Question

For an acceptance sampling plan with n = 29 and c = 0, find the probability of accepting a lot that has a defect rate of 2%. (Round your answer to four decimal places.)

What is the probability of accepting the lot if the defect rate is 8%? (Round your answer to four decimal places.)

Answer 1

a)

Let X denote the number of defectives in the sample. Then $X\sim Bin(n=29,p=0.02)$

29 P(X = z) = (0.02)'(1 – 0.02)29-1, 1 = 0,1, ...., 29

Required probability =

$\\P(X\leq c)=P(X\leq 0)=P(X=0)=\binom{29}{0}(0.02)^{0}(1-0.02)^{29-0}\\ \\=0.5566$

b)

Let X denote the number of defectives in the sample. Then $X\sim Bin(n=29,p=0.08)$

29 P(X = z) = (0.08)"(1 – 0.08)29-1, 1 = 0,1, ...., 29

Required probability =

$\\P(X\leq c)=P(X\leq 0)=P(X=0)=\binom{29}{0}(0.08)^{0}(1-0.08)^{29-0}\\ \\=0.0891$