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The class scores of a history test have a normal distribution with a mean [tex]\mu=79[/tex] and a standard deviation [tex]\sigma=7[/tex]. If Opal's test score was 72, which expression would she write to find the [tex]z[/tex]-score of her test score?

A. [tex]z = \frac{72 - 79}{7}[/tex]
B. [tex]z = \frac{72 - 7}{7}[/tex]
C. [tex]z = \frac{79 - 72}{7}[/tex]
D. [tex]z = \frac{7 - 79}{7}[/tex]


Sagot :

To find the z-score of Opal's test score given a mean and standard deviation, we use the formula for the z-score:

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

where:
- [tex]\(x\)[/tex] is the test score,
- [tex]\(\mu\)[/tex] is the mean score,
- [tex]\(\sigma\)[/tex] is the standard deviation.

Given:
- [tex]\( x = 72 \)[/tex]
- [tex]\( \mu = 79 \)[/tex]
- [tex]\( \sigma = 7 \)[/tex]

We substitute these values into the z-score formula:

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

Thus, Opal would write the expression [tex]\( z = \frac{72 - 79}{7} \)[/tex] to find the z-score of her test score.

### Choice Explanation
- [tex]\( z = \frac{72 - 79}{7} \)[/tex] correctly follows the formula for finding the z-score using the values for x, μ, and σ.
- [tex]\( z = \frac{72 - 7}{7} \)[/tex] is incorrect because it subtracts the standard deviation from the score rather than the mean.
- [tex]\( z = \frac{79 - 72}{7} \)[/tex] is incorrect because it swaps the test score and the mean in the numerator.
- [tex]\( Z = \frac{7 - 79}{7} \)[/tex] is incorrect because it places the standard deviation in the numerator and subtracts the mean from it, which does not follow the z-score formula.

Therefore, the correct expression Opal should use is:

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