If the mean IQ is 100 and the standard deviation of the IQ scores is 11,then an IQ of 122 will have a z-score of:
A) 1.0
B) 0.0
C) 2.0
D) 2.5
E) -2.0

Question

Quizplus · Accepted Answer

The z-score is calculated by subtracting the mean from the individual score and then dividing by the standard deviation: $z = \frac{(X - \mu)}{\sigma}$. Here, $X = 122$, $\mu = 100$, and $\sigma = 11$, so $z = \frac{(122 - 100)}{11} = \frac{22}{11} = 2.0$.

If the Mean IQ Is 100 and the Standard Deviation