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To find the mean and the standard deviation for the sample size of $50$. Where, the mean price of new mobile homes is $\$65,100$and the standard deviation of $\$7200$.

The mean $\stackrel{¯}{x}$indicates the price of a sample of new mobile homes.

The mean price of new mobile homes is $\$65,100$. And the standard deviation is $\$7200$.

So, $\mathrm{μ}=\$65,100$and $\mathrm{σ}=\$7200$.
Then the Size of the sample,$n=50$.
Mean of $\stackrel{¯}{x}$,
${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$

${\mathrm{μ}}_{\stackrel{¯}{x}}=\$65,100$

Standard deviation of $\stackrel{¯}{x}$,

$\begin{array}{r}{\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}\end{array}$
$=\frac{7200}{\sqrt{50}}$
$=\frac{7200}{7.071068}$
$=1018.2$
The mean and standard deviation of all possible samples mean prices for $50$ new mobile homes are $\$65,100$and $\$1018.2$ respectively.

To find the mean and the standard deviation for the sample size of $100$. Where, the mean price of new mobile homes is $\$65,100$and the standard deviation of $\$7200$.

Standard deviation of $\stackrel{¯}{x}$is $\begin{array}{r}{\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}\end{array}$

Where the sample size is $n=100$.
Mean of $\stackrel{¯}{x}$is:

${\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}$
${\mathrm{μ}}_{\overline{x}}=65,100$
Then,

$\begin{array}{r}{\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}\end{array}$
$=\frac{7200}{\sqrt{100}}$
$=\frac{7200}{10}$
$=720$
The mean and standard deviation of all possible samples mean prices for $100$new mobile homes are$\$65,100$and $\$720$, respectively.