Problem 1 Samples of size n=5 are taken from a process every half hour. After m-
ID: 3060733 • Letter: P
Question
Problem 1 Samples of size n=5 are taken from a process every half hour. After m-25 samples have collected, we calculate been A Shewhart control chart is used, Assume that both charts indicates that the process is in control and that the quality characteristics is independent and normally distributed Estimate the process standard deviation (5 pts) Find the control limits of the X and R charts (10 pts) Assume that both charts exhibit control. If the specification limits are 26.4+/- 0.50, estimate the fraction non-conforming. (10 pts) b)Explanation / Answer
Estimate of process mean = (Sample mean)/Number of samples = 662.5/25 = 26.5
Average process range = Rbar = Ri/m = 9/25 = 0.36
UCLR = D4 * Rbar = 2.114 * 0.36 = 0.76104
CLR =Rbar = 0.36
LCLR =D3 * Rbar = 0 * 0.36 = 0
Control Charts for Mean:
UCLX = Process Mean + A2 Rbar = 26.5 + 0.577 * 0.36 = 26.70772
CLX = 26.5
LCLX = Process Mean - A2 Rbar = 26.5 - 0.577 * 0.36 = 26.29228
USL = 26.9 and LSL = 25.9
Then fraction of non-conforming items
= 1 -P[LSL < X < USL]
= 1- P[25.9 < X < 26.9]
= 1- P[(25.9-26.5)/0.154772 < Z < (26.9-26.5)/0.154772]
= 0.00493 ~ 0.005
Hence there may be 5 in 1000 items that are non-conforming
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