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If test scores are approximately normally distributed with mean 82 and standard deviation 8, what score would correspond to the 80th percentile

User Ilkar
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4 votes

Answer:

A score of 88.72 corresponds to the 80th percentile.

Explanation:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the zscore of a measure X is given by:


Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mean 82 and standard deviation 8

This means that
\mu = 82, \sigma = 8

What score would correspond to the 80th percentile

This is X when Z has a pvalue of 0.8, so X when Z = 0.84.


Z = (X - \mu)/(\sigma)


0.84 = (X - 82)/(8)


X - 82 = 8*0.84


X = 88.72

A score of 88.72 corresponds to the 80th percentile.

User MathBunny
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