AP Stats Chapter 6 Practice Test

A psychologist studied the number of puzzles that subjects were able to solve in a 5-minute period while listening to soothing music. Let 𝑋 be the number of puzzles completed successfully by a randomly chosen subject. The psychologist found that 𝑋 had the following probability distribution. What is the probability that a randomly chosen subject completes more than the expected number of puzzles in the 5-minute period while listening to soothing music?

• Cannot be determined

• 0.1

• 0.8

• 1

• 0.4

Answer explanation

𝜇𝑋=∑𝑥𝑖𝑝𝑖 then sum up the probabilities that are greater than 𝐸(𝑋).

The expected number of puzzles is 𝐸(𝑋) = 2.3. 𝑃(𝑋>2.3)=0.3+0.1=0.4

A psychologist studied the number of puzzles that subjects were able to solve in a 5-minute period while listening to soothing music. Let 𝑋 be the number of puzzles completed successfully by a randomly chosen subject. The psychologist found that 𝑋 had the following probability distribution. The standard deviation of 𝑋 is 0.9. Which of the following is the best interpretation of this value?

• About 68% of subjects solved between 0.9 puzzles less and 0.9 puzzles more than the mean.

• About 90% of subjects solved 3 or fewer puzzles.

• The number of puzzles solved by subjects typically differed from the mean by about 0.9 puzzles.

• The number of puzzles solved by subjects typically differed from one another by about 0.9 puzzles.

• The typical subject solved an average of 0.9 puzzles.

Answer explanation

The Empirical rule only applies when the distribution is approximately Normal.

A psychologist studied the number of puzzles that subjects were able to solve in a 5-minute period while listening to soothing music. Let 𝑋 be the number of puzzles completed successfully by a randomly chosen subject. The psychologist found that 𝑋 had the following probability distribution. The standard deviation of 𝑋 is 0.9. Let 𝐷 be the difference in the number of puzzles solved by two randomly selected subjects in a 5-minute period. What is the standard deviation of 𝐷?

• 1.8

• 0.81

• 0

• 1.27

• 0.9

Suppose a student is randomly selected from your school. Which of the following pairs of random variables are most likely independent?

• 1.8𝑋 = average amount of homework the student does per night; 𝑌 = student’s GPA

• 0.81𝑋 = student’s height; 𝑌 = student’s weight

• 𝑋 = student’s PSAT Math score; 𝑌 = student’s PSAT Verbal score

• 𝑋 = student’s IQ; 𝑌 = student’s GPA

• 0.9𝑋 = average amount of homework the student does per night; 𝑌 = student’s height

Answer explanation

The student's height is not affected by the amount of homework he or she does per night.

A certain vending machine offers 20-ounce bottles of soda for $1.50. The number of bottles 𝑋 bought from the machine on any day is a random variable with mean 50 and standard deviation 15. Let the random variable 𝑌 equal the total revenue from this machine on a randomly selected day. Assume that the machine works properly and that no sodas are stolen from the machine.

What are the mean and standard deviation of 𝑌?

• 𝜇𝑌 = $75,

  𝜎𝑌 = $18.37

• 𝜇𝑌 = $75,

  𝜎𝑌 = $22.50

• 𝜇𝑌 = $1.50,

  𝜎𝑌 = $22.50

• 𝜇𝑌 = $75,

  𝜎𝑌 = $33.50

• 𝜇𝑌 = $1.50,

  𝜎𝑌 = $33.75

Answer explanation

Y=1.5X and we multiply the mean and standard deviation by 1.5 to obtain the mean and standard deviation of 𝑌.

The weight of tomatoes chosen at random from a bin at the farmer’s market follows a Normal distribution with mean 𝜇 = 10 ounces and standard deviation 𝜎 = 1 ounce. Suppose we pick four tomatoes at random from the bin and find their total weight 𝑇. The random variable 𝑇 is

• Normal, with mean 10 ounces and standard deviation 1 ounce.

• Normal, with mean 40 ounces and standard deviation 4 ounces.

• Binomial, with mean 40 ounces and standard deviation 2 ounces.

• Normal, with mean 40 ounces and standard deviation 2 ounces.

• Binomial, with mean 40 ounces and standard deviation 4 ounces.

Answer explanation

Remember that you must find the square root of the variance after adding the variances!

Which of the following random variables is geometric?

• The number of digits I read in a randomly selected row of the random digits table to get a 7

• The number of cards I deal from a well-shuffled deck of 52 cards to get a heart

• The number of 7s in a row of 40 random digits

• The number of 6s I get if I roll a die 10 times

• The number of times I have to roll a single die to get two 6s

Answer explanation

The trials are not independent as the cards are not put back after dealing.

Seventeen people have been exposed to a particular disease. Each one independently has a 40% chance of contracting the disease. A hospital has the capacity to handle 10 cases of the disease.

What is the probability that the hospital’s capacity will be exceeded?

• 0.989

• 0.011

• 0.965

• 0.035

• 0.092

Answer explanation

𝑃(𝑋>10)=1−𝑃(𝑋≤10)=

1−𝑏𝑖𝑛𝑜𝑚𝑐𝑑𝑓(17,0.4,10)=0.035

The figure shows the probability distribution of a discrete random variable 𝑋. Which of the following best describes this random variable?

• Binomial with 𝑛 = 8, 𝑝 = 0.8

• Binomial with 𝑛 = 8, 𝑝 = 0.1

• Binomial with 𝑛 = 8, 𝑝 = 0.3

• Geometric with 𝑝 = 0.1

• Geometric with 𝑝 = 0.2

Answer explanation

We can see the probability of X = 0 is 0.05, but X = 0 is impossible in a geometric distribution.

Also, we know this must be binomial because 𝑋 = 1 is not the tallest bar.

Also, we are given that 𝑛 = 8. We can estimate the mean at about 2.4 and using 𝑛𝑝 = 8𝑝= 2.4, we find that 𝑝 = 0.3.

A test for extrasensory perception (ESP) involves asking a person to tell which of 5 shapes—a circle, star, triangle, diamond, or heart—appears on a hidden computer screen. On each trial, the computer is equally likely to select any of the 5 shapes. Suppose researchers are testing a person who does not have ESP and so is just guessing on each trial. What is the probability that the person guesses the first 4 shapes incorrectly but gets the fifth one correct?

Answer explanation

Notice that we are looking for the first correct guess. That means this is a geometric setting.

This is a geometric random variable and we are looking for 𝑃(𝑋=5).