Question

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?

Answer #1

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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