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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?.

Respuesta :

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

[tex]z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1[/tex]

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 =[tex]\mu=79[/tex]

and standard deviation =[tex]\sigma = 7[/tex]

X=73

we know that formula of z-score is

[tex]z=\frac{X-\mu}{\sigma}[/tex]

Hence Z-score can be calculated as

[tex]z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1[/tex]

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

[tex]z= \frac{72-29}{7} \\\\z=-7/7\\\\z=-1[/tex]

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

Answer:

72-29/7

Step-by-step explanation: i just took the test it's A

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