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22 January, 21:46

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

+1
Answers (1)
  1. 23 January, 00:35
    0
    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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