Question

Using MIL STD 105 E, probability of accepting a lot acceptance number when the lot size is 600,000 units, the inspection is normal general level I. Acceptance quality level is 0.025% and the proportion of defective product in the lots is 0.4%.

This is the question: what is the probability of accepting.

Answer 1

Solution

Identifying the single sampling plan to the given stipulations

Step 1: Locate the Sample Size Code Letter

For the lot size 600,000 units, the inspection normal general level I, the code letter is N.

Step 2: Locate the sampling plan against N and AQL 0.025%

The chart indicates: 500 , 0, 1

Step 3: Interpret Step 2 result

500 , 0, 1 => Take a random sample size 500, inspect and no defective is found accept the lot; if 1 or more defectives, reject the lot.

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then, probability mass function (pmf) of X is given by

p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n …………………..(1)

[This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST……………………………………….(1a)

Mean (average) of X = E(X) = µ = np………………………………………………..(2)

Variance of X = V(X) = σ² = np(1 – p)…………………………………………………..(3)

Standard Deviation of X = SD(X) = σ = √{np(1 – p)} ………………………………………...(4)

If X ~ B(n, p), np ≥ 10 and np(1 - p) ≥ 10, then Binomial probability can be approximated by Standard Normal probabilities by Z = (X – np)/√{np(1 - p)} ~ N(0, 1) ……………………..(5)

Now, to work out the solution,

Probability of acceptance of a lot with proportion of defective as 0.4%. P(A).

If X = number of defectives in a sample of 500 units from a lot with proportion of defective as 0.4%, then X ~ B(500, 0.004)

So, P(A) = P(X = 0)

= (⁵⁰⁰C₀)(0.04⁰)(0.96)⁵⁰⁰ [vide (1)]

= 0.96⁵⁰⁰

= 1.367 x 10^-9 = 0.000000001367 Answer

Since n is large, the above probability can also be evaluated by Normal approximation – vide (5)

DONE