In hypergeometric distribution, n= population of n elements, r = sample size, n_{1 =} elements recognized as having some defined criteria, n_{2 =} n - n₁ =  remaining elements other than n₁ . We seek the probability q_k such that the sample size r contain exactly k recognized elements provided [math]k \geq 0, k \leq n_1 [/math] if n₁ is smaller  or [math] k \leq r[/math] if r is smaller. In such a case the probability [math]q_k =\frac{\binom{n_1}{k}\binom{n- n_1}{r - k}}{\binom{n}{r}} \tag {1}[/math]

Now, I want to calculate n=8500, n₁ =1000, r = 1000, k = 0 to 100.  Inserting these values in (1), calculation of summation is very difficult.  In such case, how can I use normal approximation to binomial distribution to find q_k ?

Do you have any clue or hint?

     

Edited April 1, 20223 yr by Dhamnekar Win,odd

I don't know about the normal approximation in this case, but Excel has HYPGEOMDIST() function...

On 4/1/2022 at 3:59 PM, Dhamnekar Win,odd said:

 In hypergeometric distribution, n= population of n elements, r = sample size, n_{1 =} elements recognized as having some defined criteria, n_{2 =} n - n₁ =  remaining elements other than n₁ . We seek the probability q_k such that the sample size r contain exactly k recognized elements provided k≥0,k≤n1 if n₁ is smaller  or k≤r if r is smaller. In such a case the probability qk=(n1k)(n−n1r−k)(nr)(1)

Now, I want to calculate n=8500, n₁ =1000, r = 1000, k = 0 to 100.  Inserting these values in (1), calculation of summation is very difficult.  In such case, how can I use normal approximation to binomial distribution to find q_k ?

Do you have any clue or hint?

     

Minecraft random block ticks used for crop ripening sequences etc. run on the hypergeometric distribution.

For many trials of low probability events, the Poisson Approximation is remarkably accurate.