Question

question
on a standardized exam, the scores are normally distributed with a mean of 300 and a standard deviation of 20. find the z - score of a person who scored 330 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 raw score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

Step2: Substitute the given values into the formula

Given \(x = 330\), \(\mu=300\), and \(\sigma = 20\). Then \(z=\frac{330 - 300}{20}\).

Step3: Calculate the numerator

\(330-300 = 30\). So \(z=\frac{30}{20}\).

Step4: Simplify the fraction

\(\frac{30}{20}=\frac{3}{2}=1.5\)

Answer:

\(1.5\)