Question

on a standardized exam, the scores are normally distributed with a mean of 31 and a standard deviation of 4. find the z - score of a person who scored 33 on the exam.

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step2: Identify values

We are given that $\mu = 31$, $\sigma=4$, and $x = 33$.

Step3: Substitute values into formula

$z=\frac{33 - 31}{4}$

Step4: Calculate the z - score

$z=\frac{2}{4}=0.5$

Answer:

$0.5$