Question

the scores of an eighth - grade math test have a normal distribution with a mean $mu = 83$ and a standard deviation $sigma = 5$. if dins test score was 92, which expression would she write to find the $z$-score of her test score?
$z=\frac{92 - 83}{93}$
$z=\frac{83 - 92}{5}$
$z=\frac{92 - 83}{5}$
$z=\frac{5 - 83}{92}$

Explanation:

Step1: Recall z - score formula

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

Step2: Identify values

We are given that $x = 92$ (Din's test score), $\mu=83$ (mean of the test scores), and $\sigma = 5$ (standard deviation of the test scores).

Step3: Substitute values into formula

Substitute $x = 92$, $\mu = 83$, and $\sigma=5$ into the z - score formula: $z=\frac{92 - 83}{5}$.

Answer:

$z=\frac{92 - 83}{5}$ (corresponding to the third option in the multiple - choice list)