AP Statistics Quiz: Parameters For A Binomial Distribution

What this quiz covers

This quiz focuses on Parameters For A Binomial Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 2
2. 1.34
3. 1.8 (correct answer)
4. 0.1

Explanation: The scenario is a binomial setting with n=20 and p=0.1. The question asks for the variance, not the standard deviation. The variance is calculated as $\sigma^2 = np(1-p) = 20(0.1)(0.9) = 1.8$.

Question 2

1. The mean is the most likely number of successes in one set of n trials, and the standard deviation is the maximum possible deviation from this value.
2. The mean is the probability of success on a single trial, and the standard deviation is the measure of variability of that probability over n trials.
3. The mean is the long-run average number of successes over many sets of n trials, and the standard deviation measures the typical variation of the number of successes around this average. (correct answer)
4. The mean is the number of trials n, and the standard deviation measures how much the probability of success p can be expected to vary from trial to trial.

Explanation: The mean or expected value of a random variable is defined as the long-run average of its outcomes over many repetitions. The standard deviation measures the typical or average distance of the outcomes from that mean. The other options misinterpret the definitions of mean, standard deviation, n, and p.

Question 3

1. The standard deviation of X is greater than the standard deviation of Y.
2. The standard deviation of X is less than the standard deviation of Y.
3. The standard deviations of X and Y are equal. (correct answer)
4. The relationship cannot be determined without more information.

Explanation: The standard deviation of a binomial variable is $\sigma = \sqrt{np(1-p)}$. For X, $\sigma_X = \sqrt{50(0.3)(1-0.3)} = \sqrt{50(0.3)(0.7)} = \sqrt{10.5}$. For Y, $\sigma_Y = \sqrt{50(0.7)(1-0.7)} = \sqrt{50(0.7)(0.3)} = \sqrt{10.5}$. The standard deviations are equal because the product p(1-p) is the same for p=0.3 and p=0.7.

Question 4

1. Expected Value = 6, Standard Deviation = 2.4
2. Expected Value = 4, Standard Deviation = 1.55
3. Expected Value = 6, Standard Deviation = 1.55 (correct answer)
4. Expected Value = 4, Standard Deviation = 2.4

Explanation: The scenario describes a binomial setting with n=10 and p=0.6. The expected value (mean) is $\mu = np = 10(0.6) = 6$. The standard deviation is $\sigma = \sqrt{np(1-p)} = \sqrt{10(0.6)(0.4)} = \sqrt{2.4} \approx 1.55$.

Question 5

1. Mean = 60, Standard Deviation = 7.62 (correct answer)
2. Mean = 15, Standard Deviation = 7.62
3. Mean = 60, Standard Deviation = 15.24
4. Mean = 60, Standard Deviation = 3.81

Explanation: The new sample size is n = 2000 and the defect rate is p = 0.03. The new mean is $\mu = np = 2000(0.03) = 60$. The new standard deviation is $\sigma = \sqrt{np(1-p)} = \sqrt{2000(0.03)(0.97)} = \sqrt{58.2} \approx 7.62$. Note that quadrupling the sample size (from 500 to 2000) quadruples the mean and doubles the standard deviation.

Question 6

1. $n=20,\ p=0.05$
2. $n=0.95,\ p=20$
3. $n=95,\ p=0.20$
4. $n=20,\ p=0.95$ (correct answer)
5. $n=20,\ p=\dfrac{1}{95}$

Explanation: This tests binomial parameter identification for valid codes in 20 boxes. $n=20$ and $p=0.95$ fit, as in choice D. Distractors: A uses complement $0.05$, B swaps, C scales wrongly, E reciprocal. Binomial needs fixed trials $n$, success $p$ per trial, independence. This counts valid codes. High $p$ suggests many successes expected.

Question 7

1. $n=10,\ p=0.35$
2. $n=0.65,\ p=10$
3. $n=10,\ p=0.65$ (correct answer)
4. $n=7,\ p=0.65$
5. $n=65,\ p=0.10$

Explanation: This AP Statistics question evaluates understanding of binomial parameters, where n denotes the number of independent trials and p the probability of success on each. The researcher tests 10 seeds, establishing n = 10 as the trial count. Success is germination, given p = 0.65 under the conditions. Choice A serves as a distractor by using p = 0.35, which is the failure probability, if success is misidentified. A mini-lesson reminds us that binomial settings involve n fixed trials with success probability p, and independence with constant p; X counts successes like germinated seeds. The correct modeling uses n = 10 and p = 0.65, aligning with choice C.

