2. A process is being controlled with an np-chart, and it is known that when the
ID: 3311291 • Letter: 2
Question
2. A process is being controlled with an np-chart, and it is known that when the A process is in control, the proportion ofdefective products is 0.02. Three-sigma control limits are used and the procedure calls for taking daily samples of n=100 items a) (10 pts) What are the 3-sigma control limits this control chart? b) (10pts) What is for this control chart? c) (10 pts) If the proportion of defective products were to suddenly shift to 0.05, what is the probability that the shift would be detected on the first subsequent sample? ARL (Spts) For the shifted proportion of part c), what is the probability it is detected within two samples? d) 02 -01)-22 LcL uuotExplanation / Answer
Here n = 100 and p = 0.02
so np = central line = 100 * 0.02 = 2
LCL = np - 3 * sqrt [0.02 * 0.98 * 100] = 2 - 4.2 = 0 [ Less than 0 value not permissble]
UCL = np + 3 * sqrt [0.02 * 0.98 * 100] = 2 + 4.2 = 6.2
(b) Here alpha or Type I error would be
Pr(np > 6.2) ; Here the p is very small so we can't use normal approximation.
so Pr(np > 6.2) = BIN (np > 6.2 ; 100 ; 0.02) = 1- BIN (np < 6.2 ; 100 ; 0.02)
we will use binomial table for calculating the cumulative value for binomial parameter n = 100 , p = 0.02 and x = 6
BIN (np < 6.2 ; 100 ; 0.02) = 0.9959
Type I error = alpha = 1 - 0.9959 = 0.0041
(c) Here if the new proportion of defective item will shift to 0.05, then we will detect the shift in first subsequent sample. That would happen when we will get the defective proportion above 0.062 or np > 6.2
so Pr(Detection of process change) = Pr(np > 6.2) = 1 - BIN (np < 6.2 ; 100 ; 0.05)
again here p is small so we can to normal approximation here.
BIN (np < 6.2 ; 100 ; 0.05) = BIN (np =< 6 ; 100 ; 0.05) = 0.7660
so Pr(Detection of process change) = Pr(np > 6.2) = 1 - BIN (np < 6.2 ; 100 ; 0.05) = 1- 0.7660 = 0.2340
(d) Here that detection would be done withing two samples.
Pr(Detected withing two samples) = Pr(detection in first sample ) + Pr(no detection in first sample) * Pr(detection in second sample) = 0.234 + 0.766 * 0.234 = 0.4132
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