91影视

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 $蟽=0.04$.

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 $渭$, based on the sampling distribution information provided.
Population standard deviation $=\frac{\text{Standarddeviation}}{\text{Square rootof samplesize}}$
${蟽}_{\stackrel{炉}{x}}=\frac{蟽}{\sqrt{n}}$
$=\frac{0.4}{\sqrt{50}}$
$鈮�0.057$
As a result, the shape is approximately normally distributed with center $渭$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 $蟽=0.04$.

The sample mean $\stackrel{炉}{x}$is normally distributed with the mean $渭$ and the standard deviation as:
$\frac{蟽}{\sqrt{n}}=\frac{0.4}{\sqrt{50}}$

$鈮�0.057$

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 $蟽=0.04$

And the population standard deviation ${蟽}_x=0.057$.

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 $渭$.

Let, the number of trucks tested for $NOX$emission test is $n=50$.
And the standard deviation $蟽=0.04$.
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 $渭$.