Question

When a process is in control, a variable can be taken to be normally distributed with mean of µ0 = 80 and standard deviation of σ = 10. A control chart is to be implemented by plotting the average of n = 16 observations at each time point. Using the formula µ0 ± 3σ/√ n, we have obtained an upper control limit of 87.5 and a lower control limit of 72.5. Suppose the mean of the variable now shifts to µ = 77.2 and the standard deviation remains σ = 10. What is the average run length?

Answer 1

1 Solution Given dalce: process to control, a variable: Mo=80 6 > 10 m = 16 UCL = 87.5 LcL=72.5 ARL - I where q=P[ , = pr plots outside condrol limit As the mean of the Vocriceble shifts do M2 77.2 cend 6 =10 remains Same I-P [ Mo lo-36 in -M LX-ll, 90+36-1 6/15 Elin olun = I-P I-P[ 80- 3x10 TT .-77.2 62% 80+ 3x10 -772 101056

AP[ -1.88<< < 4.12] re H- [P [OLZ 41.68] + P [02224.12]] -1- 0.9699 z 0.030) - AKL 0.0301 = 33.22 AKL 33.22 Run is 33.22 so the average Length The average run Length sen Length is 33:22 it means that it average to defect the misadjustrut takes 33.22. samples on an of size 16 please give me thumb up