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OpenStudy (anonymous):
If a normal distribution has a mean of 132 and a standard deviation of 20, what is the value that has a z-score of 2.4? A. 180 B. 168 C. 144 D. 204
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OpenStudy (agent0smith):
\[\Large z = \frac{ x - \mu }{ \sigma }\] sigma is the S.D., mu is the mean.
OpenStudy (anonymous):
132-20 ________ 2.4 ?
OpenStudy (agent0smith):
Plugging in the numbers z=2.4, mean = 132, SD = 20\[\Large 2.4 = \frac{ x - 132 }{ 20}\]
OpenStudy (anonymous):
A!
OpenStudy (agent0smith):
Correct!
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