Question

Given a mean of 9 and a standard deviation of 2.8, what is the z-score of the value 11 rounded to the nearest tenth?

–0.9

–0.7

0.9

0.7

Answer 1

The answer is 0.7. Since the formula of the z-score is the difference of the score (x) and the mean divided by the standard deviation (SD)...

z= [ (x-mean)/SD ]

You simply substitute the values given.

z = (11-9)/2.8
z=2/2.8
z=0.714

Then round it to the nearest tenth...

z- = 0.7