Question

Suppose that the mean and standard deviation of the scores on a statistics exam are 78 and 6.11, respectively, and are approximately normally distributed. Calculate the proportion of scores above 74.

Answer 1

0.7422 = 74.22% of scores are above 74.

Explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

`\mu = 78, \sigma = 6.11`

Calculate the proportion of scores above 74.

This is 1 subtracted by the pvalue of Z when X = 74. So

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

`Z = (74 - 78)/(6.11)`

`Z = -0.65`

`Z = -0.65` has a pvalue of 0.2578

So 1-0.2578 = 0.7422 = 74.22% of scores are above 74.