Binomial Distribution: Conditions, Formula, Mean, and SD
Recognize binomial settings from their design, calculate exact and cumulative probabilities, and interpret the model’s mean and standard deviation.
Binomial Distribution: direct answer
A binomial distribution counts successes across a fixed number of trials with two outcome categories, independence, and constant success probability. After checking BINS, use the binomial probability formula or cumulative probabilities and interpret μ=np and σ=√(np(1−p)).
This page keeps the lesson centered on binomial distribution. Practice is included only after the method, assumptions, interpretation, and common decision points are explained.
Quick reference: Binomial Distribution: Conditions, Formula, Mean, and SD
| Exact probability | P(X=x)=C(n,x)pˣ(1−p)ⁿ⁻ˣ |
|---|---|
| Mean | np |
| SD | √np(1−p) |
Concept mastery: Binomial Distribution: Conditions, Formula, Mean, and SD
Use the BINS conditions before the formula
A binomial random variable counts successes in a fixed number n of trials, each trial has two outcome categories for the event of interest, trials are independent or approximately independent, and the success probability p remains constant. Naming BINS prevents using the formula in a setting where n, independence, or p changes.
The binomial probability formula counts arrangements
P(X=k)=C(n,k)p^k(1−p)^(n−k). The probability part describes one particular ordering with k successes and n−k failures; the combination C(n,k) counts how many different orderings contain the same numbers of successes and failures.
Cumulative language determines the calculator command
“At most k” means P(X≤k), “fewer than k” means P(X≤k−1), “at least k” means 1−P(X≤k−1), and “more than k” means 1−P(X≤k). Translate the inequality before using binomialcdf so an off-by-one error does not survive unnoticed.
The mean and standard deviation describe repeated binomial counts
For X~Bin(n,p), μ=np and σ=√(np(1−p)). The mean is the long-run average number of successes across repetitions of the entire n-trial process. Standard deviation describes typical count-to-count variation around np, not variation in the underlying success probability.
Sampling without replacement needs an independence check
When sampling without replacement from a finite population, trials are not exactly independent. The 10% condition is commonly used as an approximation: if the sample size is no more than about 10% of the population, the changing success probability is often small enough for a binomial model to be reasonable.
Binomial and geometric questions ask different random-variable questions
A binomial variable counts successes in a fixed number of trials. A geometric variable counts trials until the first success; geometric distribution has been removed from the revised AP Statistics tested course, so current exam preparation should prioritize the binomial structure while using geometric comparisons only as legacy enrichment.
12 worked binomial distribution cases
Worked case 1: BINS check
Scenario. A fixed 20 independent items each have defect probability .04.
Reasoning. The count of defects is binomial with n=20 and p=.04 because BINS is satisfied.
Case 1: BINS check check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 1; do not replace it with a memorized generic sentence.
Worked case 2: Changing p
Scenario. A player’s success probability increases after each successful attempt.
Reasoning. The count is not binomial under a constant-p model because the success probability changes across trials.
Case 2: Changing p check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 2; do not replace it with a memorized generic sentence.
Worked case 3: Random n
Scenario. Calls continue until the first sale.
Reasoning. The number of calls is not binomial because the number of trials is not fixed in advance.
Case 3: Random n check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 3; do not replace it with a memorized generic sentence.
Worked case 4: Exactly k
Scenario. For n=10, p=.3, and k=4, use C(10,4)(.3)^4(.7)^6.
Reasoning. The combination factor counts which four of the ten positions are successes.
Case 4: Exactly k check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 4; do not replace it with a memorized generic sentence.
Worked case 5: At most
Scenario. For X~Bin(12,.2), “at most 3” means P(X≤3).
Reasoning. Use a cumulative probability through 3; do not subtract from 1.
Case 5: At most check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 5; do not replace it with a memorized generic sentence.
Worked case 6: At least
Scenario. For X~Bin(12,.2), “at least 3” means 1−P(X≤2).
Reasoning. The complement ends at 2 because values 3 and above belong to the target event.
Case 6: At least check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 6; do not replace it with a memorized generic sentence.
Worked case 7: More than
Scenario. For X~Bin(12,.2), “more than 3” means 1−P(X≤3).
Reasoning. The strict inequality shifts the cumulative boundary compared with “at least 3.”