Question 8

1. $n=35,\ p=0.92$
2. $n=35,\ p=0.08$ (correct answer)
3. $n=0.08,\ p=35$
4. $n=8,\ p=0.35$
5. $n=35,\ p=0.08\times 35$

Explanation: This medical screening scenario tests understanding of binomial parameters when success is a positive test result. With 35 patients screened (n = 35) and success defined as testing positive with probability 0.08 (p = 0.08), the correct answer is B. Students might be tempted to use 0.92 (probability of negative result) if they misunderstand what's being counted. The distractors include various misplacements of n and p values. In medical contexts, carefully identify whether you're counting positive or negative outcomes—here, success explicitly means a positive test, so p = 0.08. The binomial model applies because we have independent patient results with constant probability.

Question 9

1. $n=20,\ p=0.19$
2. $n=81,\ p=0.20$
3. $n=0.81,\ p=20$
4. $n=20,\ p=0.81$ (correct answer)
5. $n=20,\ p=0.81\times 20$

Explanation: This delivery audit problem illustrates binomial parameter identification when success is on-time delivery. The company audits 20 packages (n = 20), and success is defined as on-time delivery with probability 0.81 (p = 0.81), making D the correct answer. A common error would be using 0.19 (probability of late delivery) for p, but since we're counting on-time deliveries, we need p = 0.81. The distractors attempt various confusions including using 81 as n or unnecessary calculations. Remember: in a binomial distribution, n is the fixed number of trials and p is the probability of the specific outcome defined as success.

Question 10

1. $n=18,\ p=0.30$ (correct answer)
2. $n=18,\ p=0.70$
3. $n=0.30,\ p=18$
4. $n=0.30\times 18,\ p=0.30$
5. $n=2,\ p=0.30$

Explanation: This genetics problem tests understanding of binomial parameters in a scientific context. The lab runs 18 independent trials (n = 18), and success is defined as producing a visible color change with probability 0.30 (p = 0.30), making A the correct answer. Choice B incorrectly uses the complement probability (0.70), while other choices confuse the roles of n and p or perform unnecessary calculations. When identifying binomial parameters, n is always the fixed number of trials, and p is always the probability of the outcome you're counting. The independence assumption and fixed probability across trials confirm this is a binomial setting.

Question 11

1. $n=16,\ p=0.90$
2. $n=10,\ p=0.16$
3. $n=16,\ p=0.10$ (correct answer)
4. $n=0.10,\ p=16$
5. $n=0.10\times 16,\ p=0.10$

Explanation: This fish-tagging problem demonstrates binomial parameters in ecological sampling. The biologist catches 16 fish with replacement (n = 16), and success is defined as catching a tagged fish with probability 0.10 (p = 0.10), making C the correct answer. The "with replacement" detail ensures independence between catches, confirming the binomial model is appropriate. Choice A incorrectly uses the complement probability (0.90 for untagged fish), while other options confuse n and p. Remember that in binomial distributions, n must be the number of trials (catches) and p must be the probability of the outcome you're counting (tagged fish).

Question 12

1. $n=25,\ p=0.96$
2. $n=25,\ p=0.04$ (correct answer)
3. $n=0.04,\ p=25$
4. $n=2,\ p=0.04$
5. $n=25,\ p=0.04\times 25$

Explanation: This question tests your ability to identify the parameters n and p in a binomial distribution. In a binomial setting, n represents the number of independent trials (here, 25 lightbulbs inspected), and p represents the probability of success on each trial. Since we're counting defective bulbs and each bulb has a 0.04 probability of being defective, we have n = 25 and p = 0.04. Choice B correctly identifies these parameters. Common mistakes include confusing p with the probability of non-defective bulbs (0.96) or misunderstanding what n represents. Remember: n is always the number of trials, and p is always the probability of the specific outcome you're counting.

Question 13

1. $n=60,\ p=0.07$
2. $n=0.93,\ p=60$
3. $n=60,\ p=0.93$ (correct answer)
4. $n=93,\ p=0.60$
5. $n=60,\ p=0.93\times 60$

Explanation: This manufacturing quality control problem requires careful attention to how success is defined. The inspector observes 60 jars (n = 60), and success is defined as a properly sealed lid with probability 0.93 (p = 0.93), making C the correct answer. Students might mistakenly use 0.07 (the probability of improper sealing) if they misread what counts as success. The key lesson is to always identify what outcome is being counted before determining p. In binomial settings, p represents the probability of the specific outcome defined as success, not its complement. The fixed number of observations and constant probability confirm this follows a binomial distribution.

