Question

A company makes a chemical product, and based on extensive experience it is thought that the critical-to-quality variable for this product has mean 300 and standard deviation 20.

(a) Suppose the process output is Normally distributed, and the lower and upper specification limits on this variable are LSL = 250 and USL = 310, respectively. What proportion of manufactured products will not conform to specifications?

(b) If the standard deviation remains constant, where should the process operators change the mean of the process to minimize the proportion of products that do not conform to specifications?

(c) Suppose the process operators are able to change the mean of the process to the value that you identified above. What is the proportion of products that do not conform to specifications in this case? Also, how should the process operators change the standard deviation of the process so as to further reduce by half the proportion of products that do not conform to specifications?

Answer 1

Given process mean 300, and standard deviation ơ-20 (a) The process output is Normally distributed, and the lower and upper specification limits on this variable are LSL 250 and USL 310 Manufactured products not conform to specifications would be the one who has falls less than 250 (LSL) or more than 310 (USL) Proportion of manufactured products will not conform to specifications i.e P(X < 250) + P(X> 310-P(X < 250) + [1-P(X-310] We can use NORM.DIST(x, mean, standard_dev, cumulative) excel function to calculate left tailed (<) normal distribution probabilities For PIX < 250) write-NORM.DIST(250, 300, 20, TRUE) which gives 0.0062 (0.62%), similarly for P(X-310)-NORM. DIST(310, 300, 20, TRUE) which gives 0.6915 So P(X > 310-1-0.6915-0.3085 (30.85%) Proportion of products below LSL are 0.62% and that of above USL are 30.85% So, total 0.3147 31.47% products will not conform to specifications (b) With the mean 300, the most of the product does not conform specification are above USL 30.85% 0.62% 250 LSL 300 310 mean USL The proportion of product does not conform specification would be minimum if the mean is at the center of LSL and USL 250+310 280 2 so, if the standard deviation remains constant, the process operators should change the mean of the process to 280 to minimize the proportion of products that do not conform to specifications

(c)The proportion of products that do not conform to specifications in case of part b would -NORM. DIST(250, 280, 20, TRUE) which gives 0.0668 (6.68%), and-1-NORM.DIST(310, 280, 20, TRUE) which gives 0.0668 (6.68%) So total 0.1336-13.36% To find out the standard deviation of the process so as to further reduce by half the proportion (0.1336/2 0.0668) of products that do not conform to specifications 0.0668/2 0.0334 below LSL and 0.0334 above USL We can make use of z score in this case 0.0334 0.0334 250 280 310 Using NORM.S.INV(probability) function we can get the z value for which the left tailed probability P(Z <z) 0.0334 -NORM.S.INV(0.0334) gives z -1.8330 which mean P(Z< -1.8330) 0.0334...(1) We know the relation between X and Z, convert z score into x score (with mean 280 and unknown ơ) but we want to find ơ such that P(X < 250) 0.0334 X-u 250-280 0.0334 P(Z< - 30) 0.0334 ...(2) Comparing (1) and (2) 30 1.8330 30 -16.3669 1.8330 So, the process operators should change the standard deviation of the process to 16.3669 to further reduce by half the proportion of products that do not conform to specifications