Question

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 48 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 2 batteries do not meet specifications. A shipment contains 3000 ​batteries, and 1​% of them do not meet specifications.
a. What is the probability that this whole shipment will be​ accepted?

b. Will almost all such shipments be​ accepted, or will many be​ rejected?

Answer 1

Answer:

a) the probability that this whole shipment will be​ accepted is 0.8731

b) Almost all shipments (87.31% of the shipments) would be accepted

Step-by-step explanation:

a) Given that:

p = 1% = 0.01

n = 48

Binomial probability states that

$P(X=x) = \frac{n!}{x!(n-x)!} .p^{x}(1-p)^{n-x}$

Addition rule for mutually exclusive events is:

P(A or B) = P(A) + P(B)

To get the probability that this whole shipment will be​ accepted we evaluate at x = 0, 1, 2

$P(X=0) = \frac{48!}{0!(48-0)!} .0.01^{0}(1-0.01)^{48-0}=0.6173\\P(X=1) = \frac{48!}{1!(48-1)!} .0.01^{1}(1-0.01)^{48-1}=0.1848\\P(X=2) = \frac{48!}{2!(48-2)!} .0.01^{2}(1-0.01)^{48-2}=0.071$

Using addition rule:

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.6173 + 0.1848 + 0.071 = 0.8731

the probability that this whole shipment will be​ accepted is 0.8731

b) Almost all shipments (87.31% of the shipments) would be accepted