Question 14

1. $n=25,\ p=0.60$
2. $n=0.40,\ p=25$
3. $n=25,\ p=0.40$ (correct answer)
4. $n=0.60,\ p=25$
5. $n=10,\ p=0.40$

Explanation: This problem requires identifying n (number of trials) and p (probability of success). The organization calls 25 voters (n = 25), and a "success" is answering the call, which happens with probability 0.40, so p = 0.40. Choice A incorrectly uses p = 0.60, which would be the probability of NOT answering (the complement). Choices B and D swap n and p, which creates invalid parameters since n must be a positive integer. Remember that in binomial distributions, p is always the probability of the event defined as "success," not its complement.

Question 15

1. $n=18,\ p=0.75$ (correct answer)
2. $n=0.75,\ p=18$
3. $n=18,\ p=0.25$
4. $n=0.25,\ p=18$
5. $n=75,\ p=0.18$

Explanation: The skill here is recognizing the binomial parameters n and p for modeling successes in repeated independent trials, like free throws. With 18 shots and a 0.75 probability of making each, n=18 and p=0.75 are the correct values. Distractors include swapping n and p in choice B, or using the failure probability 0.25 in choices C and D. Choice E mistakenly alters both values, perhaps confusing percentages. In a binomial distribution, n is always the integer number of trials, and p is the success probability between 0 and 1. This models the count of made shots, assuming independence.

Question 16

1. $n=0.92,\ p=30$
2. $n=30,\ p=0.92$ (correct answer)
3. $n=30,\ p=\dfrac{1}{30}$
4. $n=0.08,\ p=30$
5. $n=30,\ p=0.08$

Explanation: This question assesses your skill in identifying the parameters for a binomial distribution, which is used to model the number of successes in a fixed number of independent trials. In this scenario, the engineer tests 30 batteries, so n, the number of trials, is 30, and each has a success probability p of 0.92, making the correct parameters n=30 and p=0.92. A common distractor is reversing n and p, as in choice A, which swaps them, or using the failure probability 0.08, as seen in choices D and E. Another misleading option is choice C, which incorrectly uses the reciprocal of n as p. To clarify, a binomial distribution requires a fixed n (trials) and constant p (success probability per trial), with independence assumed. This setup fits perfectly here, as each battery test is independent with the same success probability.

Question 17

1. $n=50,\ p=0.96$
2. $n=4,\ p=0.50$
3. $n=0.04,\ p=50$
4. $n=50,\ p=0.04$ (correct answer)
5. $n=96,\ p=0.50$

Explanation: The skill is selecting binomial parameters for defective bulbs in 50 inspections. With n=50 and p=0.04 for defects, choice D is accurate. Distractors include choice A with failure probability 0.96, C swapping n and p, and B or E altering values arbitrarily. No reciprocal is relevant here. Binomial distributions model success counts with fixed n, constant p, binary outcomes, and independence. This tracks defects efficiently.

Question 18

1. $n=12,\ p=0.90$
2. $n=0.10,\ p=12$
3. $n=12,\ p=0.10$ (correct answer)
4. $n=10,\ p=0.12$
5. $n=12,\ p=\dfrac{1}{10}$

Explanation: This question focuses on binomial parameters for positive test results in 12 independent patients. Thus, n=12 and p=0.10 are the parameters, as in choice C. Watch for distractors like choice A using the complement 0.90, or B swapping n and p. Choice D changes both values slightly, and E uses a fraction equivalent to 0.10 but pairs it wrongly. Key lesson: n is the trial count, p the success chance; binomial counts successes under independence. This applies to genetic markers here.

Question 19

1. $n=15,\ p=0.78$ (correct answer)
2. $n=15,\ p=0.22$
3. $n=0.78,\ p=15$
4. $n=0.78,\ p=0.22$
5. $n=14,\ p=0.78$

Explanation: To identify binomial parameters, we need the number of trials n and the probability of success p. The player takes 15 free throws (n = 15), and since "success" means making a free throw with probability 0.78, we have p = 0.78. Choice B incorrectly uses p = 0.22, which would be the probability of missing (failure), not making (success) the shot. Choice C swaps the roles of n and p, which is impossible since n must be a counting number. In binomial distributions, always identify what counts as "success" first, then use that probability for p.

Question 20

1. $n=10,\ p=0.25$ (correct answer)
2. $n=10,\ p=0.75$
3. $n=4,\ p=0.10$
4. $n=0.25,\ p=10$
5. $n=9,\ p=0.25$

Explanation: For binomial parameters, n equals the number of trials and p equals the success probability. The student answers 10 questions (n = 10), and with 4 choices per question where exactly one is correct, random guessing gives p = 1/4 = 0.25 probability of success. Choice B incorrectly uses p = 0.75 (the probability of guessing wrong). Choice C has the wrong n value. In multiple-choice scenarios with random guessing, the success probability equals 1 divided by the number of answer choices.