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

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