Question

If a student scored $82$ points on a test where the mean score was $78.2$ and the standard deviation was $6.7$. the student's $z$ score is .

Answer 1

Find the z-score:

The raw score is, $x=82$.

The mean score is, $\mu =78.2$.

The standard deviation is, $\sigma =6.7$

The formula of the $z-$score is, $z=\frac{x-\mu }{\sigma }$.

Substitute $82$ for $x$ and $78.2$ for $\mu$ in the equation $z=\frac{x-\mu }{\sigma }$.

$$\begin{gathered} z=\frac{82-78.2}{6.7} \\ \therefore z=0.567 \end{gathered}$$

Hence, the required $z$-score is $0.567.$