the class scores of a history test have a normal distribution with a mean $mu = 79$ and a standard deviation…

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

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

Answer

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

Here, $x = 72$ (Opal's score), $\mu=79$ (mean score), and $\sigma = 7$ (standard deviation).

Step3: Substitute values into formula

Substitute the values into the z - score formula: $z=\frac{72 - 79}{7}$.

Answer:

$z=\frac{72 - 79}{7}$ (First option)