Parameters for a Binomial Distribution
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AP Statistics › Parameters for a Binomial Distribution
A shipment contains many boxes of cereal. A consumer group randomly selects 20 boxes to check a promotional code inside each box. A success is finding a valid code, and the probability a box contains a valid code is $0.95$ (assume independence). Which values of $n$ and $p$ correctly model the number of successes with a binomial distribution?
$n=20,\ p=\dfrac{1}{95}$
$n=20,\ p=0.05$
$n=0.95,\ p=20$
$n=95,\ p=0.20$
$n=20,\ p=0.95$
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.
A researcher tests 10 seeds by planting each one and recording whether it germinates (success) within 7 days. For this type of seed, the probability of germination is 0.65 under the same conditions, and the researcher treats each seed’s outcome as independent. Let $X$ be the number of seeds that germinate. Which values correctly model this situation?
$n=65,\ p=0.10$
$n=0.65,\ p=10$
$n=10,\ p=0.65$
$n=10,\ p=0.35$
$n=7,\ p=0.65$
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.
A hospital screens 35 patients for a particular infection using a rapid test. A success is defined as a positive test result. For this group, the probability a patient tests positive is 0.08, and patient results are treated as independent. If $X$ is the number of positive test results, which values correctly model this situation (identify $n$ and $p$)?
$n=8,\ p=0.35$
$n=35,\ p=0.08\times 35$
$n=35,\ p=0.92$
$n=35,\ p=0.08$
$n=0.08,\ p=35$
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.
A delivery company audits 20 packages to see whether each arrives on time. A success is defined as an on-time delivery. Based on recent performance, the probability a package arrives on time is 0.81, and package outcomes are treated as independent. If $X$ is the number of on-time deliveries in the audit, which values correctly model this situation (identify $n$ and $p$)?
$n=81,\ p=0.20$
$n=20,\ p=0.19$
$n=20,\ p=0.81$
$n=0.81,\ p=20$
$n=20,\ p=0.81\times 20$
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.
A genetics lab runs 18 independent trials of a chemical reaction. A success is defined as the reaction producing a visible color change. Under current conditions, the probability of a color change on any trial is 0.30. If $X$ is the number of trials with a color change, which values correctly model this situation (identify $n$ and $p$)?
$n=18,\ p=0.70$
$n=18,\ p=0.30$
$n=0.30,\ p=18$
$n=0.30\times 18,\ p=0.30$
$n=2,\ p=0.30$
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.
A biologist tags 16 fish and releases them into a lake. Later, the biologist catches 16 fish (one at a time with replacement modeled as independent draws) and records whether each fish is tagged. A success is defined as catching a tagged fish. The probability a caught fish is tagged is 0.10 under this model. If $X$ is the number of tagged fish caught, which values correctly model this situation (identify $n$ and $p$)?
$n=16,\ p=0.10$
$n=0.10\times 16,\ p=0.10$
$n=0.10,\ p=16$
$n=16,\ p=0.90$
$n=10,\ p=0.16$
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).
A quality-control engineer inspects a shipment by randomly selecting 25 lightbulbs and checking whether each one is defective. Historical data suggest that any inspected bulb has a 0.04 probability of being defective, and each inspection is treated as independent. If $X$ is the number of defective bulbs found, which values correctly model this situation with a binomial distribution (identify $n$ and $p$)?
$n=2,\ p=0.04$
$n=25,\ p=0.04$
$n=25,\ p=0.04\times 25$
$n=0.04,\ p=25$
$n=25,\ p=0.96$
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.
A manufacturing robot places lids on jars. An inspector observes the next 60 jars and records whether each lid is properly sealed. A success is defined as a properly sealed lid. The probability a lid is properly sealed is 0.93, and each jar’s sealing outcome is treated as independent. If $X$ is the number of properly sealed lids, which values correctly model this situation (identify $n$ and $p$)?
$n=0.93,\ p=60$
$n=60,\ p=0.07$
$n=60,\ p=0.93$
$n=60,\ p=0.93\times 60$
$n=93,\ p=0.60$
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.
A polling organization calls 25 randomly selected registered voters in a large city. For each person, a success is defined as “the voter answers the call.” Based on past data, the probability a randomly selected voter answers is $0.40$, and calls are treated as independent. Which values correctly model this situation with a binomial random variable $X$ = number of successes?
$n=0.40,\ p=25$
$n=25,\ p=0.60$
$n=0.60,\ p=25$
$n=10,\ p=0.40$
$n=25,\ p=0.40$
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.
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?
$n=75,\ p=0.18$
$n=18,\ p=0.75$
$n=0.75,\ p=18$
$n=0.25,\ p=18$
$n=18,\ p=0.25$
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.