Z-scores and normal distribution practice problems

By Jude Wallis · Published July 28, 2026

This set covers z-score conversions in both directions, reading areas from a standard normal table, finding percentiles, and applying the empirical rule. Try each problem yourself first, then check your arithmetic against the step-by-step solution.

AP Statistics: Unit 2 (topics 1.9 Comparisons of the Distributions for One Quantitative Variable, 2.11 The Normal Distribution). In the Fall 2026 AP Statistics course, calculating a z-score with population parameters is Unit 1, topic 1.9.D (Comparisons of the Distributions for One Quantitative Variable). Reading areas, percentiles, and empirical-rule percentages from a normal distribution is Unit 2, topic 2.11 (The Normal Distribution). These problems combine both.

How to use this set and what it builds

These 8 problems build fluency with the two-way link between raw values and z-scores, and with turning z-scores into areas and percentiles using a standard normal table. Every context gives you a mean $\mu$ (mu) and a standard deviation $\sigma$ (sigma), and you practice both directions: raw value to z-score to percentile, and percentile back to a raw value. The last problems mix the empirical rule with exact table lookups, the way an AP free-response question does.

Work each one with pencil and paper before opening the solution, and keep your z-scores to two decimal places so they match the table. If you want the method first, see how to find a z-score and the empirical rule. To check a single calculation, use the z-score calculator or the normal distribution calculator, and read areas from the z-table.

Problem 1

An ultralight tent brand finds its tent weights are approximately normal with a mean of $\mu = 1.8$ kg and a standard deviation of $\sigma = 0.3$ kg, where $\mu$ (mu) is the mean and $\sigma$ (sigma) is the standard deviation. One model weighs 2.4 kg. Find its z-score and state how many standard deviations it sits from the mean.

Show the worked solution

Write the z-score formula, which counts standard deviations from the mean: $z = \frac{x - \mu}{\sigma}$ , with $x = 2.4$ , $\mu = 1.8$ , and $\sigma = 0.3$ .
Subtract the mean from the value: $2.4 - 1.8 = 0.6$ kg.
Divide by the standard deviation: $z = \frac{0.6}{0.3} = 2.00$ .
Read the sign and size: a positive z of 2.00 means the tent is 2 standard deviations above the mean weight.

$z = 2.00$ , so the tent weighs 2 standard deviations above the mean.

Problem 2

A bakery's sourdough loaves rise to a height that is approximately normal with a mean of $\mu = 12$ cm and a standard deviation of $\sigma = 1.6$ cm. One loaf rises to 14 cm. What percent of loaves rise less than 14 cm?

Show the worked solution

Standardize the value with $z = \frac{x - \mu}{\sigma}$ , using $x = 14$ , $\mu = 12$ , and $\sigma = 1.6$ .
Subtract the mean: $14 - 12 = 2$ cm.
Divide by the standard deviation: $z = \frac{2}{1.6} = 1.25$ .
Look up $z = 1.25$ in the standard normal table: the area to its left is 0.8944.
Convert the area to a percent: $0.8944 \times 100 = 89.44\%$ , the percent of loaves below 14 cm.

About 89.44% of loaves rise less than 14 cm, roughly the 89th percentile.

Problem 3

Finish times in a trail-running club are approximately normal with a mean of $\mu = 54$ minutes and a standard deviation of $\sigma = 6$ minutes. The club gives a medal to the fastest 10% of runners. What finish time is the cutoff for a medal?

Show the worked solution

The fastest runners have the lowest times, so the fastest 10% sit in the bottom 10% of the distribution, meaning an area of 0.10 to the left.
Find the z-score with 0.10 of the area to its left. The closest table area is 0.1003 at $z = -1.28$ , so use $z = -1.28$ .
Solve the z-score formula for the raw value: $x = \mu + z\sigma$ .
Substitute the numbers: $x = 54 + (-1.28)(6)$ .
Multiply, then add: $(-1.28)(6) = -7.68$ , so $x = 54 - 7.68 = 46.32$ minutes.

About 46.32 minutes; runners who finish faster than roughly 46.3 minutes earn a medal.

Problem 4

A dairy fills yogurt cups to a weight that is approximately normal with a mean of $\mu = 150$ g and a standard deviation of $\sigma = 4$ g. What percent of cups weigh between 145 g and 156 g?

Show the worked solution

Standardize both boundaries with $z = \frac{x - \mu}{\sigma}$ .
Lower boundary: $z = \frac{145 - 150}{4} = \frac{-5}{4} = -1.25$ .
Upper boundary: $z = \frac{156 - 150}{4} = \frac{6}{4} = 1.50$ .
Read areas to the left from the table: 0.1056 for $z = -1.25$ and 0.9332 for $z = 1.50$ .
Subtract the smaller area from the larger to get the area between: $0.9332 - 0.1056 = 0.8276$ .
Convert to a percent: $0.8276 \times 100 = 82.76\%$ .

