StatsDrillFree guides

College practice · Worked answers

Z score practice problems

A z score tells you how far a value is from a mean in standard-deviation units. Start with the sign, then calculate the distance. These college statistics problems separate a single observation’s z score from the standard error used for a sample mean.

One formula, two directions

z = (x − μ)/σ   and   x = μ + zσ

Subtract the mean first, then divide by a positive standard deviation. A negative result means the value is below the mean, zero means it is at the mean, and a positive result means it is above. The units cancel, so a z score has no physical units.

You can standardize a value even when its distribution is not normal. You need a normal-distribution assumption to turn that z score into a standard-normal percentile. A z score by itself is not a percentile.

Which denominator belongs here?

Observation versus sampling distribution
You are standardizingDenominatorMeaning
A single observation xσSpread of individual observations
A sample mean x̄, with σ knownσ/√nSpread of sample means under independent sampling
An observation using sample summariess, with mean x̄A descriptive sample standardized score

Using s for a sample description does not automatically make the result a hypothesis-test statistic. A one-sample t test for a mean uses s/√n and its own assumptions.

Check the direction before reaching for a table

If x is below μ, a positive answer is a warning that you reversed the subtraction. If x is exactly two standard deviations above μ, your answer should be +2 regardless of the original units. For a standard-normal table, find out whether it reports area to the left, area between zero and z, or a right-tail area.

Work through two examples

Standardize an exam score

  1. A hypothetical score is 74, with mean 80 and standard deviation 4.
  2. Compute the difference: 74 − 80 = −6. Divide by 4 to get z = −1.5.
  3. The score is 1.5 standard deviations below the mean. This does not mean the score is at the 1.5th percentile.

Recover the original value

  1. A measurement has z = 1.2 in a distribution with μ = 30 mm and σ = 5 mm.
  2. Use x = μ + zσ = 30 + 1.2 × 5.
  3. The measurement is 36 mm. Substituting back gives (36 − 30)/5 = 1.2.

Try it yourself

Choose an answer, then check your reasoning. You can retry any question or open its solution.

1. x = 90, μ = 75, σ = 10. What is z?
Show solution 1

1.5. z = (90 − 75)/10 = 1.5. The observation is above the mean, so the sign is positive.

Why the other choices are wrong:

  • 15: 15 is the raw difference. Divide it by σ = 10.
  • −1.5: The sign is backward. 90 is above the mean, so z is positive. Compute x − μ, not μ − x.
  • 9: 90/10 skips subtracting the mean first.
2. x = 42, μ = 50, σ = 4. What is z?
Show solution 2

−2. z = (42 − 50)/4 = −2. The value is two standard deviations below the mean.

Why the other choices are wrong:

  • 2: Right size, wrong sign. 42 is below the mean, so z is negative.
  • −8: −8 is x − μ before dividing by σ = 4.
  • 10.5: 42/4 skips subtracting the mean.
3. μ = 100, σ = 12, z = 0.5. What is x?
Show solution 3

106. x = μ + zσ = 100 + 0.5 × 12 = 106.

Why the other choices are wrong:

  • 100.5: That adds z itself. Multiply by σ first: 0.5 × 12 = 6.
  • 94: That subtracts. A positive z means the value is above the mean.
  • 112: That is one full standard deviation above. z = 0.5 is half of one.
4. Which statement follows from z = −1?
Show solution 4

The value is one standard deviation below its mean. A negative z describes position relative to the mean. The original value can still be positive, and a percentile requires distribution information.

Why the other choices are wrong:

  • The value is negative: z is negative, but x itself can be positive, like a score of 70 when the mean is 80.
  • The value is at the first percentile: A z score isn’t a percentile. Under a normal model, z = −1 is near the 16th percentile.
  • The value is 1 unit below its mean: It is one standard deviation below, not one unit. If σ = 5, that is 5 units.
5. For a sample mean with known σ = 15 and n = 25 independent observations, what denominator is used?
Show solution 5

3. The standard error is σ/√n = 15/5 = 3. This is for the sample mean, not one observation.

Why the other choices are wrong:

  • 15: σ describes single observations. Sample means vary less, so use σ/√n.
  • 25: 25 is the sample size, not a measure of spread.
  • 0.6: 15/25 divides by n. Use √n = 5.
6. A population has μ = 70 and σ = 8. A sample of n = 16 independent values has x̄ = 74. What is z for the sample mean?
Show solution 6

2. The standard error is 8/√16 = 2. Then z = (74 − 70)/2 = 2.

Why the other choices are wrong:

  • 0.5: That divides by σ = 8, which is for one observation. A sample mean uses σ/√n = 2.
  • 4: 4 is x̄ − μ before dividing by the standard error.
  • 0.25: That divides by n = 16. The standard error is σ/√n = 8/4 = 2.
7. Under a normal model, about what percent of values lie below z = 1?
Show solution 7

84%. About 50% lies below the mean and about 34% lies between z = 0 and z = 1, so roughly 84% lies below z = 1.

Why the other choices are wrong:

  • 34%: 34% is the area between the mean and z = 1. Add the 50% that lies below the mean.
  • 68%: 68% is the area between z = −1 and z = 1.
  • 16%: 16% is the area above z = 1, not below it.
8. Student A scores 82 on a test with mean 74 and SD 4. Student B scores 90 on a test with mean 80 and SD 8. Who did better relative to their own class?
Show solution 8

Student A: z = 2 versus z = 1.25. A: (82 − 74)/4 = 2. B: (90 − 80)/8 = 1.25. A sits further above the class mean. This compares position within each class, not skill across tests.

Why the other choices are wrong:

  • Student B, because 90 is the higher score: Raw scores on different tests aren’t comparable. Standardize each one first.
  • Student B, because 10 points above the mean beats 8: Points above the mean ignore the spread. On B’s test, 10 points is only 1.25 standard deviations.
  • Neither; both scored above the mean: Both are above the mean, but by different amounts in standard-deviation units.

Answer key

Open all answers
  1. 1.5. z = (90 − 75)/10 = 1.5. The observation is above the mean, so the sign is positive.
  2. −2. z = (42 − 50)/4 = −2. The value is two standard deviations below the mean.
  3. 106. x = μ + zσ = 100 + 0.5 × 12 = 106.
  4. The value is one standard deviation below its mean. A negative z describes position relative to the mean. The original value can still be positive, and a percentile requires distribution information.
  5. 3. The standard error is σ/√n = 15/5 = 3. This is for the sample mean, not one observation.
  6. 2. The standard error is 8/√16 = 2. Then z = (74 − 70)/2 = 2.
  7. 84%. About 50% lies below the mean and about 34% lies between z = 0 and z = 1, so roughly 84% lies below z = 1.
  8. Student A: z = 2 versus z = 1.25. A: (82 − 74)/4 = 2. B: (90 − 80)/8 = 1.25. A sits further above the class mean. This compares position within each class, not skill across tests.

Common questions

Does calculating z require a normal distribution?

No. Standardizing is arithmetic. Using the standard normal curve to calculate probabilities requires an appropriate normal model.

Is a negative z score bad?

It only means below the reference mean. Whether that is desirable depends on what is being measured.

Can I compare scores from different tests?

Z scores compare relative position within each reference distribution. They do not show that the tests measure the same skill.

Keep practicing

Try 3 days free

3-day free trial, no card needed.

Sources and fact checks

Definitions and methods checked against these textbook references. Practice scenarios and worked numbers were written for this guide.