Question

Composite scores on the ACT are normally distributed with a mean of 21 and a standard deviation of 5. find the Z score for a student that gets a 31

Answer 1

`Z = 2`

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.

Mean of 21 and a standard deviation of 5.

This means that
`\mu = 21, \sigma = 5`

Find the Z score for a student that gets a 31

This is Z when X = 31. So

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

`Z = (31 - 21)/(5)`

`Z = 2`