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Sindrei [870]
3 years ago
12

Scores on an English test are normally distributed with a mean of 31.5 and a standard deviation of 7.3. Find the score that sepa

rates the top 59% from the bottom 41%. Round your answer to the nearest tenth.
Mathematics
1 answer:
vladimir1956 [14]3 years ago
6 0

If X is a random variable representing test scores, then the 41st percentile is the score x such that P(X\le x)=0.41. Transform X to the random variable Z that follows the standard normal distribution via X=31.5+7.3Z:

P(X\le x)=P\left(\dfrac{X-31.5}{7.3}\le\dfrac{x-31.5}{7.3}\right)=P\left(Z\le\dfrac{x-31.5}{7.3}\right)=0.41

The z-score for the 41st percentile is about -0.2275, so

\dfrac{x-31.5}{7.3}\approx-0.2275\implies\boxed{x\approx29.8}

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