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 calculating 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: Identify values

Here, \(x = 72\), \(\mu=79\), and \(\sigma = 7\).

Step3: Substitute values into formula

Substituting these values into the z - score formula, we get \(z=\frac{72 - 79}{7}\).

Answer:

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