- Population Mean ( $\mu$ ): 36 mph
- Population Standard Deviation ( $\sigma$ ): 2 mph
- Sample Size ( $n$ ): 20
According to the Central Limit Theorem, the distribution of the sample mean $(\bar{X})$ will also be normal.
- The mean of the sample mean ( $\mu_{\bar{X}}$ ) is the same as the population mean:
$$
\mu_{\bar{X}}=\mu=36
$$
- The standard deviation of the sample mean, also known as the standard error ( $\sigma_{\bar{X}}$ ), is calculated as:
$$
\sigma_X=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{20}}
$$
We need to find the probability $P(35<\bar{X}<38)$. To do this, we convert the boundary values of 35 and 38 into $z$-scores using the formula:
$$
Z=\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}
$$
$$
Z=\frac{35-36}{2 / \sqrt{20}}=\frac{-1}{2 / \sqrt{20}}=\frac{-\sqrt{20}}{2}=-\sqrt{\frac{20}{4}}=-\sqrt{5}
$$
$$
Z=\frac{38-36}{2 / \sqrt{20}}=\frac{2}{2 / \sqrt{20}}=\sqrt{20}
$$
So, the original probability is equivalent to finding the probability for the standard normal variable Z:
$$
P(35<\bar{X}<38)=P(-\sqrt{5}<Z<\sqrt{20})
$$
The probability of Z falling between two values, $a$ and $b$, can be expressed using the Cumulative Distribution Function (CDF) as:
$$
P(a<Z<b)=P(Z \leq b)-P(Z \leq a)=F_Z(b)-F_Z(a)
$$
Substituting our calculated z-scores:
$$
P(-\sqrt{5}<Z<\sqrt{20})=F_Z(\sqrt{20})-F_Z(-\sqrt{5})
$$
This matches option D.
