91影视

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

Let鈥檚 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 ${渭}_{\stackrel{炉}{x}}=渭and{蟽}_{\stackrel{炉}{x}}=\frac{蟽}{\sqrt{n}}$.

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

a.

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

$\begin{array}{rcl}{蟽}_{\stackrel{炉}{x}}& =& \frac{蟽}{\sqrt{n}}\\ & =& \frac{0.001}{\sqrt{25}}\\ & =& 0.0002\\ & & \end{array}$

The standard deviation is 0.0002.

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

That is,

$$\begin{gathered} \left|\stackrel{炉}{x}-渭\right|猢�0.0001 \\ 鈬�-0.0001猢�\stackrel{炉}{x}-渭猢�0.0001 \end{gathered}$$

Now, the probability is,

$\begin{array}{rcl}\mathrm{Pr}\left(-0.0001猢�\stackrel{炉}{x}-渭猢�0.0001\right)& =& \mathrm{Pr}\left(\frac{-0.0001}{\frac{蟽}{\sqrt{n}}}猢�\frac{\stackrel{炉}{x}-渭}{\frac{蟽}{\sqrt{n}}}猢�\frac{0.0001}{\frac{蟽}{\sqrt{n}}}\right)\\ & =& \mathrm{Pr}\left(\frac{-0.0001}{0.0002}猢�z猢�\frac{0.0001}{0.0002}\right)\\ & =& \mathrm{Pr}\left(-0.50猢�z猢�0.50\right)\\ & =& \mathrm{Pr}\left(z猢�0.50\right)-\mathrm{Pr}\left(z猢�-0.50\right)\\ & =& 0.50+0.1915-0.50+0.1915\\ & =& 0.383\\ & & \end{array}$

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

b.

Let鈥檚 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.