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5.2.36-T

When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: Randomly select and test 41 batteries and determine whether each is within specifications. The entire shipment is

accepted if at most 3 batteries do not meet specifications. A shipment contains 7000 batteries, and 3% 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?

The probability that this whole shipment will be accepted is 0.9660

(Round to four decimal places as needed)

The company will accept 96.6 % of the shipments and will reject 34% of the shipments, so almost all of the shipments will be accepted

(Round to two decimal places as needed.)

Answer 1

Probability that the whole shipment will be accepted = 0.9660

96.60% of the shipment will be accepted

Almost all the shipments will be accepted

Explanation:

We are given;

n = 41

p = 3% = 0.03

x = 3

This is a binomial probability question and we will use the formula;

P(X = x) = [n!/(x!(n - x)!)] × p^(x) × (1 - p)^(n - x)

The probability that the whole shipment will be accepted is given by the form;

P(X ≤ 3) = P(0) + P(1) + P(2) + P(3)

P(3) = [41!/(3!(41 - 3)!)] × 0.03³ × (1 - 0.03)^(41 - 3)

P(3) = 0.09046

P(2) = [41!/(2!(41 - 2)!)] × 0.03² × (1 - 0.03)^(41 - 2)

P(2) = 0.22496

P(1) = [41!/(1!(41 - 1)!)] × 0.03¹ × (1 - 0.03)^(41 - 1)

P(1) = 0.36373

P(0) = [41!/(1!(41 - 0)!)] × 0.03^(0) × (1 - 0.03)^(41 - 0)

P(0) = 0.28684

Thus;

P(X ≤ 3) = 0.28684 + 0.36373 + 0.22496 + 0.09046

P(X ≤ 3) = 0.96599 ≈ 0.9660

It means that 96.60% will be accepted