forming control chart with three-sigma limits, - 100, S. 0ps4 cessis cowith a ta
ID: 3061637 • Letter: F
Question
forming control chart with three-sigma limits, - 100, S. 0ps4 cessis cowith a tac (20 pts) A process i UCL 0.172, center line 0.080, and LCL-0.000 (a) Find the equivalent control chart for the number nonconforming, and ARL. (5 pts) (b) Find the probability of a type II error, if the process fraction nonconforming shifts to 0.2.(10 pts) (c) What is the probability of detecting the shift in part (b) by at most the third sample after the shift? (5 pts) ontrol chart with thre-sigma limits, n - 100, Cumulative Binomial Distribution (p -0.2,p-0.08,n- 100) 100, e-0.2 n-100, p- 0.08 P(X-x) 0.0002392 0.5926280 0.0000000 0.0008554 0.0056964 0.0253288 0.1923380 0.2711890 0.3620870 10 12 16 0.9975910 0.9990690 18 20 18 20 0.9999630Explanation / Answer
Solution:
a. np = 100 (0.080) =8
UCL = np + 3 np (1 – p) = 8 + 3 8 (1 – 0.080) = 16.14
LCL = np – 3 np (1 – p) = 8 – 3 8 (1 – 0.080) = -0.1388 à0
b. np(new) = 100*0.20 = 20 > 15, so use normal approximation to binomial.
P (Type II error) =
= P (p < UCL|pnew) – P (p < LCL|pnew)
= P [(UCL –pnew)/p (1 – p)/n] – P [(LCL- pnew)/p (1 – p)/n]
= P [(0.172 – 0.20)/0.08 (1 – 0.08)/100] – P [(0 – 0.20)/0.08 (1 – 0.08)/100]
= P (-1.03) – P (-7.37)
= 0.1515 – 0
= 0.1515
c. P (detect shift by at most 4th sample) = 1 – P (not detect by 4th)
= 1 – (0.1515)^4
= 0.9995
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.