Making auto parts $A$grinding machine in an auto parts plant prepares axles with a target diameter $\mu =40.125$millimeters (mm). The machine has some variability, so the standard deviation of the diameters is $\sigma =0.002mm$. The machine operator inspects a random sample of $4$axles cach hour for quality control purposes and records the sample mean diameter $\overline{x}$,how many axles would you need to sample if you wanted the standard deviation of the sampling distribution $\overline{x}$to be$0.0005mm$?justify your answer.

Given in the question that,

Mean $=40.125mm$

Standard deviation$=0.002mm$

Number of axles$=4$we have to find that how many axles would you need to sample if you wanted the standard deviation of the sampling distribution $\overline{x}$to be$0.0005$.

The mean of the sampling distribution of $\overbrace{p}is{\mu }_{\overbrace{p}}=p$

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

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

Here, number of axles is the sample size.

Also,

Population standard deviation,$\sigma =0.002$

Standard deviation of the sample,${\sigma }_{\overline{x}}=0.0005$

It is known that the standard deviation of a sample distribution of the sample mean if the population standard deviation is can be written as,

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

To find the sample size, use the above formula.

Write the formula as:

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

squares both side

$n=\frac{{\sigma }^2}{{\sigma }_{\overline{x}}2}$

substitute $0.002for\sigma and0.0005for{\sigma }_{\overline{x}}$

$n=\frac{0.{002}^2}{0.{0005}^2}=16$

hence the required number of axles to sample is$16$