Answer to Question #154682 in Statistics and Probability for Eugine Haweza

Question

1. A truck manufacturer monitors the width of the door seam as vehicles come off its assembly line. The seam width is the distance between the edge of the door and the truck body, in inches. These data are 62 days of measurements of a passenger door seam, with 10 trucks measured each day. It has been claimed that the process has an average width 0.275 with  = 0.1.

(a) If the seam widths at this assembly line are normally distributed, then what is the probability of finding a seam wider than 1/2 inch?

(b) If the process is under control, what is the probability of finding the mean of a daily sample of 10 widths more than 3 standard errors away from = 0.275?

(c) Ten measurements are averaged each day. Is this a large enough sample size to justify using a normal model to set the limits in the X-bar chart? Do you recommend changes in future testing?

Answer 1

(a) The normal random variable, denoted by X ~ N(0.275, 0.1²), is the width of the door seam as vehicles come off its assembly line. Therefore, the probability of finding a seam wider than 0.5 inch is:

$P(X>0.5) = P(Z> \frac{0.5-0.275}{0.1}) \\ = P(Z>2.25) \\ = 1 -P(Z<-2.25) \\ = 0.01222$

The probability of finding a seam wider than 0.5 inch is 0.01222.

(b) If the process is under control, the probability of finding the mean of a daily sample of 10 widths more than 3 standard errors away from μ = 0.275 is:

P(Z>3) = 0.00135

The required probability is 0.00135.

(c) By using excel function =KURT(data range), we get K₄ = -0.168. By using excel function =KURT(data range), we get $\bar{X}$ if the sample size n is larger than 10 times the absolute value of the kurtosis, n>10|K₄|.

10|K₄| = 10|-0.168|

= 1.68

<10

Sample size 10 is larger than 10 times the absolute value of the kurtosis, it is okay to use a normal distribution to set the control limits in the X-bar chart. No need to change sample size for future testing.