ACCEPTANCE SAMPLING Company A purchases large shipments of c

ACCEPTANCE SAMPLING: Company A purchases large shipments of computer chips and uses this acceptance sampling plan: randomly select and test 10 computer chips, then accept the whole batch if there is less than two defective; that is, at most one chip does not work. If a particular shipment of thousands of computer chips actually has a 2% rate of defects, what is the probability that this whole shipment is accepted?

A)

P(Acceptable) = 1/10 + 2/10 = 0.1 + 0.2 = 0.3

B)

P(Acceptable) = 0.02

C)

P(Acceptable) = 10C0(0.02)0(0.98)10 + 10C1(0.02)1(0.98)9 + 10C2(0.02)2(0.98)8= (0.98)10 + 10(0.02)(0.98)9 + 45(0.02)2(.98)8= 0.9991

D)

P(Acceptable) = 10C0(0.02)0(0.98)10 + 10C1(0.02)1(0.98)9 = (0.98)10 + 10(0.02)(0.98)9 = 0.9838

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

P( X < 2) = P(X=1) + P(X=0)
= ( 10 1 ) * 0.02^1 * ( 1- 0.02 ) ^9 + ( 10 0 ) * 0.02^0 * ( 1- 0.02 ) ^10   
= 0.9838

ANSWER : P(Acceptable) = 10C0(0.02)0(0.98)10 + 10C1(0.02)1(0.98)9 = (0.98)10 + 10(0.02)(0.98)9 = 0.9838