Parameters for a Binomial Distribution - AP Statistics

If the number of trials n is increased by a factor of 9 while p remains constant, how does the standard deviation of X change?
Answer: It increases by a factor of 3. The original standard deviation is $\sigma_{old} = \sqrt{np(1-p)}$ . The new number of trials is 9n. The new standard deviation is $\sigma_{new} = \sqrt{(9n)p(1-p)} = \sqrt{9} \times \sqrt{np(1-p)} = 3 \times \sigma_{old}$ . Therefore, the standard deviation increases by a factor of 3 (is tripled).
A quality engineer tests a new type of battery by selecting 30 batteries at random from a large shipment. A battery is counted as a success if it meets the company's minimum capacity standard, and past data suggest each battery meets the standard with probability $0.92$ (assume independence). Which values of $n$ and $p$ correctly model the number of successes with a binomial distribution?
Answer: $n=30,\ p=0.92$ 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.
A teacher gives a 10-question true/false quiz. A success is defined as answering a question correctly. A student guesses on every question, so the probability of a correct answer on any question is 0.50, and answers are treated as independent. If $X$ is the number of correct answers, which values correctly model this situation (identify $n$ and $p$ )?
Answer: $n=10,\ p=0.50$ This true/false quiz scenario perfectly illustrates binomial parameters in an educational context. The student answers 10 questions (n = 10), and with random guessing on true/false questions, the probability of success (correct answer) is 0.50 (p = 0.50), making A the correct answer. This is a classic binomial situation because we have a fixed number of independent trials with constant probability of success. The distractors incorrectly use values like 2 (perhaps thinking of true/false options) or 0.10 (perhaps confusing with 10 questions). For binomial distributions, n counts the trials and p is the probability of success on each trial.
A fair die is rolled 12 times; success means rolling 5 or 6. Find the variance of the number of successes.
Answer: 2.67 For each roll, p = 2/6 = 1/3 and q = 2/3. The count of successes is binomial with n = 12, so variance = npq = 12(1/3)(2/3) = 8/3 = 2.67. The tempting wrong value, 4.00, is the mean np, not the variance.
A teacher gives a 10-question true/false quiz to a student who guesses randomly on every question. A success is a correct answer, and the probability of a correct guess on each question is $0.50$ (assume independence). Which values of $n$ and $p$ correctly model the number of successes using a binomial distribution?
Answer: $n=10,\ p=0.50$ Practice identifying binomial parameters for correct answers in 10 true/false questions. n=10 and p=0.50 are correct, per choice A. Swapping appears in B, while C, D, E alter values oddly, perhaps misreading. No complement or reciprocal fits. Lesson: binomial models guesses with fair p=0.50, independent trials. Counts correct answers.
A genetic test is administered to 12 patients. A success is defined as the test returning a positive result for a particular marker, and for the population being tested the probability of a positive result is $0.10$ for each patient (assume independence). Which values of $n$ and $p$ correctly model the number of successes using a binomial distribution?
Answer: $n=12,\ p=0.10$ 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.
A hospital screens 18 patients for a certain condition. Each patient's test result is recorded as positive or negative. Suppose the probability a randomly selected patient tests positive is $0.12$ , and test outcomes for different patients are treated as independent. Let $X$ be the number of positive tests. Which values correctly model this situation with a binomial distribution (identify $n$ and $p$ )?
Answer: $n=18,\ p=0.12$ This medical screening problem tests parameter identification with positive test results. The hospital screens n = 18 patients (number of trials), and X counts positive tests with probability p = 0.12 per patient. Since we're counting positive tests and that probability is 0.12, we have n = 18 and p = 0.12. Students might confuse this with p = 0.88 (negative tests), but p must match what X counts. Choices B and E incorrectly place decimal probabilities in the n position. In medical contexts, clearly identify whether you're counting positive or negative outcomes, as this determines which probability to use for p in your binomial model.
A student flips a biased coin 15 times. A success is defined as getting heads, and the probability of heads on each flip is $0.60$ (assume independence). Which values of $n$ and $p$ correctly model the number of successes using a binomial distribution?
Answer: $n=15,\ p=0.60$ Identify binomial parameters for heads in 15 coin flips, where n=15 and p=0.60. Choice C correctly reflects this. Common pitfalls are choice A using 0.40 (failure), B swapping parameters, and D scaling values incorrectly. Choice E uses a tiny reciprocal. Mini-lesson: binomial requires independent trials, each with success p; it counts total heads here. Bias doesn't change the structure.
A seed company claims that a particular type of seed germinates with probability 0.85 when planted under standard conditions. A gardener plants 40 seeds, and each seed either germinates or does not germinate; assume independence is reasonable. Let $X$ be the number of seeds that germinate (successes). Which values correctly model this situation?
Answer: $n=40,\ p=0.85$ For seed germination, we count germinated seeds as successes, giving us n = 40 (seeds planted) and p = 0.85 (germination probability). Choice A incorrectly uses p = 0.15, which would be the probability of NOT germinating. Choice B swaps n and p values, which is invalid since n must be a positive integer. Choice E incorrectly computes p as 0.85 × 40 = 34, which exceeds 1 and cannot be a probability. In biological applications like this, "success" typically means the desired outcome occurred (germination, survival, etc.). The binomial model assumes each seed germinates independently with the same probability - a reasonable approximation when seeds are planted under identical conditions.
A factory inspects 50 light bulbs from a production line. A bulb is a success if it is defective (the factory is tracking defects), and the long-run defect rate is $0.04$ per bulb (assume independence). Which values of $n$ and $p$ correctly model the number of successes with a binomial distribution?
