91Ó°ÊÓ

It is given that there are two distributions that is,

$$\begin{gathered} DistributionM:Normalwith{\mathrm{μ}}_M=69.3,{\mathrm{σ}}_M=2.8 \\ DistributionW:Normalwith{\mathrm{μ}}_W=64.5,{\mathrm{σ}}_W=2.5 \end{gathered}$$

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:

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

Then we obtain:

$$\begin{gathered} {\mathrm{μ}}_{M-W}={\mathrm{μ}}_M-{\mathrm{μ}}_W \\ =69.3-64.5 \\ =4.8 \\ {\mathrm{σ}}_{M-W}=\sqrt{{{\mathrm{σ}}^2}_M+{{\mathrm{σ}}^2}_W} \\ =\sqrt{2.8^2+2.5^2} \\ =3.7537 \end{gathered}$$

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

$$\begin{gathered} {\mathrm{μ}}_{M-B}=4.8 \\ {\mathrm{σ}}_{M-B}=3.7537 \end{gathered}$$

$$\begin{gathered} {\mathrm{μ}}_1=69.3 \\ {\mathrm{μ}}_2=64.5 \\ {n}_1=16 \\ {n}_2=9 \\ {\mathrm{σ}}_1=2.8 \\ {\mathrm{σ}}_2=2.5 \end{gathered}$$

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

$$\begin{gathered} {\mathrm{μ}}_{x_i-x_2}={\mathrm{μ}}_1-{\mathrm{μ}}_2 \\ =69.3-64.5 \\ =4.8 \end{gathered}$$

Thus, we conclude that the mean of the sampling mean is$4.8$inches.

Since the sample $16$young men is less than $10\%$of all young men and since the sample of $9$young women is less than $10\%$of all young women. Thus, the standard deviation is as follows:

$$\begin{gathered} {\mathrm{σ}}_{x_1-x_2}=\sqrt{\frac{{\mathrm{σ}}_1^2}{n_1}+\frac{{\mathrm{σ}}_2^2}{n_2}} \\ =\sqrt{\frac{2.8^2}{16}+\frac{2.5^2}{9}} \\ =1.0883 \end{gathered}$$

Thus, we conclude that the difference in the sample means are expected to be vary by$1.0883$ inches from the mean difference of$4.8inches$.