A test score of 48.4 on a test having a mean of 66 and a standard deviation of 11. Find the z_ score corresponding to the given value and use the z score to determine whether the value is
Significant. Consider a score to be significant if its z score is less than -2.00 or greater than
2.00. Round the z score to the nearest tenth if necessary.
A)-17.6; significant
B)-1.6; significant
C)-1.6; not significant
D)1.6; not significant

Question

Quizplus · Accepted Answer

The z-score is calculated using the formula $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Plugging in the given values, $z = \frac{(48.4 - 66)}{11} = \frac{-17.6}{11} = -1.6$. A z-score of -1.6 does not meet the criteria for significance, as it is not less than -2.00 or greater than 2.00.

A Test Score of 48