AP Statistics Binomial and Geometric Distributions Worksheet Solutions (PDF)
This binomial geometric answers for AP Statistics covers binomial and geometric distributions (Common Core AP STAT 4.10, AP STAT 4.11, AP STAT 4.12, HSS-MD.A.3, HSS-CP.B.9), with a free PDF to print.
AP Statistics · Common Core State Standards: AP STAT 4.10, AP STAT 4.11, AP STAT 4.12, HSS-MD.A.3, HSS-CP.B.9
Binomial geometric answers: exercises 1 to 10
Exercise 1 – Check BINS
(a) Yes. (b) No: the number of trials is not fixed (it is geometric-like, and cards are drawn without replacement). (c) Yes: the sample is far less than 10% of the output. (d) No: the trials are not independent.
(a) yes; (b) no; (c) yes; (d) no.
Exercise 2 – Binomial formula
\(P(X=3)=\frac{4!}{3!\,1!}(0.7)^{3}(0.3)^{1}=4(0.343)(0.3)=0.4116\).
0.4116.
Exercise 3 – binompdf
X is binomial with n = 8 and p = 0.25.
\(P(X=2)=\text{binompdf}(8,0.25,2)=0.3115\). \(P(X=5)=0.0231\).
0.3115 and 0.0231.
Exercise 4 – binomcdf
\(P(X\le 3)=\text{binomcdf}(8,0.25,3)=0.8862\).
0.8862.
Exercise 5 – At least
\(P(X\ge 4)=1-P(X\le 3)=1-0.8862=0.1138\).
0.1138.
Exercise 6 – Between
\(P(2\le X\le 4)=P(X\le 4)-P(X\le 1)=0.9727-0.3671=0.6056\).
0.6056.
Exercise 7 – Mean and standard deviation
\(\mu=np=25(0.8)=20\). \(\sigma=\sqrt{25(0.8)(0.2)}=\sqrt{4}=2\).
Mean 20, standard deviation 2.
Exercise 8 – Interpret the mean
The mean is a long-run average.
Over many sets of 25 attempts, she makes an average of 20 shots per set.
Exercise 9 – Shape
With p = 0.5 the distribution is symmetric and mound-shaped, centered at 5.
Symmetric; with p = 0.1 it is skewed right with its peak at 1.
Exercise 10 – Unusual result
\(P(X=10)=0.7^{10}=0.0282\).
0.0282, which is less than 5%, so making all ten is unusual.
Exercises 11 and up
Exercise 11 – 10 percent condition
The sample (4) is more than 10% of the population (30), and the probability changes after each draw.
No: the 10% condition fails (4 is about 13% of 30).
Exercise 12 – Geometric setting
(a) Yes. (b) No: fixed number of trials (binomial). (c) No: the probability changes, so trials are not independent.
(a) geometric; (b) binomial; (c) neither.
Exercise 13 – Geometric probability
\(P(X=4)=(0.8)^{3}(0.2)=0.1024\).
0.1024.
Exercise 14 – More than n trials
More than 5 boxes means the first 5 are failures: \(0.8^{5}=0.3277\).
0.3277.
Exercise 15 – Geometric cumulative
\(P(X\le 3)=1-0.8^{3}=1-0.512=0.488\).
0.488.
Exercise 16 – Expected trials
\(\mu=\frac{1}{0.125}=8\).
You expect about 8 tests until the first pass; in the long run, it takes an average of 8 trials.
Exercise 17 – Geometric sd
\(\sigma=\frac{\sqrt{1-p}}{p}=\frac{\sqrt{0.8}}{0.2}\approx 4.472\).
About 4.47 boxes.
Exercise 18 – Exactly and at least
\(P(X=0)=0.95^{20}=0.3585\). \(P(X\ge 1)=1-0.3585=0.6415\).
0.3585 and 0.6415.
Exercise 19 – Compare binomial and geometric
(a) \(\text{binompdf}(5,0.6,3)=0.3456\). (b) \(0.4(0.6)=0.24\).
0.3456 and 0.24.
Exercise 20 – Multi-step
\(\mu=12(0.04)=0.48\). \(\sigma=\sqrt{12(0.04)(0.96)}=0.679\).
\(P(X > 2)=1-\text{binomcdf}(12,0.04,2)=0.0107\).
Mean 0.48; sd 0.679; P = 0.0107.
Key facts behind this binomial geometric answers
- Binomial conditions: Binary, Independent, fixed Number, Same probability.
- Binomial: mean = np, standard deviation = sqrt(np(1-p)); use binompdf and binomcdf.
- At least k successes: 1 – binomcdf(n, p, k – 1).
- Geometric: trials until the first success; P(X = n) = (1-p)^(n-1) p; mean 1/p.
- Without replacement, treat trials as independent only if n is at most 10% of the population.
How to use this binomial geometric answers
Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.
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Download the binomial geometric answers as a PDF with the button below and keep it with your notes.
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AP Statistics Binomial and Geometric Distributions Worksheet Solutions
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