Hypergeometric Distribution

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Hypergeometric Distribution

Hypergeometric Distribution

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  • 1. 1.11 Hypergeometric Distribution
  • 2. Hypergeometric Distribution Suppose we are interested in the number of defectives in a sample of size n units drawn from a lot containing N units, of which a are defective. Each object has same chance of being selected, then the probability that the first drawing will yield a defective unit = a/N but for the second drawing, probability a 1 if first unit is defective, N 1 a if first unit is not defective. N 1
  • 3. Hypergeometric Distribution • The trials here are not independent and hence the fourth assumption underlying the binomial distribution is not fulfilled and therefore, we cannot apply binomial distribution here. • Binomial distribution would have been applied if we do sampling with replacement, viz., if each unit selected from the sample would have been replaced before the next one is drawn.
  • 4. Hypergeometric Distribution Sampling without replacement Number of ways in which x successes (defectives) can be chosen is a x Number of ways in which n – x failures (non defectives) be chosen is N a n x Hence number of ways x successes and n – x failures can be chosen is a N a x n x
  • 5. Hypergeometric Distribution Number of ways n objects can be chosen from N objects is N If all the possibilities are equally likely then for sampling n without replacement the probability of getting “x successes in n trials” is given by a N a x n x h ( x; n, a , N ) for x 0 , 1,...., n N n where x a, n x N a.
  • 6. Hypergeometric Distribution • The solution of the problem of sampling without replacement gave birth to the above distribution which we termed as hypergeometric distribution. • The parameters of hypergeometric distribution are the sample size n, the lot size (or population size) N, and the number of “successes” in the lot a. • When n is small compared to N, the composition of the lot is not seriously affected by drawing the sample and the binomial distribution with parameters n and p = a/N will yield a good approximation.
  • 7. Hypergeometric Distribution The difference between the two values is only 0.010. In general it can be shown that h( x; n, a, N) b( x; n, p) with p = (a/N) when N ∞. A good rule of thumb is to use the binomial distribution as an approximation to the hyper-geometric distribution if n/N ≤0.05
  • 8. The Mean and the Variance of a Probability Distribution Mean of hypergeometric distribution a n sample size n N population size N a number of success Proof: a N a n n x n x x .h ( x ; n , a , N ) x. N x 0 x 1 n
  • 9. The Mean and the Variance of a Probability Distribution a a! a (a 1)! a a 1 x x! ( a x )! x (x 1)! ( a x )! x x 1 a 1 N a n n a 1 N a x 1 n x a a. N N x 1 n x x 1 x 1 n n
  • 10. The Mean and the Variance of a Probability Distribution Put x – 1= y k n 1 n 1 a a 1 N a m a 1 N y 0 y n 1 y r y n s N a k m s m s Use the identity r k r k r 0
  • 11. The Mean and the Variance of a Probability Distribution We get a N 1 N n 1 n a n N
  • 12. Variance of hypergeometric distribution 2 n a (N a) (N n) 2 N (N 1) Proof: a N a n n 2 2 x n x 2 x .h ( x ; n , a , N ) x . x 0 N x 1 n
  • 13. a 1 N a n x 1 n x 2 a x. N x 1 n n a 1 N a a (x 1 1). N x 1 n x x 1 n
  • 14. n a 2 N a a (a 1) 2 . N x 2 n x x 2 n n a 1 N a a N x 1 n x x 1 n Put x – 2 = y in 1st summation and x – 1 = z in 2nd one
  • 15. n 2 a 2 N a a (a 1) 2 . N y n 2 y y 0 n n 1 a 1 N a a N z n 1 z z 0 n k m s m s Use the identity r k r k r 0 k n 2, m a 2 k n 1, m a 1 r y, s N a r z, s N a
  • 16. a (a 1) N 2 a N 1 2 N n 2 N n 1 n n n(n 1) n a (a 1) a N (N 1) N 2 2 n a (N a) (N n) 2 2 N (N 1)