Question

When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: Randomly select and test 47 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 4000 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 .. (Round to four decimal places as needed.) The company will accept nothing % of the shipments and will reject nothing % of the shipments, so ▼ almost all of the shipments will be accepted. many of the shipments will be rejected. (Round to two decimal places as needed.)

Answer 1

Here percentage of shipment that do not meet specificantions = 3%

so, proportion of shipment that do not meet specificantions = 0.03

Here population size = N = 4000

and we will test 47 batteries that means n = 47

so here n/N = 47/4000 < 0.05 so we will assume independence if we select a battery to test.

Now if we assume that x is the number of battery out of 47 which do not meeti specifications. So here x will have binomial distribution with parameter n = 47 and p = 0.03

p(x) = ⁴⁷C_x (0.03)^x(0.97)^(47-x)

Whole shipment wil be accepted if at most 3 batteries do not meet specifications that means x 3

Now we have to find

Pr(this whole shipment will be accepted) = Pr(x 3) = Pr(x = 0) + Pr(x = 1) + Pr(x = 2) + Pr(x = 3)

= ⁴⁷C₀ (0.03)⁰(0.97)⁴⁷ + ⁴⁷C₁ (0.03)²(0.97)⁴⁶ + ⁴⁷C₂ (0.03)²(0.97)⁴⁵ + ⁴⁷C₃ (0.03)³(0.97)⁴⁴

= 0.2389 + 0.3473 + 0.2471 + 0.1146 = 0.9479

The probability that this whole shipment will be accepted is 0.9479. The company will accept 94.79% of the shipments and will reject 5.21% of the shipments, so many of the shipments will be rejected.