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A set of test scores is normally distributed with a mean of 145 and a standard deviation of 11. Seif received a score of 167. What was his percentile score


97.5%


84%


95%


63.5%

Verified Answer :

LammettHash

Since \(145+2\times11=167\), which means the corresponding z-score for Seif's test score is \(z=2\). The empirical rule asserts that for any normal distribution, approximately 95% of it lies within two standard deviations of the mean, i.e. \(\mathbb P(|Z|<2)=0.95\), which leaves 5% outside of that range, and specifically 2.5% to either side.

This means

\(\mathbb P(Z<2)=\mathbb P(Z<-2)+\mathbb P(|Z|<2)=0.025+0.95=0.975\)

In other words, a test score of 167, which corresponds to a z-score of 2, marks the 97.5th percentile.

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