AP Statistics Sampling Distributions and the Central Limit Theorem Worksheet Solutions (PDF)

AP Statistics math worksheet solutions: Sampling Distributions and the Central Limit Theorem – Maths PDF

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This sampling distributions answers for AP Statistics covers sampling distributions and the central limit theorem (Common Core AP STAT 5.1, AP STAT 5.2, AP STAT 5.3, AP STAT 5.4, AP STAT 5.5, AP STAT 5.6, AP STAT 5.7, HSS-IC.A.1, HSS-IC.A.2), with a free PDF to print.

Each solution lists the mean, the standard deviation, and the conditions, then standardizes and finds the area with the normal table or normalcdf. Probabilities are rounded to four decimals.

AP Statistics · Common Core State Standards: AP STAT 5.1, AP STAT 5.2, AP STAT 5.3, AP STAT 5.4, AP STAT 5.5, AP STAT 5.6, AP STAT 5.7, HSS-IC.A.1, HSS-IC.A.2

Sampling distributions answers: exercises 1 to 10

Exercise 1 – Parameter or statistic

The parameter describes the population; the statistic describes the sample.

Parameter: p = 0.56; statistic: p-hat = 0.53.

Exercise 2 – Sampling variability

300 simulated sample proportions: centered at 0.40, approximately normal.

The distribution is roughly normal and centered near 0.40, with most values between about 0.26 and 0.54.

Approximately normal, centered at 0.40, spread about 0.07. A single sample proportion will usually fall within about 0.14 of p.

Exercise 3 – Mean and SD of p-hat

$\mu_{\hat{p}}=0.30$. $\sigma_{\hat{p}}=\sqrt{\frac{0.30(0.70)}{150}}=0.0374$.

Mean 0.30; standard deviation 0.0374.

Exercise 4 – Check conditions

$np=60(0.08)=4.8$ is less than 10.

No: np = 4.8 < 10, so the sampling distribution is not approximately normal (the 10% condition holds since 60 < 90).

Exercise 5 – Probability for p-hat

Sampling distribution of p-hat for n = 50: mean 0.40, standard deviation 0.069.

$z=\frac{0.50-0.40}{0.0693}=1.44$. $P(Z > 1.44)=0.0749$.

About 0.0749.

Exercise 6 – Effect of sample size

$n=50:\ 0.0693$. $n=200:\ 0.0346$. The ratio is $2.00$.

Quadrupling n cuts the standard deviation in half.

Exercise 7 – Unbiased estimator

The center of the sampling distribution should equal the parameter.

Biased (it overestimates by 2). A larger sample reduces variability but does not remove the bias.

Exercise 8 – Mean and SD of x-bar

$\mu_{\overline{x}}=1200$. $\sigma_{\overline{x}}=\frac{300}{\sqrt{36}}=50$.

Mean $1,200; standard deviation $50.

Exercise 9 – Probability for x-bar

$\sigma_{\overline{x}}=\frac{3}{\sqrt{9}}=1$. $z=\frac{71-70}{1}=1$. $P(Z > 1)=0.1587$.

0.1587.

Exercise 10 – Individual versus mean

$z=\frac{71-70}{3}=0.33$. $P(Z > 0.33)=0.3707$.

0.3707, much larger than 0.1587; sample means vary less than individuals.

Exercises 11 and up

Exercise 11 – Central limit theorem

$\mu=5,\ \sigma_{\overline{x}}=\frac{5}{\sqrt{100}}=0.5$. Since n = 100 is large, the CLT applies.

Approximately normal with mean 5 and standard deviation 0.5 minutes.

Exercise 12 – Is n large enough?

The CLT needs large n; for strong skew, n of at least 30 is a common guideline.

n = 8: no. n = 50: yes, approximately normal.

Exercise 13 – Difference of proportions

Mean: $0.40-0.30=0.10$.

SD: $\sqrt{\frac{0.4(0.6)}{100}+\frac{0.3(0.7)}{120}}=\sqrt{0.0024+0.00175}=0.0644$.

Mean 0.10; standard deviation 0.0644.

Exercise 14 – Probability of a difference

$z=\frac{0-0.10}{0.0644}=-1.55$. $P(Z < -1.55)=0.0606$.

About 0.0606: a rare outcome.

Exercise 15 – Difference of means

Mean: $250-240=10$. SD: $\sqrt{\frac{30^{2}}{40}+\frac{25^{2}}{50}}=\sqrt{22.5+12.5}=\sqrt{35}=5.916$.

Mean 10 ms; standard deviation 5.916 ms.

Exercise 16 – Sample size needed

$\sqrt{\frac{0.5(0.5)}{n}}\le 0.025$, so $n\ge\frac{0.25}{0.000625}=400$.

n = 400.

Exercise 17 – Interpret the standard deviation

It measures typical sample-to-sample variation.

Sample proportions from random samples of 400 voters typically differ from the true proportion by about 0.025 (2.5 percentage points).

Exercise 18 – Within two standard deviations

$\sigma_{\hat{p}}=\sqrt{\frac{0.6(0.4)}{100}}=0.0490$.

$0.6\pm 2(0.0490)=0.6\pm 0.098$.

About 0.502 to 0.698.

Exercise 19 – Multi-step

$\sigma_{\overline{x}}=\frac{0.4}{4}=0.1$. $z=\frac{16-16.1}{0.1}=-1$. $P(Z < -1)=0.1587$.

0.1587.

Exercise 20 – Poll accuracy

Sampling distribution of p-hat, n = 400.

$\sigma_{\hat{p}}=\sqrt{\frac{0.25(0.75)}{400}}=0.0217$. $z=\frac{0.22-0.25}{0.0217}=-1.39$. $P(Z < -1.39)=0.0823$.

About 0.0823.

Key facts behind this sampling distributions answers

Key facts

A sampling distribution describes how a statistic varies from sample to sample.
p-hat: mean p, sd sqrt(p(1-p)/n); approximately normal if np >= 10 and n(1-p) >= 10.
x-bar: mean mu, sd sigma/sqrt(n); normal if the population is normal or n is large (CLT).
Differences of independent statistics: means subtract, variances add.
Larger n reduces variability but never corrects bias; the 10% condition is needed for independence.

How to use this sampling distributions 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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