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

Answer:

He needs a score of 28.7315 to qualify for the scholarship

Step-by-step 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:

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

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 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.

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

$1.645 = \frac{X - 21}{4.7}$

$X - 21 = 4.7*1.645$

$X = 28.7315$

He needs a score of 28.7315 to qualify for the scholarship