91Ó°ÊÓ

To find the mean of the four times. Note that the mean is $437$days and the standard deviation is $399$days.

Earthquakes having a magnitude of $7.5$ or larger on the Richter scale make up the majority of the population.
The mean time between two earthquakes in the population days $\mathrm{μ}=437$.
Population for standard deviation is $399$ days.
Sample size $n=4$

Since, $\stackrel{¯}{x}$be the sample mean time between two earthquakes.

In other words, because the mean of all possible sample means is equal to population mean $\mathrm{μ}$, will expect the mean of the four times in the sample to be equal to $437$ days on average.
As a result, on an average can expect that mean of the four times in the sample will be equal to $437$ days.

To find the expected variation from the answer in part (a) by using the three-standard-deviations rule.

Let, the mean is $=437$ days.
And the standard deviation $=399$ days.
Since the standard deviation of all possible sample mean is ${\mathrm{σ}}_{\stackrel{¯}{x}}$ as:
$=\frac{\mathrm{σ}}{\sqrt{n}}$
$=\frac{399}{\sqrt{4}}$$=199.5$days.
Expect $Â\pm 3{\mathrm{σ}}_{\stackrel{¯}{x}}$ variation from the mean of $\stackrel{¯}{x}$which is from population mean $\mathrm{μ}$.
Expected amount of variation from $\stackrel{¯}{x}=Â\pm 3{\mathrm{σ}}_{\stackrel{¯}{x}}$
$\begin{array}{r}\overline{x}=Â\pm 3\frac{\mathrm{σ}}{\sqrt{n}}\end{array}$

$=Â\pm 3×199.5$

$=Â\pm 598.5$

Asa result, the expected variation is $Â\pm 598.5$.