Question

Scores on a recent national statistics exam were normally distributed with a mean of 82.2 and a standard deviation of 5.If the top 2.5% of test scores receive merit awards, what is the lowest score eligible for an award

Answer 1

The lowest score eligible for an award is 92.

Explanation:

When the distribution is normal, we use 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.

In this question:

`\mu = 82.2, \sigma = 5`

If the top 2.5% of test scores receive merit awards, what is the lowest score eligible for an award

The lowest score is the 100 - 2.5 = 97.5th percentile, which is X when Z has a pvalue of 0.975. So X when Z = 1.96. Then

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

`1.96 = (X - 82.2)/(5)`

`X - 82.2 = 5*1.96`

`X = 92`

The lowest score eligible for an award is 92.