Case 7: More than check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 7; do not replace it with a memorized generic sentence.
Worked case 8: Mean
Scenario. For n=50 and p=.08, μ=np=4.
Reasoning. Across repeated 50-trial experiments, the average success count approaches 4.
Case 8: Mean check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 8; do not replace it with a memorized generic sentence.
Worked case 9: SD
Scenario. For n=50 and p=.08, σ=√(50×.08×.92)≈1.918.
Reasoning. The standard deviation describes typical variation in the success count across repeated 50-trial runs.
Case 9: SD check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 9; do not replace it with a memorized generic sentence.
Worked case 10: 10% condition
Scenario. A sample of 40 is drawn without replacement from 200 individuals.
Reasoning. Because 40 exceeds 10% of 200, independence is not well approximated by the usual 10% condition.
Case 10: 10% condition check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 10; do not replace it with a memorized generic sentence.
Worked case 11: Rare success
Scenario. n is large and p is small.
Reasoning. A small p does not invalidate a binomial model; BINS still controls whether the exact binomial distribution applies.
Case 11: Rare success check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 11; do not replace it with a memorized generic sentence.
Worked case 12: Legacy contrast
Scenario. A question counts trials until first success.
Reasoning. That is geometric rather than binomial; geometric distribution is legacy enrichment for the revised 2027 AP Statistics exam.
Case 12: Legacy contrast check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 12; do not replace it with a memorized generic sentence.
Deep-dive notes for Binomial Distribution: Conditions, Formula, Mean, and SD
BINS is a model-selection checklist, not a slogan
Fixed number of trials, binary success classification, independence, and constant success probability must describe the random process. The word “success” is only a label; it can represent a defect, failure, yes response, or any event being counted. Define success explicitly so p and X stay consistent.
The combination term prevents undercounting
The product p^k(1−p)^(n−k) gives the probability of one specified ordering with k successes. There are C(n,k) different orderings with the same counts. The binomial coefficient multiplies that single-order probability by the number of arrangements, which is why it appears in the exact probability formula.
Cumulative boundaries cause many calculator errors
“At least 5” includes 5, so its complement ends at 4. “More than 5” begins at 6, so its complement ends at 5. “Fewer than 5” ends at 4. Translate the words into an inequality for X before typing a cumulative command; doing so prevents off-by-one mistakes that calculators cannot diagnose.
Mean and standard deviation are count summaries
For X~Bin(n,p), np is the expected number of successes in a full n-trial run, not a probability. The standard deviation √(np(1−p)) is also measured in successes. Keeping the units as counts helps distinguish the random variable X from the underlying trial success probability p.
The 10% condition is an approximation for sampling without replacement
Trials drawn without replacement are technically dependent because the population composition changes. When the sample is no more than about 10% of the population, the change is often small enough for independence to be approximated. If the sample fraction is large, use a model that respects the finite-population dependence rather than forcing binomial assumptions.
Small p does not invalidate a binomial model
Rare-event settings are often excellent binomial applications when n is fixed and p remains constant. The model can produce a distribution concentrated near zero with a long right tail. What matters is the data-generating process, not whether p is close to one-half.
Binomial probability and normal approximation are separate ideas
The exact binomial model can be valid even when a normal approximation to it is poor. Conditions such as sufficiently large np and n(1−p) concern approximation shape, not the validity of BINS itself. Keep exact-model conditions separate from approximation conditions so one failed shortcut does not invalidate the original binomial distribution.
Geometric comparison is legacy enrichment for 2027
A binomial variable counts successes in a predetermined number of trials. A geometric variable counts trials until the first success. College Board removed geometric distribution from the revised 2026–27 AP Statistics tested course, so the practical exam priority is to recognize and use the binomial model correctly while treating geometric comparisons as background enrichment.
Additional method notes
Define the random variable before calculating
Write X as the number of successes in the fixed n trials and state what counts as a success. This prevents p from being attached to the wrong event and keeps cumulative statements interpretable. If the question asks for failures instead, either redefine success consistently or transform the requested count carefully rather than mixing labels mid-calculation.
Calculator output still needs probability language
A binomial calculator can produce an exact decimal but not the interpretation. Report the probability as the chance that the random count satisfies the stated event under the binomial model. Avoid saying that a probability is the “number of successes expected”; expected count is np, whereas the binomial probability answers how likely a particular count or range of counts is.
