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? z = StartFraction 72 minus 79 Over 7 EndFraction z = StartFraction 72 minus 7 Over 7 EndFraction z = StartFraction 79 minus 72 Over 7 EndFraction z = StartFraction 7 minus 79 Over 7 EndFraction.

Answer 1

The value of z score for the history class test of Opal’s test score is negative one (-1).

What is normally distributed data?

Normally distributed data is the distribution of probability which is symmetric about the mean.

The mean of the data is the average value of the given data. The standard deviation of the data is the half of the difference of the highest value and mean of the data set.

The z score for the normal distributed data can be given as,

$Z=\dfrac{X-\mu}{\sigma}$

The mean value of the scores of history class is,

$\mu=79$

The standard deviation value of the scores of history class is,

$\sigma=7$

Here, the test score of the Opal’s is 72. Thus the z score for her test score can be given as,

$z=\dfrac{72-79}{7}\\z=-1$

Hence, the value of z score for the history class test of Opal’s test score is negative one (-1).

Learn more about the normally distributed data here;

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