Question

 A shipment contains 200 items of which 50 are defective. A sample of 16 items from the shipment is selected at random without replacement. We accept the shipment if at most 3 items in the sample

 are defective.

 (a) Write down (but do not evaluate) an exact formula for the probability of acceptance.

 (b) Use a Table to give the decimal value for the binomial approximation of the probability of acceptance. Show your work.

 (c) Suppose instead that the shipment contains 500 items of which 50 are defective. We still sample 16 items at random without replacement and accept the shipment if at most 3 items in the sample are

 defective. Repeat part (b) for the new data.

Answer 1

Back-up Theory

Hyper-geometric distribution

If a sample n is taken without replacement from a finite (small) population of size N in which M have an attribute (and hence N – M do not possess that attribute), the number of sample units, X possessing that attribute follows a Hyper-geometric distribution with parameters N, M, n.

Probability mass function, pmf is: p(x) = P(X = x) = {(^MC_x)(^{N - M}C_{n – x})/(^NC_n)} …………………...............................…. (1)

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 …………...................................……………………..………..(2)

Now, to work out the solution,

Let X = Number of defectives in a sample of 16. ………………...............................……………………………………… (3)

Part (a)

This probability is based on Hyper-geometric distribution, vide (1), where in our case,

N = 200, n = 16, M = 50. So, the required probability is:

P(X ≤ 3)

= Σ(x = 0 to 3){(⁵⁰C_x)(¹⁵⁰C_{16 – x})/(²⁰⁰C₁₆)} Answer

Part (b)

For Binomial approximation, p = 50/200 = 0.25. So, vide (2),

x	p(x) - formula	p(x) –value from Table
0	(15C0)(0.250)(0.75)15	0.0100
1	(15C1)(0.251)(0.75)14	0.0535
2	(15C2)(0.252)(0.75)13	0.1336
3	(15C3)(0.253)(0.75)12	0.2079
	Total	0.4050

Thus, probability of acceptance = 0.4050 Answer

Part (c)

Here, p = 50/500 = 0.1. And working is identical to the working of Part (b)

x	p(x) - formula	p(x) –value from Table
0	(15C0)(0.10)(0.9)15	0.1853
1	(15C1)(0.11)(0.9)14	0.3294
2	(15C2)(0.12)(0.9)13	0.2745
3	(15C3)(0.13)(0.9)12	0.1423
	Total	0.9316

Thus, probability of acceptance = 0.9316   Answer

DONE