Chapter 7.3, Problem 50E

To determine

Tocalculate:The mean and the standard deviation of the sampling distribution of $\overline{x}$ .

Answer to Problem 50E

The required mean and standard deviation are 40.125 millimeters and 0.001 millimeters respectively.

Explanation of Solution

Given information:

Mean = 40.125 millimeters

Standard deviation = 0.002 millimeters

Number of axles = 4

Formula used:

The mean of the sampling distribution of $\widehat{p}$is, ${\mu }_{\widehat{p}}=p$ .

The standard deviation of a sample distribution of the sample mean $\overline{x}$if the population standard deviation is $\sigma$can be written as,

  ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Calculation:

Here, given data can be written as:

Mean, $\mu =40.125$

Population standard deviation, $\sigma =0.002$

Sample size, n = 4

It is known that the mean of the sampling distribution of $\widehat{p}$is, ${\mu }_{\widehat{p}}=p$ .

So, the mean of the sampling distribution of the sample mean $\overline{x}$is,

  $\begin{array}{l}{\mu }_{\overline{x}}=\mu \\ {\mu }_{\overline{x}}=40.125\end{array}$

Also, the standard deviation of a sample distribution of the sample mean $\overline{x}$if the population standard deviation is $\sigma$can be written as,

  ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Substitute 0.002 for $\sigma$and 4 for n in the above formula and simplify.

  $\begin{array}{c}{\sigma }_{\overline{x}}=\frac{0.002}{\sqrt{4}}\\ =\frac{0.002}{2}\\ =0.001\end{array}$

Hence, the required mean and standard deviation are 40.125 millimeters and 0.001 millimeters respectively.