Question

Let N=15,000 and a double sampling plan with n1=40, n2= 80, c1=1, c2=2, r1=r2=c2+1. The lotOs fraction defective is 2%. Compute the probability of accepting lot in the first sample.

Answer 1

Solution

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}, …………...........................................................................….................……..(1)

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

Now to work out the solution,

‘double sampling plan with n1=40, n2= 80, c1=1, c2=2, r1=r2=c2+1 means:

Take the first sample of 40 units. If the number of defectives is less than or equal to 1,

accept the lot; if it is 2 take the second sample of 40 units; if it is 3 or more, reject the lot.

Since we are interested in the first sample only, let

X = number of defectives in a sample of 40 units.

Then, X ~ B(40, p), where p = lot fraction defective......................................................................................................(2)

Given p = 0.02 [i.e., 2%],

Probability of accepting lot in the first sample

= P(X ≤ 1)

= P(X = 0) + P(X = 1)

= (⁴⁰C₀)(0.02⁰)(0.98)⁴⁰ + (⁴⁰C₁)(0.02¹)(0.98)³⁹ [vide (1)]

= (0.98)⁴⁰ + (0.8 x 0.98³⁹)

= 0.8095 Answer

DONE