Check tail direction against the observed count
Before trusting a cumulative probability, ask whether the event is a lower tail, upper tail, or middle range. If an observed count is unusually high relative to np, the relevant upper-tail probability should be small. A large reported upper-tail probability in that setting is a warning that the complement or inequality boundary was entered incorrectly.
Exact binomial probabilities remain valid when the distribution is skewed
A binomial distribution can be strongly skewed when p is near 0 or 1, especially for moderate n. That shape does not invalidate the exact binomial formula or calculator. It matters only when a question proposes a normal approximation or when you are interpreting the distribution visually. Keep the model-selection question—does BINS hold?—separate from the approximation question—does the binomial shape have enough expected successes and failures to resemble a normal curve?
Interpret a binomial result in the original count context
After calculating a binomial probability, return to the count being modeled. A value such as 0.042 is the probability that the number of successes falls in the specified count event under the stated n and p; it is not 4.2 successes and it is not automatically evidence that the model is wrong. If the probability is being used to judge unusualness, state the tail event explicitly and compare the probability with the criterion supplied by the question.
Binomial Distribution: Conditions, Formula, Mean, and SD: 36 multiple-choice questions
These questions stay within this page’s topic. Work them after the concept and worked-case sections so practice reinforces the method rather than replacing instruction.
Question 1. Binomial Distribution
For 8 independent installations at a solar installer in Coastal Plains during a yearly program evaluation, each meets the daily energy output criterion with probability 0.64. Find P(X=3) and give the binomial mean and SD.
Answer: D
X~Bin(8,0.64). P(X=3)=C(8,3)(0.64)3(0.36)5=0.0888. μ=np=5.12 and σ=√np(1−p)=1.358.
Question 2. Binomial Distribution
For 23 independent learners at a digital learning platform in Lakeside district during a weekday operations study, each meets the lesson completion criterion with probability 0.487. Find P(X=10) and give the binomial mean and SD.
Answer: A
X~Bin(23,0.487). P(X=10)=C(23,10)(0.487)10(0.513)13=0.1463. μ=np=11.201 and σ=√np(1−p)=2.397.
Question 3. Binomial Distribution
For 23 independent applications at a housing authority in Lakeside district during a summer implementation review, each meets the application processing time criterion with probability 0.608. Find P(X=2) and give the binomial mean and SD.
Answer: D
X~Bin(23,0.608). P(X=2)=C(23,2)(0.608)2(0.392)21=0.00000027.00000027. μ=np=13.984 and σ=√np(1−p)=2.341.
Question 4. Binomial Distribution
For 24 independent accounts at a municipal water office in Sunbelt district during a multiweek validation study, each meets the monthly household use criterion with probability 0.551. Find P(X=9) and give the binomial mean and SD.
Answer: A
X~Bin(24,0.551). P(X=9)=C(24,9)(0.551)9(0.449)15=0.0372. μ=np=13.224 and σ=√np(1−p)=2.437.
Question 5. Binomial Distribution
For 15 independent applications at a housing authority in Cedar Grove during a semester-long cohort study, each meets the application processing time criterion with probability 0.598. Find P(X=2) and give the binomial mean and SD.
Answer: B
X~Bin(15,0.598). P(X=2)=C(15,2)(0.598)2(0.402)13=0.0003. μ=np=8.97 and σ=√np(1−p)=1.899.
Question 6. Binomial Distribution
For 20 independent samples at a food safety laboratory in Prairie District during a fall 2026 audit, each meets the sample concentration criterion with probability 0.451. Find P(X=14) and give the binomial mean and SD.
Answer: A
X~Bin(20,0.451). P(X=14)=C(20,14)(0.451)14(0.549)6=0.0153. μ=np=9.02 and σ=√np(1−p)=2.225.
Question 7. Binomial Distribution
For 9 independent learners at a digital learning platform in Westview during a multiweek validation study, each meets the lesson completion criterion with probability 0.511. Find P(X=6) and give the binomial mean and SD.
Answer: C
X~Bin(9,0.511). P(X=6)=C(9,6)(0.511)6(0.489)3=0.1749. μ=np=4.599 and σ=√np(1−p)=1.5.