About 82.76% of cups weigh between 145 g and 156 g.

Problem 5

Migrating geese at a refuge have body masses that are approximately normal with a mean of $\mu = 3.6$ kg and a standard deviation of $\sigma = 0.4$ kg. Using the empirical rule, find (a) the percent of geese between 2.8 kg and 4.4 kg, and (b) the percent heavier than 4.0 kg.

Show the worked solution

Mark the standard-deviation cutoffs from the mean. One step: $3.6 \pm 0.4$ gives 3.2 and 4.0. Two steps: $3.6 \pm 0.8$ gives 2.8 and 4.4.
(a) The interval 2.8 to 4.4 runs from 2 standard deviations below the mean to 2 above, since $\frac{2.8 - 3.6}{0.4} = -2$ and $\frac{4.4 - 3.6}{0.4} = 2$ .
By the empirical rule, about 95% of values fall within 2 standard deviations of the mean, so the answer to (a) is about 95%.
(b) The value 4.0 kg is 1 standard deviation above the mean, since $\frac{4.0 - 3.6}{0.4} = 1$ .
The empirical rule puts about 68% within 1 standard deviation, leaving $100\% - 68\% = 32\%$ in the two tails, split evenly: $\frac{32\%}{2} = 16\%$ above the upper cutoff.

(a) about 95%; (b) about 16%.

Problem 6

A job applicant takes two standardized assessments, each approximately normal. On the coding test, applicant scores have a mean of $\mu = 70$ and a standard deviation of $\sigma = 8$ , and she scores 82. On the design test, scores have a mean of $\mu = 24$ and a standard deviation of $\sigma = 5$ , and she scores 32. On which test did she perform better relative to other applicants?

Show the worked solution

Relative position is measured by the z-score $z = \frac{x - \mu}{\sigma}$ , so standardize each score.
Coding test: $z = \frac{82 - 70}{8} = \frac{12}{8} = 1.50$ .
Design test: $z = \frac{32 - 24}{5} = \frac{8}{5} = 1.60$ .
Compare the z-scores: 1.60 is larger than 1.50, so the design score sits farther above its mean.
For context, the table gives 0.9332 to the left of $z = 1.50$ (about the 93rd percentile) and 0.9452 to the left of $z = 1.60$ (about the 95th percentile).

She did better on the design test ( $z = 1.60$ , about the 95th percentile) than on the coding test ( $z = 1.50$ , about the 93rd percentile).

Problem 7

A climbing gym rates members with a standardized grade index that is approximately normal with a mean of $\mu = 21$ and a standard deviation of $\sigma = 4$ . The gym labels the top 5% of members as advanced. What grade index is the cutoff to be labeled advanced?

Show the worked solution

The top 5% lie above the 95th percentile, so the cutoff has an area of 0.95 to its left.
Find the z-score for 0.95. The table shows 0.9495 at $z = 1.64$ and 0.9505 at $z = 1.65$ , so 0.95 falls halfway between, giving $z \approx 1.645$ .
Solve the z-score formula for the raw value: $x = \mu + z\sigma$ .
Substitute the numbers: $x = 21 + (1.645)(4)$ .
Multiply, then add: $(1.645)(4) = 6.58$ , so $x = 21 + 6.58 = 27.58$ .

About 27.58; a grade index above roughly 27.58 puts a member in the top 5%.

Problem 8

A drone's battery flight time is approximately normal with a mean of $\mu = 15$ minutes. A reviewer reports that the 84th percentile of flight time is 18 minutes. (a) Estimate the standard deviation $\sigma$ . (b) What percent of flights last more than 20 minutes? (c) Use the empirical rule to find the percent of flights between 12 and 21 minutes.

Show the worked solution

(a) The 84th percentile has an area of 0.84 to its left. The table area closest to 0.84 is 0.8413 at $z = 1.00$ , so 18 minutes is about 1 standard deviation above the mean.
Solve $x = \mu + z\sigma$ for the standard deviation: $18 = 15 + (1.00)\sigma$ , so $\sigma = 18 - 15 = 3$ minutes.
(b) Standardize 20 minutes with $\sigma = 3$ : $z = \frac{20 - 15}{3} = \frac{5}{3} = 1.67$ , rounded to two decimals.
The table gives 0.9525 to the left of $z = 1.67$ , so the area above is $1 - 0.9525 = 0.0475$ , which is $4.75\%$ .
(c) Locate the boundaries in standard deviations: $\frac{12 - 15}{3} = -1$ and $\frac{21 - 15}{3} = 2$ , so the interval runs from 1 standard deviation below the mean to 2 above.
By the empirical rule, the mean to 1 standard deviation below holds about 34% and the mean to 2 standard deviations above holds about 47.5%, giving $34\% + 47.5\% = 81.5\%$ .

(a) $\sigma \approx 3$ minutes; (b) about 4.75%; (c) about 81.5%.