Question

# A control chart indicates that the current process fraction nonconforming is 0.02. If 50 items are...

A control chart indicates that the current process
fraction nonconforming is 0.02. If 50 items are

inspected each day, what is the probability of detect-
ing a shift in the fraction nonconforming to 0.04 on

the first day after the shift? By the end of the third
day following the shift?

Given that probapility of frction non conforming (p¯ )= 0.02 and total sample inspected (n )= 50, we compute the control limits,
UCL = 0.02+3×(0.02(1−0.02))÷50)1/2
UCL = 0.07939697
Central Line(CL) = 0.02
LCL = 0.02−3×(0.02(1−0.02))÷50)1/2
LCL = −0.03939697 ⇒ 0( Due to negative limit)
Since p(new) = 0.04 < 0.1 and n = 50 is large, we use the Normal approximation to the binomial with mean p(new )= 0.04
and variance p(new)(1 − p(new))/n = (0.04*(1-0.04))/÷50)=0.000768. P(detect|shift) = P(D/n > UCL) + P(D/n < LCL) = 1 − Φ((0.07939697 −
0.04)/p
0.000768)+Φ((0−0.04)/p
0.000768) ≈ 0.1520263. P(detected by 3rd day) = 1−(1−0.1520263)3 ≈ 0.39025

P(detecting by 3rd day)= 0.39025

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