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Z-Scores on a Normal Distribution

A public statistics question that shows how to standardize a raw score into a z-score using the mean and standard deviation.

Question

Scores on a placement exam are normally distributed with mean 150150 and standard deviation 88. A student scores 162162. What is the student's z-score?

  1. a.−1.5-1.5
  2. b.0.670.67
  3. c.1.51.5
  4. d.1212

Correct answer

C. 1.51.5

Full reasoning

  1. 1Find the deviation of the score from the mean: 162−150=12162 - 150 = 12.
  2. 2Divide the deviation by the standard deviation: z=12÷8=1.5z = 12 \div 8 = 1.5.
  3. 3The positive sign confirms the score lies 1.51.5 standard deviations above the mean of 150150.

Why each choice is right or wrong

Choice A

This reverses the subtraction, computing (150−162)/8(150 - 162)/8, which would describe a score below the mean rather than above it.

Choice B

This inverts the final division, computing 8/128/12 instead of dividing the deviation 1212 by the standard deviation 88.

Choice C

Correct. The deviation from the mean is 162−150=12162 - 150 = 12, and 12÷8=1.512 \div 8 = 1.5 standard deviations above the mean.

Choice D

This stops at the raw deviation 162−150162 - 150 and never divides by the standard deviation, so it is not a standardized score.

Related formula

Z-score of a value

z=x−μσz = \frac{x - \mu}{\sigma}

Assumptions: The standard deviation is positive so the division is defined.

Related topic and practice

Descriptive StatisticsProbability and CountingAddition Rule of ProbabilityCombinationsMedian of a Data SetTwo draws without replacementCreate a free account to practice

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Sources

Authored for this public explainer registry. It has no database question ID and is not copied from a protected or official exam bank.

  1. Introductory Statistics 2e, Section 6.1: The Standard Normal Distribution — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY 4.0; attribute and avoid verbatim reuse beyond short cited references.
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Sources and corrections

Sources last checked 2026-08-07

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