Question 8. Binomial Distribution
For 11 independent installations at a solar installer in Cedar Grove during a school-year data collection, each meets the daily energy output criterion with probability 0.632. Find P(X=8) and give the binomial mean and SD.
Answer: D
X~Bin(11,0.632). P(X=8)=C(11,8)(0.632)8(0.368)3=0.2093. μ=np=6.952 and σ=√np(1−p)=1.599.
Question 9. Binomial Distribution
For 17 independent bus trips at a city transit agency in Desert County during a winter readiness review, each meets the on-time arrival criterion with probability 0.309. Find P(X=11) and give the binomial mean and SD.
Answer: D
X~Bin(17,0.309). P(X=11)=C(17,11)(0.309)11(0.691)6=0.0033. μ=np=5.253 and σ=√np(1−p)=1.905.
Question 10. Binomial Distribution
For 20 independent households at a recycling program in Riverbend during a six-week field trial, each meets the weekly material weight criterion with probability 0.293. Find P(X=15) and give the binomial mean and SD.
Answer: D
X~Bin(20,0.293). P(X=15)=C(20,15)(0.293)15(0.707)5=0.00002758.00002758. μ=np=5.86 and σ=√np(1−p)=2.035.
Question 11. Binomial Distribution
For 17 independent appointments at a university advising center in Westview during a spring 2027 pilot, each meets the appointment wait time criterion with probability 0.392. Find P(X=15) and give the binomial mean and SD.
Answer: C
X~Bin(17,0.392). P(X=15)=C(17,15)(0.392)15(0.608)2=0.00003987.00003987. μ=np=6.664 and σ=√np(1−p)=2.013.
Question 12. Binomial Distribution
For 14 independent accounts at a municipal water office in Metro East during a regional benchmarking study, each meets the monthly household use criterion with probability 0.507. Find P(X=6) and give the binomial mean and SD.
Answer: B
X~Bin(14,0.507). P(X=6)=C(14,6)(0.507)6(0.493)8=0.178. μ=np=7.098 and σ=√np(1−p)=1.871.
Question 13. Binomial Distribution
For 21 independent visitors at a county library in Desert County during a regional benchmarking study, each meets the weekly program attendance criterion with probability 0.482. Find P(X=11) and give the binomial mean and SD.
Answer: B
X~Bin(21,0.482). P(X=11)=C(21,11)(0.482)11(0.518)10=0.16. μ=np=10.122 and σ=√np(1−p)=2.29.
Question 14. Binomial Distribution
For 10 independent customers at a grocery cooperative in North Valley during a six-week field trial, each meets the checkout time criterion with probability 0.259. Find P(X=8) and give the binomial mean and SD.
Answer: C
X~Bin(10,0.259). P(X=8)=C(10,8)(0.259)8(0.741)2=0.0005. μ=np=2.59 and σ=√np(1−p)=1.385.
Question 15. Binomial Distribution
For 16 independent installations at a solar installer in New England network during a spring 2027 pilot, each meets the daily energy output criterion with probability 0.468. Find P(X=10) and give the binomial mean and SD.
Answer: C
X~Bin(16,0.468). P(X=10)=C(16,10)(0.468)10(0.532)6=0.0915. μ=np=7.488 and σ=√np(1−p)=1.996.
Question 16. Binomial Distribution
For 8 independent visitors at a county library in North Valley during a weekday operations study, each meets the weekly program attendance criterion with probability 0.565. Find P(X=3) and give the binomial mean and SD.
Answer: A
X~Bin(8,0.565). P(X=3)=C(8,3)(0.565)3(0.435)5=0.1573. μ=np=4.52 and σ=√np(1−p)=1.402.
Question 17. Binomial Distribution
For 14 independent visitors at a state park in Coastal Plains during a randomized pilot period, each meets the trail-use duration criterion with probability 0.228. Find P(X=4) and give the binomial mean and SD.
Answer: D
X~Bin(14,0.228). P(X=4)=C(14,4)(0.228)4(0.772)10=0.2034. μ=np=3.192 and σ=√np(1−p)=1.57.
Question 18. Binomial Distribution
For 8 independent learners at a digital learning platform in Westview during a service-improvement study, each meets the lesson completion criterion with probability 0.381. Find P(X=6) and give the binomial mean and SD.
Answer: A
X~Bin(8,0.381). P(X=6)=C(8,6)(0.381)6(0.619)2=0.0328. μ=np=3.048 and σ=√np(1−p)=1.374.
