Question

Surface roughness of pipe. Refer to the Anti-CorrosionMethods and Materials(Vol. 50, 2003) study of the surface roughness of oil field pipes, Exercise 2.46 (p. 96). Recall that a scanning probe instrument was used to measure thesurface roughness x(in micrometers) of 20 sampled sectionsof coated interior pipe. Consider the sample mean,$\overline{\mathbf{X}}$.

1. Assume that the surface roughness distribution has a mean of = 1.8 micrometers and a standard deviation of = .5 micrometer. Use this information to find theprobability thatexceeds 1.85 micrometers.
2. The sample data are reproduced in the following table.Compute.
3. Based on the result, part b, comment on the validity ofthe assumptions made in part a.

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

Short answer

1. The probability that $\overline{X}$exceeds 1.85 is 0.3274.
2. $\overline{X}=1.88$
   
3. The assumptions made in part (a) is wrong.

Step-by-step solution

Given information

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,

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

Determine the probability

a.

Let us assume that the distribution has a mean $\mu =1.8$micrometer and a standard deviation$\sigma =0.5$micrometer.

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

Now, the corresponding Z-value is,

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

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

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

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

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

Calculate the sample mean

b.

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

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

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

Comment on the validity of assumption 

c.

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.