91Ó°ÊÓ

To describe the shape, center, and spread of the sampling distribution of $\overline{x}$.

Let, the sample mean as $\stackrel{¯}{x}=280$.

The standard deviation as $\mathrm{σ}=60$
And the Sample size as localid="1652786402483" $n=840$
Use the Central Limit to determine the shape of the sampling distribution.
As a result, the sampling distribution of the sample mean has a normal shape.
The mean of the sampling distribution of the sample mean$\mathrm{μ}=280$ is calculated using the information provided.
So, the center equal to the mean $\mathrm{μ}=280$
And for the spread, use the formula
Population standard deviation localid="1653041512150" $=\frac{\text{Standarddeviation}}{\text{Squarerootofsamplesize}}$

${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$

$=\frac{60}{\sqrt{840}}$

$≈2.0702$

As a result, the shape is approximately normal with mean $280$as center and the spread is the standard deviation which is equal $2.0702$.

To sketch the sampling distribution of $\stackrel{¯}{x}$. Then to mark its mean and the values one, two, and three standard deviations on either side of the mean.

Let, the Sample mean of the score$\stackrel{¯}{x}=280$
The standard deviation $\mathrm{σ}=60$
And the sample size of the people from the large population $n=840$.

The sample mean $\stackrel{¯}{x}$is then normally distributed with mean $\mathrm{μ}=280$ and standard deviation is calculated as:
$\frac{\mathrm{σ}}{\sqrt{n}}=\frac{60}{\sqrt{840}}$

$≈2.070$.

To determine the distance$m$ of the mean of the sampling distribution.
Then shade the region on the axis of the sketch that is within $m$ of the mean.

Let, the sample mean is $\stackrel{¯}{x}=280$.

The standard deviation is $\mathrm{σ}=60$.
And the Sample size is $n=840$
Determine the standard deviation as follows:
${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$
$=\frac{60}{\sqrt{840}}$
$≈2.0702$
According to the $68-95-00.7$percent rule, around $95$percent of all $\overline{x}$values are within $m$ of the mean, or twice the population standard deviations of the mean.
Hence,

$m=2\mathrm{σ}$
$=2×2.070$
$=4.140$

Shade the region on the axis of the sketch that is within $m$of the mean as follows:

As a result, the distance $m$of the mean of the sampling distribution is $4.140$.

The population mean $\mathrm{μ}$lies in the confidence interval $\overline{x}Â\pm m$. To determine the percent of all possible samples $\mathrm{μ}$captured by the interval.

Since, the sample mean of the score is $\stackrel{¯}{x}=280$.
The standard deviation is $\mathrm{σ}=60$.
And the sample size of the people from the large population is $n=840$.
The mean of the sampling distribution is distance $m$.
Approximately $95$percent of all $\overline{x}$values are within $m$of the sampling distribution's mean.
As a result, $\mathrm{μ}$ captures $95\%$ of all possible samples.