Question 19. Binomial Distribution
For 17 independent visitors at a county library in New England network during a weekday operations study, each meets the weekly program attendance criterion with probability 0.204. Find P(X=12) and give the binomial mean and SD.
Answer: B
X~Bin(17,0.204). P(X=12)=C(17,12)(0.204)12(0.796)5=0.00001027.00001027. μ=np=3.468 and σ=√np(1−p)=1.661.
Question 20. Binomial Distribution
For 14 independent enrolled learners at a community college in South Harbor during a service-improvement study, each meets the course completion criterion with probability 0.362. Find P(X=4) and give the binomial mean and SD.
Answer: A
X~Bin(14,0.362). P(X=4)=C(14,4)(0.362)4(0.638)10=0.1921. μ=np=5.068 and σ=√np(1−p)=1.798.
Question 21. Binomial Distribution
For 9 independent participants at a city recreation department in Metro East during a monthly quality review, each meets the program satisfaction criterion with probability 0.408. Find P(X=2) and give the binomial mean and SD.
Answer: D
X~Bin(9,0.408). P(X=2)=C(9,2)(0.408)2(0.592)7=0.1527. μ=np=3.672 and σ=√np(1−p)=1.474.
Question 22. Binomial Distribution
For 13 independent parts at a regional manufacturer in Sunbelt district during a fall 2026 audit, each meets the part diameter criterion with probability 0.469. Find P(X=10) and give the binomial mean and SD.
Answer: D
X~Bin(13,0.469). P(X=10)=C(13,10)(0.469)10(0.531)3=0.022. μ=np=6.097 and σ=√np(1−p)=1.799.
Question 23. Binomial Distribution
For 22 independent appointments at a university advising center in Cedar Grove during a weekday operations study, each meets the appointment wait time criterion with probability 0.688. Find P(X=8) and give the binomial mean and SD.
Answer: B
X~Bin(22,0.688). P(X=8)=C(22,8)(0.688)8(0.312)14=0.0013. μ=np=15.136 and σ=√np(1−p)=2.173.
Question 24. Binomial Distribution
For 23 independent visitors at a county library in Sunbelt district during a randomized pilot period, each meets the weekly program attendance criterion with probability 0.469. Find P(X=13) and give the binomial mean and SD.
Answer: A
X~Bin(23,0.469). P(X=13)=C(23,13)(0.469)13(0.531)10=0.1083. μ=np=10.787 and σ=√np(1−p)=2.393.
Question 25. Binomial Distribution
For 16 independent bus trips at a city transit agency in Central County during a multiweek validation study, each meets the on-time arrival criterion with probability 0.658. Find P(X=11) and give the binomial mean and SD.
Answer: C
X~Bin(16,0.658). P(X=11)=C(16,11)(0.658)11(0.342)5=0.2046. μ=np=10.528 and σ=√np(1−p)=1.898.
Question 26. Binomial Distribution
For 20 independent travelers at a regional airport authority in Great Lakes during a six-week field trial, each meets the security wait time criterion with probability 0.272. Find P(X=16) and give the binomial mean and SD.
Answer: A
X~Bin(20,0.272). P(X=16)=C(20,16)(0.272)16(0.728)4=0.00000122.00000122. μ=np=5.44 and σ=√np(1−p)=1.99.
Question 27. Binomial Distribution
For 15 independent participants at a city recreation department in New England network during a six-week field trial, each meets the program satisfaction criterion with probability 0.333. Find P(X=7) and give the binomial mean and SD.
Answer: C
X~Bin(15,0.333). P(X=7)=C(15,7)(0.333)7(0.667)8=0.1145. μ=np=4.995 and σ=√np(1−p)=1.825.
Question 28. Binomial Distribution
For 21 independent bus trips at a city transit agency in Pacific Northwest during a semester-long cohort study, each meets the on-time arrival criterion with probability 0.656. Find P(X=8) and give the binomial mean and SD.
Answer: D
X~Bin(21,0.656). P(X=8)=C(21,8)(0.656)8(0.344)13=0.0066. μ=np=13.776 and σ=√np(1−p)=2.177.
Question 29. Binomial Distribution
For 17 independent visitors at a state park in Sunbelt district during a community outreach cycle, each meets the trail-use duration criterion with probability 0.552. Find P(X=11) and give the binomial mean and SD.
