Question

Voltage sags and swells. Refer to the Electrical Engineering (Vol. 95, 2013) study of the power quality (sags and swells) of a transformer, Exercise 2.76 (p. 110). For transformers built for heavy industry, the distribution of the number of sags per week has a mean of 353 with a standard deviation of 30. Of interest is , that the sample means the number of sags per week for a random sample of 45 transformers.

a. Find$\mathbf{E}\left(\overline{\mathbf{\chi }}\right)$ and interpret its value.

b. Find$\mathbf{Var}\left(\overline{\mathbf{\chi }}\right).$

c. Describe the shape of the sampling distribution of$\overline{\mathbf{\chi }}.$

d. How likely is it to observe a sample mean a number of sags per week that exceeds 400?

Short answer

The random sample is a sampling strategy in which every test has an equal probability of getting selected. A random sample is intended to provide an impartial reflection of the overall population. It guarantees that the findings obtained from the sample are close to those obtained if the complete population was tested.

Step-by-step solution

(a) The data is given below

The calculation is given below:

To find $\mathrm{E}\left(\overline{\mathrm{\chi }}\right)$

The weekly average value of sags is $\mu =353$ 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,

$\begin{array}{c}{\mu }_{\overline{\mathrm{\chi }}}=\mu \\ =353\end{array}$

The value of localid="1652096362826" $\mathrm{E}\left(\overline{\mathrm{\chi }}\right)\mathrm{is}353$

(b) The data is given below

The calculation is given below:

To find $\mathrm{Var}\left(\overline{\mathrm{\chi }}\right)$

The standard deviation of the weekly number of sags is $\sigma =30$

The standard deviation of the sampling distribution is ${\sigma }_{\overline{\mathrm{\chi }}}=\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}$

Therefore,

$\begin{array}{c}{\sigma }_{\overline{\mathrm{\chi }}}=\frac{30}{\sqrt{45}}\\ =4.4721\end{array}$

The variance is:

$\begin{array}{c}\mathrm{Var}\left(\overline{\mathrm{\chi }}\right)={\sigma }_{\overline{\mathrm{\chi }}}^2\\ ={\left(4.4721\right)}^2\\ =19.9996\\ \approx 20\end{array}$

Therefore, the value of $\mathrm{Var}\left(\overline{\mathrm{\chi }}\right)\mathrm{is}20$

(c) Sampling distribution shape

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

(d) Sampling distribution shape

Calculate the value of z at $\overline{\mathrm{\chi }}=400$

$\begin{array}{c}z=\frac{\overline{\mathrm{\chi }}-{\mu }_{\overline{\mathrm{\chi }}}}{{\sigma }_{\overline{\mathrm{\chi }}}}\\ =\frac{400-353}{19.9996}\\ =\frac{47}{19.9996}\\ =2.35\end{array}$

The value of z is 2.35

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