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 the z - score formula

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

Step2: Identify the values of \(x\), \(\mu\), and \(\sigma\)

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

Step3: Substitute the values into the formula

Substitute \(x = 33\), \(\mu = 31\), and \(\sigma=4\) into the z - score formula:
\(z=\frac{33 - 31}{4}\)

Step4: Simplify the expression

First, calculate the numerator: \(33-31 = 2\). Then, divide by the denominator: \(\frac{2}{4}=0.5\).

Answer:

The z - score is \(0.5\)