Question

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

Answer 1

Answer:

A score of 88.72 corresponds to the 80th percentile.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$0.84 = \frac{X - 82}{8}$

$X - 82 = 8*0.84$

$X = 88.72$

A score of 88.72 corresponds to the 80th percentile.