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The calculation is given below:

To find $\mathrm{E}\left(\stackrel{¯}{\mathrm{χ}}\right)$

The weekly average value of sags is $\mathrm{μ}=353$ as well as the random sample is 45, which is

When a random sampling of n observations is chosen from a standard normal distribution, the sample distribution's mean is equal to the population distribution's mean, demonstrating that the population is normal.

Therefore,

$\begin{array}{c}{\mathrm{μ}}_{\stackrel{¯}{\mathrm{χ}}}=\mathrm{μ}\\ =353\end{array}$

The value of localid="1652096362826" $\mathrm{E}\left(\stackrel{¯}{\mathrm{χ}}\right)\mathrm{is}353$

The calculation is given below:

To find $\mathrm{Var}\left(\stackrel{¯}{\mathrm{χ}}\right)$

The standard deviation of the weekly number of sags is $\mathrm{σ}=30$

The standard deviation of the sampling distribution is ${\mathrm{σ}}_{\stackrel{¯}{\mathrm{χ}}}=\frac{\mathrm{σ}}{\sqrt{\mathrm{n}}}$

Therefore,

$\begin{array}{c}{\mathrm{σ}}_{\stackrel{¯}{\mathrm{χ}}}=\frac{30}{\sqrt{45}}\\ =4.4721\end{array}$

The variance is:

$\begin{array}{c}\mathrm{Var}\left(\stackrel{¯}{\mathrm{χ}}\right)={\mathrm{σ}}_{\stackrel{¯}{\mathrm{χ}}}^2\\ ={\left(4.4721\right)}^2\\ =19.9996\\ ≈20\end{array}$

Therefore, the value of $\mathrm{Var}\left(\stackrel{¯}{\mathrm{χ}}\right)\mathrm{is}20$

The sampling distribution will be essentially normal for relatively significant sample sizes. Furthermore, the distribution has an asymmetric form.

Calculate the value of z at $\stackrel{¯}{\mathrm{χ}}=400$

$\begin{array}{c}z=\frac{\stackrel{¯}{\mathrm{χ}}-{\mathrm{μ}}_{\stackrel{¯}{\mathrm{χ}}}}{{\mathrm{σ}}_{\stackrel{¯}{\mathrm{χ}}}}\\ =\frac{400-353}{19.9996}\\ =\frac{47}{19.9996}\\ =2.35\end{array}$

The value of z is 2.35

Determine the probability that the sample mean count of sags each week is more than 400.