The level of cholesterol in the blood for all men aged $20to34$follows a Normal distribution with mean $\mu M=188$milligrams per deciliter (mg/dl) and standard deviation $\sigma M=41$mg/dl. For $14-$year-old boys, blood cholesterol levels follow a Normal distribution with mean $\mu B=170$mg/dl and standard deviation $\sigma B=30$mg/dl. Suppose we select independent SRSs of $25$men aged $20to34$and $36$boys aged $14$and calculate the sample mean cholesterol levels $\overline{x}Mand\overline{x}B$.

a. What is the shape of the sampling distribution of B${\overline{x}}_M-{\overline{x}}_B$? Why?

b. Find the mean of the sampling distribution.

c. Calculate and interpret the standard deviation of the sampling distribution.

It is given that there are two distributions that is,

$$\begin{gathered} DistributionM:Normalwith{\mu }_M=188,{\sigma }_M=41 \\ DistributionB:Normalwith{\mu }_B=170,{\sigma }_B=30 \end{gathered}$$

Now, if M and B are normally distributed then there difference$M-B$is also normally distributed.

And we know that the properties of normal are:

$$\begin{gathered} {\mu }_{aX+bY}=a{\mu }_x+b{\mu }_y \\ {\sigma }_{aX+bY}=\sqrt{a^2{{\sigma }^2}_x+b^2{{\sigma }^2}_y} \end{gathered}$$

Then we obtain:

$$\begin{gathered} {\mu }_{M-B}={\mu }_M-{\mu }_B \\ =188-170 \\ =18 \\ {\sigma }_{M-B}=\sqrt{{{\sigma }^2}_M+{{\sigma }^2}_B} \\ =\sqrt{{41}^2+{30}^2} \\ =\sqrt{2581} \\ =50.8035 \end{gathered}$$

Thus, we conclude that the distribution of $M-B$is normal with mean and standard deviation as:

$$\begin{gathered} {\mu }_{M-B}=18 \\ {\sigma }_{M-B}=50.8035 \end{gathered}$$

$$\begin{gathered} {\mu }_1=188 \\ {\mu }_2=170 \\ {n}_1=25 \\ {n}_2=36 \\ {\sigma }_1=41 \\ {\sigma }_2=30 \end{gathered}$$

The mean of the sampling distribution of the difference in sample means is the difference in the population means. This implies:

$$\begin{gathered} {\mu }_{X_1-X_2}={\mu }_1-{\mu }_2 \\ =188-170 \\ =18 \end{gathered}$$

Thus, we conclude that the mean of the sampling mean is $18$mg/dl.

Since the sample $25$men aged $20to34$is less than $10\%$of all men aged $20to34$and since the sample of $36$boys aged fourteen is less than $10\%$of all boys aged fourteen. Thus, the standard deviation is as follows:

$$\begin{gathered} {\sigma }_{x_1-x_2}=\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}} \\ =\sqrt{\frac{{41}^2}{25}+\frac{{30}^2}{36}} \\ =9.6042 \end{gathered}$$

Thus, we conclude that the difference in the sample means are expected to be vary by$9.6042$ mg/dl from the mean difference of $18$mg/dl.