Question

Producing machine bearings. To determine whether a metal lathe that produces machine bearings is properly adjusted, a random sample of 25 bearings is collected and the diameter of each is measured.

1. If the standard deviation of the diameters of the bearings measured over a long period of time is .001 inch, what is the approximate probability that the mean diameter xof the sample of 25 bearings will lie within.0001 inch of the population mean diameter of the bearings?
2. If the population of diameters has an extremely skewed distribution, how will your approximation in part a be affected?

Short answer

1. The approximate probability that the mean diameter x of the sample of the 25 bearings will lie within 0.0001 inches of the population mean diameter of the bearings is 0.383.
2. There cannot be applied the central limit theorem.

Step-by-step solution

Given information

There is a random sample of 25 bearings and the diameter of each bearing is measured.

Determine the assumption

Let’s consider that the distribution of the diameter of the bearing follows the normal distribution.

So, the sampling distribution of the sample mean is normally distributed with the mean ${\mu }_{\overline{x}}=\mu and{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

Here, the mean is unknown and the standard deviation is 0.001.

Derivation of the standard deviation

a.

The standard deviation of the sample mean$\left(\overline{x}\right)$ is,

$\begin{array}{rcl}{\sigma }_{\overline{x}}& =& \frac{\sigma }{\sqrt{n}}\\ & =& \frac{0.001}{\sqrt{25}}\\ & =& 0.0002\\ & & \end{array}$

The standard deviation is 0.0002.

Calculation of the probability

The sample mean will lie within 0.0001inches of the population mean.

That is,

$$\begin{gathered} \left|\overline{x}-\mu \right|⩽0.0001 \\ \Rightarrow -0.0001⩽\overline{x}-\mu ⩽0.0001 \end{gathered}$$

Now, the probability is,

$\begin{array}{rcl}\mathrm{Pr}\left(-0.0001⩽\overline{x}-\mu ⩽0.0001\right)& =& \mathrm{Pr}\left(\frac{-0.0001}{\frac{\sigma }{\sqrt{n}}}⩽\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}⩽\frac{0.0001}{\frac{\sigma }{\sqrt{n}}}\right)\\ & =& \mathrm{Pr}\left(\frac{-0.0001}{0.0002}⩽z⩽\frac{0.0001}{0.0002}\right)\\ & =& \mathrm{Pr}\left(-0.50⩽z⩽0.50\right)\\ & =& \mathrm{Pr}\left(z⩽0.50\right)-\mathrm{Pr}\left(z⩽-0.50\right)\\ & =& 0.50+0.1915-0.50+0.1915\\ & =& 0.383\\ & & \end{array}$

Thus, the required probability that the sample mean will lie within 0.0001 inches of the population mean is 0.383.

Approximation of part a

b.

Let’s consider that the population diameter has an extremely skewed distribution, so, the approximation which is taken in part a. will be inaccurate.

To apply the central limit theorem, the sample size need to be large but here the sample n=25, which is not so large. So, one cannot apply the central limit theorem in this case.