Question

Question:Quality control. Refer to Exercise 5.68. The mean diameter of the bearings produced by the machine is supposed to be .5 inch. The company decides to use the sample mean from Exercise 5.68 to decide whether the process is in control (i.e., whether it is producing bearings with a mean diameter of .5 inch). The machine will be considered out of control if the mean of the sample of n = 25 diameters is less than .4994 inch or larger than .5006 inch. If the true mean diameter of the bearings produced by the machine is .501 inch, what is the approximate probability that the test will imply that the process is out of control?

Short answer

The probability that the test will imply that the process is out of control is 0.97725.

Step-by-step solution

Given Information

The sample size is 25

The mean is 0.501.inch

The standard deviation is 0.001.inch

The standard deviation of $\overline{x}$is calculated as

$$\begin{gathered} {\sigma }_{\overline{X}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{0.001}{\sqrt{25}} \\ =0.0002 \end{gathered}$$

Explanation

The probability is computed as

$$\begin{gathered} p\left(\overline{x}<0.4994\right)=p\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{0.4994-0.501}{0.0002}\right) \\ =p\left(z<\frac{-0.0016}{0.0002}\right) \\ =p\left(z<-8\right) \\ =0 \end{gathered}$$

And

$$\begin{gathered} p\left(\overline{x}>0.5006\right)=p\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{0.5006-0.501}{0.0002}\right) \\ =p\left(z>\frac{-0.0004}{0.0002}\right) \\ =p\left(z>-2\right) \\ =0.97725 \end{gathered}$$

Here we used equation 1 & 2 to get required probability

$$\begin{gathered} p\left(\overline{x}<0.4994\right)+p\left(\overline{x}>0.5006\right)=0+0.97725 \\ =0.97725 \end{gathered}$$

Hence, the probability is 0.97725.