MC AP Exam Practice Part 2

According to a recent survey, 31 percent of the residents of a certain state who are age 25 years or older have a bachelor’s degree. A random sample of 50 residents of the state, age 25 years or older, will be selected. Let the random variable B represent the number in the sample who have a bachelor’s degree. What is the probability that B will equal 40 ?

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Answer explanation

Define success as selecting a person with the degree and failure as selecting a person without the degree. Random variable B has a binomial distribution, with probability of success of 0.31 and probability of failure of 0.69. The probability of 40 out of 50 people having a degree is given by

$\left(_{40}^{50}\right)\left(0.31\right)^{40}\left(0.69\right)^{10}$

A high school science teacher has 78 students. Of those students, 35 are in the band and 32 are on a sports team. There are 16 students who are not in the band or on a sports team. One student from the 78 students will be selected at random. Let event B represent the event of selecting a student in the band, and let event S represent the event of selecting a student on a sports team.

Are B and S mutually exclusive events?

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Answer explanation

The sum of 35, 32, and 16 is 83, which is 5 greater than 78. So there must be 5 students who are both in the band and on a sports team. Because the events can occur at the same time, they are not mutually exclusive.

In a certain board game, a player rolls two fair six-sided dice until the player rolls doubles (where the value on each die is the same). The probability of rolling doubles with one roll of two fair six-sided dice is 1/6.

What is the probability that it takes three rolls until the player rolls doubles?

• $\left(\frac{1}{6}\right)^3$

• $\left(\frac{5}{6}\right)^3$

• $\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^3$

• $\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^2$

• $\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^2$

Answer explanation

Let the discrete random variable X represent the number of rolls it takes until the player rolls doubles. The random variable has a geometric distribution, where $P\left(X=2\right)=\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^2$

The following table shows the probability distribution for the number of books a student typically buys at the annual book fair held at an elementary school.

Let the random variable B represent the number of books a student buys at the next book fair. What is the expected value of B ?

• 0

• 1.00

• 1.79

• 3.50

• 28

Answer explanation

The expected value of B

B is the sum of the products of the number of books and their respective probabilities. In this case, 0(0.35)+1(0.20)+2(0.15)+3(0.10)+4(0.07)+5(0.08)+6(0.04)+7(0.01)=1.79

One student from a high school will be selected at random. Let A

A be the event that the selected student is a student athlete, and let B be the event that the selected student drives to school. If P(A∩B)=0.08 and P(B|A)=0.25, what is the probability that the selected student will be a student athlete?

• 0.02

• 0.17

• 0.32

• 0.33

• 3.13

Answer explanation

By the multiplication rule, P(A∩B)=P(A)⋅P(B|A), so 0.08=P(A)⋅0.25. Solving for P(A) gives 0.32.

For which of the following conditions is it not appropriate to assume that the sampling distribution of the sample mean is approximately normal?

• A random sample of 8 taken from a normally distributed population

• A random sample of 50 taken from a normally distributed population

• A random sample of 10 taken from a population distribution that is skewed to the right

• A random sample of 75 taken from a population distribution that is skewed to the left

• A random sample of 100 taken from a population that is uniform

Answer explanation

It is not appropriate to assume the sampling distribution of the sample mean is approximately normal for this sample. Because the distribution is skewed, the sample size needs to be at least 30 to assume normality for the sampling distribution.

A certain statistic d-hat is being used to estimate a population parameter D. The expected value of d-hat is not equal to D. What property does d-hat exhibit?

• The sampling distribution of d-hat is Normal.

• The sampling distribution of d-hat is Binomial.

• The sampling distribution of d-hat is uniform.

• d-hat is unbiased

• d-hat is biased

Answer explanation

A biased estimator is one in which its value, on average, is not equal to the value of the parameter it is intended to estimate.

A bag contains chips of which 27.5 percent are blue. A random sample of 5 chips will be selected one at a time and with replacement. What are the mean and standard deviation of the sampling distribution of the sample proportion of blue chips for samples of size 5 ?

• The mean is $5\left(0.275\right)$ , and the standard deviation is $\sqrt{\left(5(0.275)(0.725\right)}$

• The mean is $0.275$ , and the standard deviation is $\sqrt{\left(5(0.275)(0.725\right)}$

• The mean is $0.275$ , and the standard deviation is $\sqrt{\frac{\left(0.275\right)\left(0.725\right)}{5}}$

• The mean is $27.5$ , and the standard deviation is $\sqrt{\left(5(0.275)(0.725\right)}$

• The mean is $27.5$ , and the standard deviation is $\sqrt{\frac{\left(27.5\right)\left(72.5\right)}{5}}$

Answer explanation

The mean is the population proportion p=0.275, and the standard deviation is $\sqrt{\frac{\left(0.275\right)\left(0.725\right)}{5}}$

Suppose the variance in trunk diameter of the giant sequoia tree species is $15.7\ m^2$  , while the variance in trunk diameter of the California redwood tree species is 
 $10.6\ m^2$  . Let  $\overline{x_1}$   represent the average trunk diameter of four randomly sampled giant sequoia trees, and let  $\overline{x_2}$   represent the average trunk diameter of three randomly sampled California redwood trees. If the random sampling is done with replacement, what is the standard deviation  $\sigma_{\left(\overline{x_1}-\overline{x_2}\right)}$  of the sampling distribution of the difference in sample means  $\overline{x_1}-\overline{x_2}$   ?

•  $\sqrt{\frac{15.7}{4}-\frac{10.6}{3}}$  

•  $\sqrt{15.7-10.6}$  

•  $\sqrt{\frac{15.7}{4}+\frac{10.6}{3}}$  

•  $\sqrt{\frac{15.7}{4}}+\sqrt{\frac{10.6}{3}}$  

•  $\frac{15.7}{4}+\frac{10.6}{3}$  

Answer explanation

The standard deviation of the sampling distribution of   $\overline{x_1}-\overline{x_2}$  is  $\sqrt{\frac{\sigma_1^2}{n_1}+}\frac{\sigma_2^2}{n_2}$  with  $\sigma^2$   being the variance of the respective distribution.

Researchers are studying populations of two squirrels, the eastern gray and the western gray. For the eastern gray squirrel, about 22 percent of the population weighs over 0.5 kilogram (kg)

(kg). For the western gray squirrel, about 36 percent of the population weighs over 0.5 kg

kg. A random sample of 60 squirrels will be selected from the population of eastern gray squirrels, and a random sample of 120 squirrels will be selected from the population of western gray squirrels. What is the mean of the sampling distribution of the difference in sample proportions (eastern minus western) ?

• $\sqrt{\frac{0.22\left(0.78\right)}{120}+\frac{0.36\left(0.64\right)}{60}}$

• $\sqrt{\frac{0.22\left(0.78\right)}{60}+\frac{0.36\left(0.64\right)}{120}}$

• $60\left(0.22\right)-129\left(0.36\right)$

• $0.36-0.22$

• $0.22-0.36$

Answer explanation

The mean of the sampling distribution of the difference in sample proportions for eastern minus western is equal to the difference in the population proportions, which equals 0.22−0.36