Question

Consider the following data regarding the acceptance sampling process:

N = 10,000, n = 89, c = 2, P = 0.01, and P_a = 0.9397

The Average Total Inspection (ATI) will be

1. 795
2. 687
3. 595
4. 487

Answer 1

Correct Answer - Option 2 : 687

Concept:

In Acceptance sampling process a small part or a fraction of the units/items is selected randomly from a lot and then selected units are inspected. The decision to accept the lot or reject the lot is based on information received by inspecting selected sample.

The average total inspection per lot is based on the incoming quality, probability of acceptance of lot (P_a) and the sample and lot sizes and is given by

Average total inspection (ATI) = n + (1 – P_a)(N – n)

Where,

N = Lot size

n = Sample size

c = Acceptance number

P_a = Probability of accepting lot

P = Quality of lot

Calculation:

Given N = 10,000, n = 89, c = 2, P = 0.01, and P_a = 0.9397

Average total inspection (ATI) = n + (1 – P_a)(N – n)

⇒ ATI = 89 + (1 – 0.9397) (10000 – 89) = 686.633 ∼ 687