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?.

Answer 1

Using the normal distribution, we got that expression for the z-score of her test score is

$z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1$

What is Normal Probability Distribution?

it's a type of probability distribution which is  symmetric about the mean, it gives data near the mean are more frequent than data far from the mean.

Given that mean =$\mu=79$

and standard deviation =$\sigma = 7$

X=73

we know that formula of z-score is

$z=\frac{X-\mu}{\sigma}$

Hence Z-score can be calculated as

$z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1$

Using the normal distribution, we got that expression for the z-score of her test score is

$z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1$

To learn more about the normal distribution visit : brainly.com/question/24663213