Question

the class scores of a history test have a normal distribution with a mean \\( \mu = 79 \\) and a standard deviation \\( \sigma = 7 \\). if opal’s test score was 72, which expression would she write to find the z - score of her test score?\\( \bigcirc \\ z = \frac{72 - 79}{7} \\)\\( \bigcirc \\ z = \frac{72 - 7}{7} \\)\\( \bigcirc \\ z = \frac{79 - 72}{7} \\)\\( \bigcirc \\ z = \frac{7 - 79}{7} \\)

Explanation:

Step1: Recall z - score formula

The formula for the z - score of a value \(x\) in a normal distribution with mean \(\mu\) and standard deviation \(\sigma\) is \(z=\frac{x - \mu}{\sigma}\).

Step2: Identify values

Here, \(x = 72\) (Opal's test score), \(\mu=79\) (mean of the class scores), and \(\sigma = 7\) (standard deviation of the class scores).

Step3: Substitute values into formula

Substitute \(x = 72\), \(\mu = 79\), and \(\sigma=7\) into the z - score formula. We get \(z=\frac{72 - 79}{7}\).

Answer:

\(z=\frac{72 - 79}{7}\) (the first option)