Suppose that X has a hypergeometric distribution with N=500, n=50

Reggie

Reggie

Answered question

2021-09-22

Suppose that X has a hypergeometric distribution with N=500,n=50, and  K=100. Find P(X=10) by combining the approximate binomial distribution with the hypergeometric distribution. Can we trust the binomial approximation?

Answer & Explanation

Usamah Prosser

Usamah Prosser

Skilled2021-09-23Added 86 answers

Step 1 
Hypergeometric Distribution: 
N: Number of items in the population 
K: Number of items in the population classified as success 
n: Number of items in the sample 
x: Number of items in the sample classified as success 
The hypergeometric probability distribution is: 
h(x,N,n,k)=(10010)(5001005010)(50050) 
=(10010)(40040)(50050) 
Use an Excel spreadsheet or an online hypergeometric calculator to determine the probability because the calculation takes a while.
Using Excel function, “=HYPGEOM.DIST(10,50,100,500,FALSE)”, the probability is 0.1474. 
Step 2 
Binomial distribution: 
n: Sample size 
p: Probability of success 
x: Number of items in the sample classified as success 
Binomial probability distribution is, 
B(x;n,p)=(nx)px(1p)nx 
where p=KN
Given: N=500,n=50,K=100 and  x=10
That is, p=100500=0.2
Thus, the Binomial distribution is to be obtained for, 
B(10;50,0.2)=(5010)0.210(10.2)5010 
=(5010)0.210(0.8)40 
=0.1398 
The hypergeometric probability of 0.1474 and the binomial estimate of 0.1398 are rather similar.
As a result, the Binomial approximation makes sense.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school probability

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?