Question

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30 What score is necessary to reach the 75th percentile? Select the closest answer

Answer 1

A score of 150.25 is necessary to reach the 75th 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.

A set of test scores is normally distributed with a mean of 130 and a standard deviation of 30.

This means that
`\mu = 130, \sigma = 30`

What score is necessary to reach the 75th percentile?

This is X when Z has a pvalue of 0.75, so X when Z = 0.675.

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

`0.675 = (X - 130)/(30)`

`X - 130 = 0.675*30`

`X = 150.25`

A score of 150.25 is necessary to reach the 75th percentile.