those samples. That is what we call the sampling distribution of (the short version: the sampling distribution of is the set of all possible values of computed from samples of size n .)
2. The distribution is skewed to the right.

(a) If n = 3, the sampling distribution of will have some right skewness, but will be more mound shaped than the parent population.

(b) If n = 40, the central limit theorem tells us that the sampling distribution of will be approximately normal.

1. If X is the count of O-rings that failed, then X has B (12, 0.003).

On the TI-83/84, the solution is given by 1–binompdf(12,0.003,0).

The clear message here is that even though the probability of any one failure seems remote (0.003), the probability of at least one failure (3.5%) is large enough to be worrisome.

1. Because you already have 2 of the 5 prizes, the probability that the next box contains a prize you don”t have is 3/5 = 0.6. If n is the number of trials until the first success, then P (X = n ) = (0.6). (0.4) n −1 .

(a) P (X = 1) = (0.6)(0.4)1−1 = (0.6)(1) = 0.6. On the TI-83/84 calculator, the answer can be found by geometpdf(0.6,1).

(b) P (X = 3) = (0.6)(0.4)3−2 = 0.096. On the calculator: geometpdf(0.6,3).

(c) The average number of boxes you will have to buy before getting the third prize is

1. 40(0.8) = 32 and 40(0.2) = 8. The rule we have given is that both np and n (1 – p ) must be greater than 10 to use a normal approximation. However, as noted in earlier in this chapter, many texts allow the approximation when np ≥ 5 and n (1 – p ) ≥ 5. Whether the normal approximation is valid or not depends on the standard applied. Assuming that, in this case, the conditions necessary to do a normal approximation are present, we have
2. If p = 0.70, then μ = 0.70 and . Thus, P ( ≤ 0.65) = . Since there is a very small probability of getting a sample proportion as small as 0.65 if the true proportion is really 0.70, it appears that the San Francisco Bay Area may not be representative of the United States as a whole (that is, it is unlikely that we would have obtained a value as small as 0.65 if the true value were 0.70).

Solutions to Cumulative Review Problems

1. The sample space for this event is {HHH, HHT , HTH , THH , HTT, HTH, THH, TTT}. One way to do this problem, using techniques developed in Chapter 9 , is to compute the probability of each event. Let X = the count of heads. Then, for example (bold faced in the list above), P (X = 2) = (0.6)(0.6)(0.4) + (0.6)(0.4)(0.6) + (0.4)(0.6)(0.6) = 3(0.6)2 (0.4) = 0.432. Another way is to take advantage of the techniques developed in this chapter (noting that the possible values of X are 0, 1, 2, and 3):

P (X = 0) = (0.4)3 = 0.064; = binompdf(3,0.6,1)= 0.288; = binompdf(3,0.6,2)= 0.432; and P(X = 3) = = binompdf(3,0.6,3)= 0.216. Either way, the probability distribution is then:

Be sure to check that the sum of the probabilities is 1 (it is!).

1. (a) Yes, it is a random sample because each student in any of the 5 classes is equally likely to be included in the sample.

(b) No, it is not a simple random sample (SRS) because not all samples of size 25 are equally likely. For example, in an SRS, one possible sample is having all 25 come from the same class. Because we only take 5 from each class, this isn”t possible.

1. The slope of the regression line is 0.8. For each additional milligram of the drug, reaction time is predicted to increase by 0.8 seconds. Or you could say for each additional milligram of the drug, reaction time will increase by 0.8 seconds, on average .
2. P (A or B) = P (A∪B) = P (A) + P (B) – P (A∩B) = 0.5 + 0.3 – P (A∩B) = 0.65 ⇒ P (A∩B) = 0.15. Now, A and B are independent if P (A∩B) = P (A) · P (B). So, P (A) · P (B) = (0.3)(0.5) = 0.15 = P (A∩B). Hence, A and B are independent.
3. (a) Quantitative

(b) Qualitative

(c) Qualitative

(d) Quantitative

(e) Qualitative

(f) Quantitative

(g) Qualitative

CHAPTER 10

Binomial Distributions, Geometric Distributions, and Sampling Distributions

1. If you roll a pair of six-sided dice, what is the probability that it will take at least 3 rolls to get a double (both dice showing the same face)?

(A)

(B)

(C)

(D)

(E)

1. Residential properties for sale in one small city have a mean price of \$180,000 and a standard deviation of \$119,000. If a random sample of 10 homes is selected, what is the probability that the mean price of the 10 homes is more than \$260,000?

(A) 0.017

(B) 0.034

(C) 0.139

(D) 0.251

(E) You cannot accurately estimate this probability because the sampling distribution may not be approximately normal.

1. Random samples are selected from a large right-skewed population, and the sample mean is calculated. As the sample size increases, the sampling distribution of the sample mean

(A) remains skewed right with the same mean and the same standard deviation.

(B) remains skewed right with the same mean and a smaller standard deviation.

(C) becomes more symmetric with a smaller mean and a smaller standard deviation.

(D) becomes more symmetric with the same mean and a smaller standard deviation.

(E) becomes more symmetric with a smaller mean and the same standard deviation.

1. The following table shows a five-number summary of prices for residential properties in a small city. If a random sample of 8 of these properties is selected, what is the approximate probability that exactly 3 of them are priced below \$60,000?

(A) 0.0037

(B) 0.2076

(C) 0.6785

(D) 0.7929

(E) 0.8862

1. When we use a normal model to approximate a binomial model, we get slightly different estimates. Which of the following statements explains why?

(A) The binomial model is discrete, and the normal model is continuous.

(B) The binomial model is skewed, and the normal model is symmetric.

(C) The binomial model assumes sampling without replacement, and the normal model assumes sampling with replacement.

(D) Both choices A and B are correct reasons for the different estimates.

(E) Both choices A and C are correct reasons for the different estimates.