IQs are known to be normally distributed with mean 100 and standard deviation 15. In a random sample of 32 people, find the probability that the average IQ is between 96 and 105.
(x-100)/15 then came up with 96<-value<-105 >>>>>>> -1<-Z<0.33 which was 0.62930 - 0.15866 = 0.4707
however when I answered as (0.4707)^32 my answer did not make sense. Can someone please explain what I should do ? Thanks
On
If $X_1,\ldots,X_{32}$ are independent and each is distributed as $\operatorname N(100,15^2),$ then $$ \overline X = \frac{X_1+\cdots+X_{32}}{32} $$ is distributed as $\operatorname N\left( 100, \dfrac{15^2}{32} \right) = \operatorname N\left( 100, \left(\dfrac{15}{\sqrt{32}}\right)^2 \right) = \operatorname N\left( 100, \left( \dfrac{15}{4\sqrt2} \right)^2 \right).$
So you need \begin{align} & \Pr\left( 96 < \overline X < 105 \right) \\[8pt] = {} & \Pr\left( \frac{96-100}{15/\sqrt{32}} < \frac{\overline X - 100}{15/\sqrt{32}} < \frac{105-100}{14/\sqrt{32}} \right) \\[8pt] = {} & \Pr\left( \frac{96-100}{15/\sqrt{32}} < Z < \frac{105-100}{14/\sqrt{32}} \right). \end{align}
$96$ is only $\frac 4{15}$ standard deviations low, so the Z-score is about $-0.2667$, not $-1$. Once you correct that, you will get the probability that one specific IQ is in that range, which is not what you need. You are supposed to look up the standard deviation of the sample mean, which is a factor $\sqrt {n}$ less than the standard deviation of the distribution, then use that in your z-score table.