Question

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. -1.6; not significant B.-17.6; significant C. -1.6, significant D. 1.6; not significant

Answer 1

Answer:

A. -1.6; not significant

Step-by-step explanation:

The z-score of a data set that is normally distributed with a mean of $\bar x$ and a standard deviation of $\sigma$, is given by:

$z=\frac{x-\bar x}{\sigma}$.

From the question, the test score is: $x=48.4$, the mean is $\bar x=66$, and the standard deviation is $\sigma =11$.

We just have to plug these values into the above formula to obtain:

$z=\frac{48.4-66}{11}$.

This simplifies to:  $z=\frac{-17.6}{11}$.

$z=-1.6$.

We can see that the z-score falls within two standard deviations of the mean.

Since $-2\le-1.6\le2$ the value is not significant.

The correct answer is A. -1.6; not significant