Question

John finds that there is a scholarship available to all personals scoring in the top 5% on the ACT test. If the mean score on the ACT is 21 with a standard deviation of 4.7, what score does he need in order to qualify for the scholarship

Answer 1

He needs a score of 28.7315 to qualify for the scholarship

Explanation:

Normal Probability Distribution:

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

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The mean score on the ACT is 21 with a standard deviation of 4.7

This means that
`\mu = 21, \sigma = 4.7`

What score does he need in order to qualify for the scholarship?

The top 5%, so the 100 - 5 = 95th percentile, which is X when Z has a p-value of 0.95, so X when Z = 1.645.

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

`1.645 = (X - 21)/(4.7)`

`X - 21 = 4.7*1.645`

`X = 28.7315`

He needs a score of 28.7315 to qualify for the scholarship