Question

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 60 batteries and determine whether each is within specifications. The entire shipment is accepted if at most batteries do not meet specifications. A shipment contains 5000 ​batteries, and ​1% of them do not meet specifications. What is the probability that this whole shipment will be​ accepted? Will almost all such shipments be​ accepted, or will many be​ rejected?

a. The probability that this whole shipment will be accepted is ??.

b. The company will accept ??% of the shipments and will reject ??% of the​ shipments, ??.

Answer 1

To determine the probability that the whole shipment will be accepted, we need to calculate the probability that at most one battery does not meet specifications.

Given that 1% of the batteries do not meet specifications, it means that out of 5000 batteries, 0.01 * 5000 = 50 batteries do not meet specifications.

Now, using the binomial probability formula, we can calculate the probability of at most one battery not meeting specifications:

P(X ≤ 1) = P(X = 0) + P(X = 1)

Where X follows a binomial distribution with parameters n = 60 (sample size) and p = 50/5000 = 0.01 (probability of a battery not meeting specifications).

P(X = 0) = (60 choose 0) * (0.01)^0 * (1 - 0.01)^(60 - 0)

P(X = 1) = (60 choose 1) * (0.01)^1 * (1 - 0.01)^(60 - 1)

Calculating these probabilities:

P(X = 0) ≈ 0.301

P(X = 1) ≈ 0.401

Therefore,

P(X ≤ 1) = P(X = 0) + P(X = 1) ≈ 0.301 + 0.401 ≈ 0.702

a. The probability that this whole shipment will be accepted is approximately 0.702.

b. The company will accept approximately 70.2% (0.702 * 100) of the shipments and will reject approximately 29.8% (100 - 70.2) of the shipments.