91Ó°ÊÓ

We are given data of two populations:

For the first population, the sample size is $n_1=9$, mean is ${\mathrm{μ}}_1=40$, and the standard deviation is ${\mathrm{σ}}_1=12$.

For the second population, the sample size is $n_2=4$, mean is ${\mathrm{μ}}_2=40$, and the standard deviation is ${\mathrm{σ}}_2=6$.

The mean for the variable $\stackrel{¯}{x_1}-\stackrel{¯}{x_2}$is given as

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}={\mathrm{μ}}_1-{\mathrm{μ}}_2 \\ {\mathrm{μ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=40-40 \\ {\mathrm{μ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=0 \end{gathered}$$

And the standard deviation is given as

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=\sqrt{\frac{{{\mathrm{σ}}_1}^2}{n_1}+\frac{{{\mathrm{σ}}_2}^2}{n_2}} \\ {\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=\sqrt{\frac{{12}^2}{9}+\frac{6^2}{4}} \\ {\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=\sqrt{16+9} \\ {\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=\sqrt{25} \\ {\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=5 \end{gathered}$$

As it is given that the variable is normally distributed on each of the two populations. So we can conclude using it that the variable $\stackrel{¯}{x_1}-\stackrel{¯}{x_2}$is normally distributed.

We have that the variable $\stackrel{¯}{x_1}-\stackrel{¯}{x_2}$is normally distributed and ${\mathrm{μ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=0,{\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=5$.

Using the $68.26-95.44-99.74$rule, we determine the percentage of all pairs of independent samples of sizes $9$and $4$respectively, from the two populations with the property that the difference role="math" localid="1652703305588" $\stackrel{¯}{x_1}-\stackrel{¯}{x_2}$between the sample means is between $-10$and $10$as:

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}-2{\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=0-2×5 \\ =-10 \\ {\mathrm{μ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}+2{\mathrm{σ}}_{\stackrel{¯}{x_1}-\stackrel{¯}{x_2}}=0+2×5 \\ =10 \end{gathered}$$

Therefore, we can conclude that $95.44\%$of all pairs of independent samples of sizes $9$and $4$respectively, from the two populations with the property that the difference $\stackrel{¯}{x_1}-\stackrel{¯}{x_2}$between the sample means is between $-10$and $10$.