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