Answer: D
X~Bin(17,0.552). P(X=11)=C(17,11)(0.552)11(0.448)6=0.1451. μ=np=9.384 and σ=√np(1−p)=2.05.
Question 30. Binomial Distribution
For 10 independent bus trips at a city transit agency in Capital Region during a quarterly performance study, each meets the on-time arrival criterion with probability 0.467. Find P(X=4) and give the binomial mean and SD.
Answer: D
X~Bin(10,0.467). P(X=4)=C(10,4)(0.467)4(0.533)6=0.229. μ=np=4.67 and σ=√np(1−p)=1.578.
Question 31. Binomial Distribution
For 16 independent ballots at a county election office in New England network during a weekday operations study, each meets the ballot-processing time criterion with probability 0.438. Find P(X=13) and give the binomial mean and SD.
Answer: A
X~Bin(16,0.438). P(X=13)=C(16,13)(0.438)13(0.562)3=0.0022. μ=np=7.008 and σ=√np(1−p)=1.985.
Question 32. Binomial Distribution
For 10 independent parts at a regional manufacturer in North Valley during a randomized pilot period, each meets the part diameter criterion with probability 0.317. Find P(X=4) and give the binomial mean and SD.
Answer: C
X~Bin(10,0.317). P(X=4)=C(10,4)(0.317)4(0.683)6=0.2153. μ=np=3.17 and σ=√np(1−p)=1.471.
Question 33. Binomial Distribution
For 11 independent plots at a farm cooperative in Coastal Plains during a baseline measurement week, each meets the crop yield criterion with probability 0.423. Find P(X=8) and give the binomial mean and SD.
Answer: D
X~Bin(11,0.423). P(X=8)=C(11,8)(0.423)8(0.577)3=0.0325. μ=np=4.653 and σ=√np(1−p)=1.639.
Question 34. Binomial Distribution
For 21 independent customers at a grocery cooperative in Mountain Region during a semester-long cohort study, each meets the checkout time criterion with probability 0.493. Find P(X=15) and give the binomial mean and SD.
Answer: A
X~Bin(21,0.493). P(X=15)=C(21,15)(0.493)15(0.507)6=0.0228. μ=np=10.353 and σ=√np(1−p)=2.291.
Question 35. Binomial Distribution
For 8 independent appointments at a university advising center in Coastal Plains during a baseline measurement week, each meets the appointment wait time criterion with probability 0.254. Find P(X=2) and give the binomial mean and SD.
Answer: C
X~Bin(8,0.254). P(X=2)=C(8,2)(0.254)2(0.746)6=0.3114. μ=np=2.032 and σ=√np(1−p)=1.231.
Question 36. Binomial Distribution
For 23 independent visitors at a county library in South Harbor during a summer implementation review, each meets the weekly program attendance criterion with probability 0.257. Find P(X=21) and give the binomial mean and SD.
Answer: B
X~Bin(23,0.257). P(X=21)=C(23,21)(0.257)21(0.743)2=5.671 × 10−11. μ=np=5.911 and σ=√np(1−p)=2.096.
Binomial Distribution: Conditions, Formula, Mean, and SD: 14 free-response questions
For each response, show the statistical reasoning, use the scenario’s language, and state only the conclusion supported by the design or probability model.
FRQ set 1: Binomial Distribution
Scenario. For 10 independent calls at a municipal emergency dispatch center in Lakeside district during a quarterly performance study, each meets the response time criterion with probability 0.553. Find P(X=7) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(10,0.553). P(X=7)=C(10,7)(0.553)7(0.447)3=0.1695. μ=np=5.53 and σ=√np(1−p)=1.572.
FRQ set 2: Binomial Distribution
Scenario. For 12 independent installations at a solar installer in Mountain Region during a two-month observation window, each meets the daily energy output criterion with probability 0.508. Find P(X=4) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(12,0.508). P(X=4)=C(12,4)(0.508)4(0.492)8=0.1132. μ=np=6.096 and σ=√np(1−p)=1.732.
FRQ set 3: Binomial Distribution
Scenario. For 9 independent patients at a regional hospital in Pacific Northwest during a weekday operations study, each meets the appointment completion criterion with probability 0.531. Find P(X=7) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(9,0.531). P(X=7)=C(9,7)(0.531)7(0.469)2=0.0943. μ=np=4.779 and σ=√np(1−p)=1.497.
