Question

Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points. Rachel scored 78 points on an exam that had a mean score of 75 points and a standard deviation of 3.7 points. Find Keenan's and Rachel’s z-scores, to the nearest hundredth.

Answer 1

Keenan's z-score was of 0.61.

Rachel's z-score was of 0.81.

Explanation:

Z-score:

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.

Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points.

This means that
`X = 80, \mu = 77, \sigma = 4.9`

So

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

`Z = (80 - 77)/(4.9)`

`Z = 0.61`

Keenan's z-score was of 0.61.

Rachel scored 78 points on an exam that had a mean score of 75 points and a standard deviation of 3.7 points.

This means that
`X = 78, \mu = 75, \sigma = 3.7`. So

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

`Z = (75 - 78)/(3.7)`

`Z = 0.81`

Rachel's z-score was of 0.81.