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IQ Scores and the Normal Distribution

A public GRE Psychology question converting a Wechsler IQ score to a percentile with a z-score, and naming the empirical-rule errors that produce the wrong percentiles.

Question

An adult intelligence test is standardized to a normal distribution with a mean of 100 and a standard deviation of 15. A test taker earns a score of 130. Approximately what percentage of the standardization sample scored below this person?

  1. a.About 68%
  2. b.About 84%
  3. c.About 95%
  4. d.About 98%
  5. e.About 99.9%

Correct answer

D. About 98%

Full reasoning

  1. 1Convert the raw score to a z-score: z = (X - mean) / SD = (130 - 100) / 15 = 2.00, so the score is exactly two standard deviations above the mean.
  2. 2For a normal distribution the cumulative proportion below z = 2.00 is about 0.9772, which is roughly the 98th percentile.
  3. 3Cross-check with the empirical rule: about 95% of the distribution lies within two standard deviations of the mean, leaving about 5% split evenly between the two tails, so about 2.5% lies above 130 and about 97.5% below. The two methods agree to within rounding.
  4. 4Note which quantity the question asks for. Percentile is a one-tailed cumulative proportion, whereas the 68% and 95% figures of the empirical rule are two-tailed central intervals; confusing the two is what produces the wrong options here.

Why each choice is right or wrong

Choice A

68% is the proportion of a normal distribution lying within one standard deviation on both sides of the mean. Reporting a central interval as if it were a percentile is the error here; it also ignores that 130 is two standard deviations out, not one.

Choice B

84% is the percentile for a score one standard deviation above the mean, which would be 115. This comes from dividing the 30-point gap by the wrong figure — treating 30 as one standard deviation rather than two.

Choice C

95% is the proportion lying within two standard deviations on both sides of the mean. Converting it to a percentile requires splitting the remaining 5% between the two tails, so only about 2.5% sits above the score, not 5%.

Choice D

Correct. z = (130 - 100) / 15 = 2.00, and the proportion of a normal distribution below z = 2 is about 0.977, so roughly 98% scored lower and about 2% scored higher.

Choice E

99.9% corresponds to about three standard deviations above the mean, which on this scale would be a score of 145. Using a standard deviation of 10 instead of 15 turns 130 into z = 3 and produces exactly this answer.

Related formula

z-score

z=X−μσz = \frac{X - \mu}{\sigma}

Assumptions: The percentile reading requires the distribution to be normal, which standardized intelligence scales are constructed to be.

Empirical rule for a normal distribution

about 68% within 1 SD of the mean, about 95% within 2 SD, about 99.7% within 3 SD

Assumptions: These are two-tailed central proportions; converting one to a percentile means adding half of the remainder below the interval.

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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. Psychology 2e, Section 7.5: Measures of Intelligence — OpenStax. Accessed 2026-08-15. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
  2. 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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