Atest 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 minus2.00 or greater than 2.00. round the z-score to the nearest tenth if necessary.

The z-score is -1.6 and it is not significant .

Step-by-step explanation:

Given : Test score = 48.4

            Mean = 66

            Standard deviation = 11

To Find :Find the z-score corresponding to the given value and use the z-score to determine whether the value is significant.

Solution:

Formula :

Substitute the values

So, the z-score is -1.6

Now we are given that Consider a score to be significant if its z-score is less than -2.00 or greater than 2.00.

Since -1.6>-2 and -1.6<2

So, It is not significant

Hence the z-score is -1.6 and it is not significant .