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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

Respuesta :

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

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