91影视

Referring to exercise 2.46 (page 96), we have to measure the surface roughness of 20 sampled sections of coated interior pipe by a scanning probe instrument. The sample data we have collected is given below,

Let us assume that the distribution has a mean $渭=1.8$micrometer and a standard deviation$蟽=0.5$micrometer.

By using this information, we have to calculate the probability that X exceeds 1.85 micrometer. That is$P\left(\stackrel{炉}{X}猢�1.85\right)$

Now, the corresponding Z-value is,

$\begin{array}{rcl}{Z}_L& =& \frac{X_1-渭}{蟽/\sqrt{n}}\\ & =& \frac{1.85-1.8}{0.5/\sqrt{20}}\\ & =& 0.45\\ & & \end{array}$

So, by the property of normal distribution we can say that,

$\begin{array}{rcl}P\left(\stackrel{炉}{X}猢�1.85\right)& =& P\left(\frac{\stackrel{炉}{X}-1.8}{0.5/\sqrt{20}}猢�\frac{1.85-1.8}{0.5/\sqrt{20}}\right)\\ & =& P\left(Z猢�Z_L\right)\\ & =& P\left(Z猢�0.45\right)\\ & & \end{array}$

Therefore, by the Z-table we got $P\left(Z猢�0.45\right)=0.3274$

Thus, by the given information we conclude that the probability that$\stackrel{炉}{X}$ exceeds 1.85 is 0.3274.

Consider the sample data. So, the sample mean is,

$$\begin{gathered} \begin{array}{rcl}\stackrel{炉}{X}& =& \frac{1}{n}\underset{i=1}{\overset{n}{鈭�}}{X}_i\\ & =& \frac{1}{20}\left(1.72+2.50+2.16+2.13+1.06+2.24+2.31+2.03+1.09+1.40+ \\ 2.57+2.64+1.26+2.05+1.19+2.13+1.27+1.51+2.41+1.95 \\ \right)\\ & =& \frac{37.62}{20}\\ & =& 1.88\\ & & \\ & & \end{array} \end{gathered}$$

Thus, the sample mean $\stackrel{炉}{X}$is 1.88 that is the roughness of the pipe is around 1.88 micrometer.

In the part (a) we assume that the mean of the roughness distribution is 1.8 and the standard deviation is 0.5. Based on the information we got the probability that exceeds 1.85 is 0.3274.

In part (b) we can see that the sample mean is 1.88. So, our assumption is wrong as the sample mean is exceeds 1.85 with a probability 1.