91Ó°ÊÓ

It is given that there are two distributions that is,

$$\begin{gathered} DistributionM:Normalwith{\mathrm{μ}}_M=188,{\mathrm{σ}}_M=41 \\ DistributionB:Normalwith{\mathrm{μ}}_B=170,{\mathrm{σ}}_B=30 \end{gathered}$$

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:

$$\begin{gathered} {\mathrm{μ}}_{aX+bY}=a{\mathrm{μ}}_x+b{\mathrm{μ}}_y \\ {\mathrm{σ}}_{aX+bY}=\sqrt{a^2{{\mathrm{σ}}^2}_x+b^2{{\mathrm{σ}}^2}_y} \end{gathered}$$

Then we obtain:

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

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

$$\begin{gathered} {\mathrm{μ}}_{M-B}=18 \\ {\mathrm{σ}}_{M-B}=50.8035 \end{gathered}$$

$$\begin{gathered} {\mathrm{μ}}_1=188 \\ {\mathrm{μ}}_2=170 \\ {n}_1=25 \\ {n}_2=36 \\ {\mathrm{σ}}_1=41 \\ {\mathrm{σ}}_2=30 \end{gathered}$$

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

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

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:

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

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.