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According to theCentral Limit Theorem,the sampling distribution of the sample means approaches a normal distribution, irrespective of the shape of population distribution if the sample size is over 30.

A statistic's standard error is the estimated standard deviation of a statistics population sample.

It is given that,${\stackrel{炉}{x}}_1=17.5$, ${蟽}_1=2$, and $n_1=58$.

The standard error is localid="1652760610266" $\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}$and localid="1652760613813" ${渭}_1={\stackrel{炉}{x}}_1$

So,

${渭}_1=17.5$

Standard error for Sample 1 =localid="1652760584440" $\frac{{蟽}_1}{\sqrt{n_1}}$

$$\begin{gathered} =\frac{2}{\sqrt{58}} \\ =0.26 \end{gathered}$$

Therefore, the standard error of the sampling distribution for Sample 1 is $0.26$.

It is given that, ${\stackrel{炉}{x}}_1=16.23$,${蟽}_1=8$, and $n_1=62$.

The standard error is $\frac{蟽}{\sqrt{n}}$and${渭}_2={\stackrel{炉}{x}}_2$

So,

${渭}_2=16.23$

The standard error for Sample 2 =role="math" localid="1652760921438" $\frac{{蟽}_2}{\sqrt{n_2}}$

$\begin{array}{l}=\frac{8}{\sqrt{62}}\\ =1.02\end{array}$

Therefore, the standard error of the sampling distribution for Sample 2 is $1.02$.

It is given that, ${渭}_1=17.5$, ${渭}_2=16.23$, ${蟽}_1=2$, ${蟽}_2=8$,$n_1=58$, and $n_2=62$

$\begin{array}{l}{\stackrel{炉}{x}}_{{}_1}鈭�{\stackrel{炉}{x}}_{{}_2}={渭}_1鈭�{渭}_2\\ =17.5鈭�16.23\\ =1.27\end{array}$

role="math" localid="1652761176006" $\begin{array}{l}{蟽}_{{}_{{\stackrel{炉}{x}}_{{}_1}鈭�{\stackrel{炉}{x}}_{{}_2}}}=\sqrt{\frac{{{蟽}_1}^2}{n_1}+\frac{{{蟽}_2}^2}{n_2}}\\ =\sqrt{\frac{4}{58}+\frac{64}{62}}\\ =\sqrt{1.1012}\\ =1.05\end{array}$

Therefore, the mean and standard deviation of the sampling distribution$({\stackrel{炉}{x}}_{{}_1}鈭�{\stackrel{炉}{x}}_{{}_2})$ are$1.27$ and $1.05$respectively.

The difference in sample means will be normally distributed according to the Central Limit Theorem as $n>30$.