Answer: $n=50,\ p=0.04$ 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.
A factory machine produces screws, and each screw is independently classified as within tolerance or out of tolerance. An engineer inspects 40 screws and defines a success as a screw being within tolerance. If the probability a screw is within tolerance is $0.93$ , let $X$ be the number within tolerance in the sample. Which values correctly model this situation with a binomial distribution (identify $n$ and $p$ )?
Answer: $n=40,\ p=0.93$ This question requires careful reading to identify what constitutes success. The engineer inspects n = 40 screws (number of trials), and success is defined as a screw being within tolerance with probability p = 0.93. Since X counts screws within tolerance (successes), we use n = 40 and p = 0.93. The problem explicitly states we're counting within-tolerance screws, so p = 0.93, not 0.07. Choices B and D incorrectly swap parameter positions, placing decimal values where integers belong. Remember: in binomial distributions, success is what you define it to be, and p is always the probability of that defined success occurring on any single trial.
A factory machine produces bottles, and historically 2% of bottles are underfilled. A supervisor inspects 80 bottles chosen at random from the day's production, classifying each as underfilled or not underfilled. Assume independence is reasonable. Let $X$ be the number of underfilled bottles (successes). Which values correctly model this situation?
Answer: $n=80,\ p=0.02$ To model the number of underfilled bottles, we identify n = 80 (bottles inspected) and p = 0.02 (probability a bottle is underfilled). Since we're counting underfilled bottles as successes, p equals the 2% underfill rate, not the 98% properly filled rate. Choice A incorrectly uses p = 0.98, which would model the number of properly filled bottles instead. Choices B and D swap n and p, violating the requirement that n be a positive integer. Choice E incorrectly computes p as 0.02 × 80 = 1.6, which exceeds 1 and cannot be a probability. Remember: p represents the probability of success on a single trial, not a count or percentage of total successes.
A website designer runs a small experiment: for each of 25 visitors shown a new landing page, a success is defined as the visitor clicking the "Sign Up" button. Previous testing suggests the probability of a click is $0.18$ for each visitor (assume independence). Which values of $n$ and $p$ correctly model the number of successes with a binomial distribution?
Answer: $n=25,\ p=0.18$ You're practicing identifying binomial parameters for clicks in 25 independent visitor trials. The number of trials n is 25, and success probability p is 0.18, matching choice D. Distractors often involve the failure probability 0.82 in choice A, or swapping values in choice B. Choice C seems unrelated, possibly a mix-up, and E uses a reciprocal of 18 erroneously. A mini-lesson: binomial models require binary outcomes per trial, fixed n, constant p, and independence. This counts sign-up clicks effectively.
A basketball player takes 18 free throws in practice. A success is making a free throw, and the player's long-run free-throw percentage is $0.75$ (assume each shot is independent). Which values of $n$ and $p$ correctly model the number of made free throws with a binomial distribution?
Answer: $n=18,\ p=0.75$ 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.
A political campaign calls 30 registered voters. For each call, define a success as the voter agreeing to volunteer. Assume calls are independent and the probability any contacted voter agrees to volunteer is $0.15$ . Let $X$ be the number of voters who agree to volunteer. Which values correctly model this situation with a binomial distribution (identify $n$ and $p$ )?
Answer: $n=30,\ p=0.15$ This political campaign scenario requires identifying parameters for volunteer recruitment. The campaign calls n = 30 voters (number of trials), and success is defined as agreeing to volunteer with probability p = 0.15. Since X counts voters who agree to volunteer, we use n = 30 and p = 0.15. Choice A incorrectly uses p = 0.85 (those who don't volunteer), while choices B and E swap parameter positions. The key insight is that p represents the probability of the specific outcome we're counting (volunteering), not its complement. In binomial distributions, always match your probability parameter to exactly what your random variable is designed to count.
For a binomial distribution with mean 12 and variance 3, what is n?
Answer: 16 For a binomial, mean = np and variance = np(1-p). Divide variance by mean: 3/12 = 1/4 = 1-p, so p = 3/4. Then np = 12 becomes n(3/4) = 12, so n = 16. The tempting wrong answer is 12, which just repeats the mean; n is the number of trials, not the mean.
Which of the following statements provides the best description of these parameters?
Answer: 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. 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.
How do the standard deviations of X and Y compare?
Answer: The standard deviations of X and Y are equal. 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.
A biologist tags 9 turtles and later checks whether each tagged turtle is recaptured during a follow-up survey. Assume each turtle's recapture status is independent of the others and the probability a tagged turtle is recaptured is $0.30$ . Let $X$ be the number of tagged turtles recaptured. Which values correctly model this situation with a binomial distribution (identify $n$ and $p$ )?
Answer: $n=9,\ p=0.30$ This question involves identifying parameters when counting a specific outcome. The biologist tags n = 9 turtles (number of trials), and X counts recaptured turtles with probability p = 0.30 each. Since we're counting recaptured turtles and that probability is 0.30, we use n = 9 and p = 0.30. A common error would be using p = 0.70 (not recaptured), but p must match the outcome X is counting. Choices B and D incorrectly swap n and p positions - remember that n must be a positive integer. The key to binomial parameter identification is matching p to exactly what your random variable X is counting, not its complement.
A basketball player takes 15 free throws during practice. Define a success as making a free throw. The player's long-run free-throw percentage is $0.78$ , and each shot is considered independent of the others. Which values correctly model this situation with a binomial random variable $X$ = number of successes?
Answer: $n=15,\ p=0.78$ 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.

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