AP Statistics · Lesson 3

Normal models and sampling distributions

Know Know it

A normal distribution N(μ,σ)N(\mu, \sigma) is symmetric and bell-shaped. A z-score z=x−μσz = \frac{x - \mu}{\sigma} counts standard deviations from the mean, so it lets you compare values on different scales.

By the empirical rule, about 68, 95 and 99.7 percent of values fall within 1, 2 and 3 standard deviations of the mean. For other cutoffs, find the area with a calculator or the standard normal table. For a percentile, go the other way: find the z with that area to its left, then x=μ+zσx = \mu + z\sigma.

A sampling distribution shows how a statistic varies over all random samples of size n. For a sample proportion p^\hat{p}: the mean is μp^=p\mu_{\hat{p}} = p; the standard deviation is σp^=p(1−p)n\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} when the sample is less than 10 percent of the population; and the shape is approximately normal when np≥10np \geq 10 and n(1−p)≥10n(1-p) \geq 10.

For a sample mean xˉ\bar{x}: the mean is μxˉ=μ\mu_{\bar{x}} = \mu and the standard deviation is σxˉ=σn\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}. The shape is normal if the population is normal, and approximately normal for large n even when the population is not, by the central limit theorem (n of at least 30 is a common rule of thumb).

Both statistics are unbiased: their sampling distributions are centered at the parameter. Larger samples give less spread, but quadrupling n only halves the standard deviation.

Calculator On the exam

A graphing calculator with statistical capabilities is expected on both sections of the exam, and the exam provides a reference booklet of formulas and tables, including a standard normal table. Use the calculator for normal and binomial areas and for percentiles, but write the distribution, its mean and standard deviation, and the boundary next to each answer so a reader can follow your work.

Worked example Easy

Lena scored 82 on a fictional history test with mean 74 and standard deviation 5. Marco scored 88 on a different fictional test with mean 80 and standard deviation 8.

Who did better compared with the other students who took the same test?

  1. Marco, because 88 is a higher score than 82
  2. They did equally well, since each scored 8 points above the mean
  3. Lena, with a z-score of 1.6 compared with Marco's 1.0Answer
  4. Marco, with a z-score of 1.6 compared with Lena's 1.0

How to solve it

  1. Lena: z=82−745=1.6z = \frac{82 - 74}{5} = 1.6.
  2. Marco: z=88−808=1.0z = \frac{88 - 80}{8} = 1.0.
  3. Lena is 1.6 standard deviations above her test's mean, more than Marco's 1.0, so she did better relative to her group.

Why each choice is right or wrong

  • A. Incorrect. Raw scores of 88 and 82 come from tests with different means and spreads, so they cannot be compared directly.
  • B. Incorrect. Both are 8 points above the mean, but 8 points is 1.6 standard deviations on Lena's test and only 1 on Marco's.
  • C. Correct. Lena's z is 82 - 74/5 = 1.6 and Marco's is 88 - 80/8 = 1.0, so Lena is further above her test's mean.
  • D. Incorrect. This swaps the z-scores: Marco's is 88 - 80/8 = 1.0, and Lena's is the 1.6.

Trap Divide by n inside the square root

The spread of p^\hat{p} is p(1−p)n\sqrt{\frac{p(1-p)}{n}}, not p(1−p)\sqrt{p(1-p)}. Leaving out n uses the spread of one individual's yes-or-no answer, which is far too large and makes every sample result look ordinary. In the same way, the spread of xˉ\bar{x} is σn\frac{\sigma}{\sqrt{n}}, not σ\sigma.

Explain Explain it

Task verbs: calculate, justify

For a probability about p^\hat{p}, work in four steps: name the sampling distribution and its mean, compute its standard deviation, check that a normal model fits (large counts, and the 10 percent condition for the standard deviation), then find the area.

Justify the normal shape with numbers, not words alone: write npnp and n(1−p)n(1-p) and say each is at least 10.

Worked example Medium

A fictional online store says that 12 percent of its orders are returned. Suppose that is true, and take a random sample of 400 of the store's many thousands of orders.

What is the approximate probability that more than 15 percent of the sampled orders are returned?

  1. 0.4632
  2. 0.0324Answer
  3. 0.9676
  4. 0.0464

How to solve it

  1. Center: μp^=0.12\mu_{\hat{p}} = 0.12. Spread: σp^=0.12(0.88)400≈0.01625\sigma_{\hat{p}} = \sqrt{\frac{0.12(0.88)}{400}} \approx 0.01625 (400 is far less than 10 percent of all orders).
  2. Shape: 400(0.12)=48400(0.12) = 48 and 400(0.88)=352400(0.88) = 352 are both at least 10, so the shape is approximately normal.
  3. z=0.15−0.120.01625≈1.846z = \frac{0.15 - 0.12}{0.01625} \approx 1.846, and P(Z>1.846)≈0.0324P(Z > 1.846) \approx 0.0324. A table rounds z to 1.85 and gives about 0.032.

Why each choice is right or wrong

  • A. Incorrect. 0.4632 uses 0.12(0.88)≈0.325\sqrt{0.12(0.88)} \approx 0.325 as the standard deviation, which leaves out the sample size of 400.
  • B. Correct. σp^=0.12(0.88)400≈0.01625\sigma_{\hat{p}} = \sqrt{\frac{0.12(0.88)}{400}} \approx 0.01625, so z=0.15−0.120.01625≈1.846z = \frac{0.15 - 0.12}{0.01625} \approx 1.846, and the area to the right is about 0.0324.
  • C. Incorrect. 0.9676 is the area to the left of 0.15, the chance of 15 percent or fewer returns.
  • D. Incorrect. 0.0464 puts 0.15 into the standard deviation formula, but the spread of p^\hat{p} uses the true proportion 0.12.

Justify Your turn

A fictional store's true return rate is p = 0.12. Justify using a normal model for the sampling distribution of the sample proportion of returns in random samples of 400 orders. Then say whether the same reasoning works for random samples of 40 orders, and why.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Checks large counts for n = 400 with numbers
  • Says a normal model is not justified for n = 40, with the number that fails

Model answer

For samples of 400, the expected counts are 400(0.12) = 48 returns and 400(0.88) = 352 non-returns. Both are at least 10, so the sampling distribution of the sample proportion is approximately normal. For samples of 40, the expected number of returns is 40(0.12) = 4.8, which is less than 10, so the sampling distribution would be skewed right and a normal model is not justified.

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