91Ó°ÊÓ

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

Let, the number of trucks tested for $NOX$emission test is $n=50$.

And the standard deviation is $\mathrm{σ}=0.04$.

Use the Central Limit Theorem to determine the shape of the sample distribution $\left(CLT\right)$. The sample mean $\overline{x}$is nearly normally distributed in terms of sampling distribution. The sample mean for the center is $\mathrm{μ}$, based on the sampling distribution information provided.
Population standard deviation $=\frac{\text{Standarddeviation}}{\text{Square rootof samplesize}}$
${\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}}$
$=\frac{0.4}{\sqrt{50}}$
$≈0.057$
As a result, the shape is approximately normally distributed with center $\mathrm{μ}$and spread is $0.057$.

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

Let, the number of trucks tested for $NOX$emission test is $n=50$.

And the standard deviation $\mathrm{σ}=0.04$.

The sample mean $\stackrel{¯}{x}$is normally distributed with the mean $\mathrm{μ}$ and the standard deviation as:
$\frac{\mathrm{σ}}{\sqrt{n}}=\frac{0.4}{\sqrt{50}}$

$≈0.057$

To determine the distance $m$. And to shade the region on the axis of the sketch that is within $m$ of the mean.

Let, the number of trucks tested for$NOX$emission test is $n=50$.

The standard deviation is $\mathrm{σ}=0.04$

And the population standard deviation ${\mathrm{σ}}_x=0.057$.

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

As a result, distance $m$of the mean of the sampling distribution is $0.114$.

To determine the percent of all possible samples does the interval capture $\mathrm{μ}$.

Let, the number of trucks tested for $NOX$emission test is $n=50$.
And the standard deviation $\mathrm{σ}=0.04$.
Since, $95$percent of the $\overline{x}$values from part c are within a distance $m$ of the sample distribution's mean.
As a result, $95\%$of all possible sample capture $\mathrm{μ}$.