AP Statistics Practice Quiz: Parameters for a Binomial Distribution
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 13 questions to check your progress.
Question 1 of 13
All Questions (13)
A) 10
B) 5
C) 20
D) 2.24
Correct Answer: A
The situation describes a binomial distribution with n = 20 trials (flips) and a probability of success p = 0.5 (getting a head). The mean of a binomial random variable is calculated as np. Therefore, the mean is 20 * 0.5 = 10.
A) 2.500
B) 1.875
C) 1.369
D) 7.500
Correct Answer: C
This is a binomial setting with n = 10 trials (questions) and a probability of success p = 0.25 (guessing correctly). The standard deviation of a binomial random variable is sqrt(np(1-p)). So, the standard deviation is sqrt(10 * 0.25 * (1 - 0.25)) = sqrt(1.875) ≈ 1.369.
A) Mean = 40, Standard Deviation = 8
B) Mean = 40, Standard Deviation = 2.828
C) Mean = 10, Standard Deviation = 2.828
D) Mean = 10, Standard Deviation = 8
Correct Answer: B
The number of trials is n = 50 and the probability of success (a non-defective chip) is p = 0.80. The mean is np = 50 * 0.80 = 40. The standard deviation is sqrt(np(1-p)) = sqrt(50 * 0.80 * 0.20) = sqrt(8) ≈ 2.828.
A) She will make exactly 21 free throws in every game where she attempts 30.
B) The most likely number of free throws she will make in a single game is 21.
C) Over many games where she attempts 30 free throws, the average number of made free throws is 21.
D) The probability of her making 21 free throws is the highest possible probability for any outcome.
Correct Answer: C
The mean of a binomial distribution, np, represents the expected value or the long-run average number of successes over many repetitions of the experiment. In this context, it's the average number of made free throws over many games with 30 attempts, not the outcome of a single game.
A) It is most likely that exactly 60 adults in the sample use public transportation.
B) In any sample of 200 adults from this city, we will find exactly 60 who use public transportation.
C) If we took many random samples of 200 adults from this city, the average number of adults who use public transportation across all samples would be about 60.
D) The probability of finding 60 adults who use public transportation in a sample of 200 is 0.30.
Correct Answer: C
The mean (or expected value) of a random variable in a sampling context describes the long-run average value we would expect to see if the sampling process were repeated many times. It does not guarantee the outcome of a single sample.
A) The number of seeds that germinate will be exactly 3 away from the mean.
B) The number of seeds that germinate in a random batch of 100 seeds will typically vary from the mean of 90 by about 3 seeds.
C) If we plant many batches of 100 seeds, the maximum difference between the number of germinated seeds and the mean will be 3.
D) The probability of a seed germinating is, on average, 3% away from 90%.
Correct Answer: B
The standard deviation of a random variable measures the typical or average distance of the outcomes from the mean. In this context, it describes how much the number of germinated seeds in a batch of 100 is expected to vary from the average of 90.
A) n = 400, p = 0.05
B) n = 400, p = 0.95
C) n = 20, p = 0.05
D) n = 0.05, p = 400
Correct Answer: A
In a binomial setting, 'n' is the number of trials, which is the sample size of 400 bulbs. 'p' is the probability of success on a single trial. Here, a "success" is defined as finding a defective bulb, so p = 0.05.
A) 10
B) 50
C) 250
D) 500
Correct Answer: C
The mean, or expected number of successes, of a binomial distribution is given by the formula np. We are given the mean (50) and the probability p (0.2). To find n, we solve the equation: n * 0.2 = 50, which gives n = 50 / 0.2 = 250.
A) n = 24, p = 0.5
B) n = 36, p = 1/3
C) n = 48, p = 0.25
D) n = 16, p = 0.75
Correct Answer: C
We are given np = 12 and sqrt(np(1-p)) = 3. Squaring the standard deviation gives the variance, np(1-p) = 9. We can substitute np=12 into the variance equation: 12(1-p) = 9. Solving for p gives 1-p = 9/12 = 0.75, so p = 0.25. Then, using np=12, we find n = 12/0.25 = 48.
A) X is a binomial random variable with a mean of 12. This means we expect a student who guesses to get about 12 questions right on average.
B) X is a binomial random variable with a mean of 30. This means the most likely score is 30 out of 60.
C) X is not a binomial random variable because the probability of guessing correctly changes from question to question.
D) X is a binomial random variable with a standard deviation of 12. This means scores will typically be 12 points away from the mean.
Correct Answer: A
The scenario fits a binomial distribution with n=60 trials (questions) and p=1/5=0.2 probability of success (guessing correctly). The mean is np = 60 * 0.2 = 12. This mean represents the expected or average number of correct answers if many students were to guess on the test.
A) The standard deviation of X is greater than the standard deviation of Y.
B) The standard deviation of X is less than the standard deviation of Y.
C) The standard deviations of X and Y are equal.
D) The relationship cannot be determined without more information.
Correct Answer: B
For a fixed number of trials n, the standard deviation of a binomial distribution, sqrt(np(1-p)), is maximized when p=0.5. The standard deviation of X is sqrt(100*0.2*0.8) = sqrt(16) = 4. The standard deviation of Y is sqrt(100*0.5*0.5) = sqrt(25) = 5. Therefore, the standard deviation of X is less than the standard deviation of Y.
A) The mean number of passengers who show up (144) is less than the number of seats (150), so the airline will never have a problem with overbooking.
B) The standard deviation of 3.79 is small, which means exactly 144 passengers will show up for every flight.
C) The mean number of passengers who show up is 144. On average, the flight will have empty seats, but the number of passengers will vary, and it is possible for more than 150 to show up.
D) The mean is 144, which is only 6 away from 150. Therefore, the flight is almost certain to be overbooked.
Correct Answer: C
The mean of 144 indicates that, on average, the flight will not be overbooked. However, the standard deviation of 3.79 indicates that the actual number of passengers will vary. An outcome of 151 passengers, which would overbook the flight, is less than 2 standard deviations from the mean (144 + 2*3.79 = 151.58), so it is not an unusual event. The parameters suggest that while the average outcome is favorable for the airline, there is a real risk of overbooking.
A) Mean = 20, Standard Deviation = 15
B) Mean = 20, Standard Deviation = 3.87
C) Mean = 60, Standard Deviation = 3.87
D) Mean = 60, Standard Deviation = 15
Correct Answer: B
The number of trials is n=80 and the probability of success (ordering a diet soda) is p=0.25. The mean is calculated as np = 80 * 0.25 = 20. The standard deviation is calculated as sqrt(np(1-p)) = sqrt(80 * 0.25 * 0.75) = sqrt(15) ≈ 3.87.