Question

Question 1 20 pts Acceptance sampling is used to monitor the quality of incoming raw materials. Suppose a purchaser of electronic components allows 1 percent of the components to be defective. To ensure the quality of incoming parts, a purchaser or manufacturer normally samples 20 parts and allows 1 defect. a. What is the likelihood of accepting a lot that is 1 percent defective? b. If the quality of the incoming lot was actually 2 percent, what is the likelihood of accepting it? c. If the quality of the incoming lot was actually 5 percent, what is the likelihood of accepting it? Please keep four decimals.

Answer 1

(a)

n = 20

p = 0.01

So,

q = 1 - p = 0.99

P(X$\leq$1) = P(X=0) + P(X=1)

P(X = 0) = x 0.9920 x 0.01° = 1 x 0.8179 x 1 = 0.8179

P(X = 1) = x 0.999 x 0.01€ = 20 x 0.8262 x 0,01 = 0.1652

So,

P(X$\leq$1) = 0.9831

So,

Answer is:

0.9831

(b)

n = 20

p = 0.02

So,

q = 1 - p = 0.98

P(X$\leq$1) = P(X=0) + P(X=1)

P(X = 0) = x 0.9820 x 0.02° = 1 x 0.6676 x 1 = 0.6676 0

P(X = 1) = x 0.9819 x 0.021 = 20 x 0.6812 x 0.02 = 0.2725

So,

P(X$\leq$1) = 0.9401

So,

Answer is:

0.9401

(a)

n = 20

p = 0.05

So,

q = 1 - p = 0.95

P(X$\leq$1) = P(X=0) + P(X=1)

20 P(X = 0) = x 0.9520 x 0.050 = 1 x 0.3585 x 1 = 0.3585 0

P(X = 1) = x 0.9519 x 0.051 = 20 x 0.3774 x 0.05 = 0.3774

So,

P(X$\leq$1) = 0.7358

So,

Answer is:

0.7358