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The class scores of a history test have a normal distribution with a mean mc010-1.jpg = 79 and a standard deviation mc010-2.jpg = 7. If Opal’s test score was 72, which expression would she write to find the z-score of her test score?

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Given:
mean = 79 
standard deviation = 7

Opal's test score = 72

Finding the z score of her test score:
 
z = X - μ / σ

z = (72 - 79) / 7
z = -7/7
z = -1

This z score means that 72 is -1 standard deviations away from the mean history test. 

Using the normal distribution, it is found that the expression for the z-score of her test score is given by:

[tex]Z = \frac{72 - 79}{7} = -1[/tex]

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

  • The mean is of 79, hence [tex]\mu = 79[/tex].
  • The standard deviation is of 7, hence [tex]\sigma = 7[/tex].
  • Her score is of 72, hence [tex]X = 72[/tex].

Then, her z-score is given by:

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

[tex]Z = \frac{72 - 79}{7} = -1[/tex]

You can learn more about the normal distribution at https://brainly.com/question/24663213

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