Question

Please Help:)

Answer 1

@mathmate please help:)

Answer 2

There are two ways to do this.

A. Approximate, but easy: use the same defect proportion as the sample.
This applies to a population of infinite size.
6/200 = x/5000
solve for x.

B. use the hypergeometric distribution equation since the population is finite (5000).
We define:
A=Total number of defectives in the population
a=number of defectives in the sample
B=Total number of non-defectives in the population
b=number of non-defectives in the sample.
For the given problem, it is obvious that A+B=5000, and a+b=200.
The probability of defectives in the population (a day's output) is given by
P(defective)=C(A,a)*C(B,b)/C(A+B,a+b)
where C(n,r) is the number of combinations of choosing r items from n, and
C(n,r)=n!/((n-r)!r!)

Substituting numbers,
P(defective)=C(200,6)*C(4800,x)/C(5000,x+6)
Since we do not know x, nor the probability, we need to use the principle of maximum likelihood, meaning we need to find a value of x which gives the maximum probability P(defective).
By trial and error, we will find that P(defective)=0.16644 is a maximum at when 143<x<144, and therefore the best estimate for the number of defectives for a day's output is n=x+6, where 149<n<150.

This may sound trivial from comparison with solution A. In fact, solution B is a tool to be used when either the sample size is large relative to population, or the value of P(defective) is high. Part B introduces the method, and confirms that the method is valid.