91影视

It is given that there are two distributions that is,

$$\begin{gathered} DistributionM:Normalwith{渭}_M=69.3,{蟽}_M=2.8 \\ DistributionW:Normalwith{渭}_W=64.5,{蟽}_W=2.5 \end{gathered}$$

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:

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

Then we obtain:

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

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

$$\begin{gathered} {渭}_{M-B}=4.8 \\ {蟽}_{M-B}=3.7537 \end{gathered}$$

$$\begin{gathered} {渭}_1=69.3 \\ {渭}_2=64.5 \\ {n}_1=16 \\ {n}_2=9 \\ {蟽}_1=2.8 \\ {蟽}_2=2.5 \end{gathered}$$

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

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

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:

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

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