FRQ set 4: Binomial Distribution
Scenario. For 23 independent visitors at a state park in Coastal Plains during a monthly quality review, each meets the trail-use duration criterion with probability 0.658. Find P(X=15) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(23,0.658). P(X=15)=C(23,15)(0.658)15(0.342)8=0.1722. μ=np=15.134 and σ=√np(1−p)=2.275.
FRQ set 5: Binomial Distribution
Scenario. For 14 independent students at a school district in Great Lakes during a fall 2026 audit, each meets the lunch-program participation criterion with probability 0.503. Find P(X=9) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(14,0.503). P(X=9)=C(14,9)(0.503)9(0.497)5=0.1251. μ=np=7.042 and σ=√np(1−p)=1.871.
FRQ set 6: Binomial Distribution
Scenario. For 21 independent customers at a community bank in North Valley during a weekday operations study, each meets the mobile-deposit adoption criterion with probability 0.212. Find P(X=5) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(21,0.212). P(X=5)=C(21,5)(0.212)5(0.788)16=0.1926. μ=np=4.452 and σ=√np(1−p)=1.873.
FRQ set 7: Binomial Distribution
Scenario. For 24 independent students at a public high school in Mountain Region during a two-month observation window, each meets the algebra benchmark completion criterion with probability 0.607. Find P(X=10) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(24,0.607). P(X=10)=C(24,10)(0.607)10(0.393)14=0.0279. μ=np=14.568 and σ=√np(1−p)=2.393.
FRQ set 8: Binomial Distribution
Scenario. For 10 independent calls at a municipal emergency dispatch center in Pine Ridge during a monthly quality review, each meets the response time criterion with probability 0.239. Find P(X=7) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(10,0.239). P(X=7)=C(10,7)(0.239)7(0.761)3=0.0024. μ=np=2.39 and σ=√np(1−p)=1.349.
FRQ set 9: Binomial Distribution
Scenario. For 15 independent learners at a digital learning platform in Midwest consortium during a winter readiness review, each meets the lesson completion criterion with probability 0.465. Find P(X=2) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(15,0.465). P(X=2)=C(15,2)(0.465)2(0.535)13=0.0067. μ=np=6.975 and σ=√np(1−p)=1.932.
FRQ set 10: Binomial Distribution
Scenario. For 10 independent parts at a regional manufacturer in Westview during a community outreach cycle, each meets the part diameter criterion with probability 0.517. Find P(X=5) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(10,0.517). P(X=5)=C(10,5)(0.517)5(0.483)5=0.2447. μ=np=5.17 and σ=√np(1−p)=1.58.
FRQ set 11: Binomial Distribution
Scenario. For 21 independent parts at a regional manufacturer in South Harbor during a pre-exam training cycle, each meets the part diameter criterion with probability 0.258. Find P(X=5) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(21,0.258). P(X=5)=C(21,5)(0.258)5(0.742)16=0.1964. μ=np=5.418 and σ=√np(1−p)=2.005.
FRQ set 12: Binomial Distribution
Scenario. For 23 independent residents at a public health department in Desert County during a quarterly performance study, each meets the vaccination appointment completion criterion with probability 0.445. Find P(X=3) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(23,0.445). P(X=3)=C(23,3)(0.445)3(0.555)20=0.0012. μ=np=10.235 and σ=√np(1−p)=2.383.
FRQ set 13: Binomial Distribution
Scenario. For 24 independent visitors at a county library in Coastal Plains during a monthly quality review, each meets the weekly program attendance criterion with probability 0.555. Find P(X=22) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(24,0.555). P(X=22)=C(24,22)(0.555)22(0.445)2=0.0001. μ=np=13.32 and σ=√np(1−p)=2.435.
FRQ set 14: Binomial Distribution
Scenario. For 8 independent samples at a food safety laboratory in Cedar Grove during a spring 2027 pilot, each meets the sample concentration criterion with probability 0.297. Find P(X=5) and give the binomial mean and SD.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
X~Bin(8,0.297). P(X=5)=C(8,5)(0.297)5(0.703)3=0.045. μ=np=2.376 and σ=√np(1−p)=1.292.
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