Question

=> Find the Binomial distribution whose mean is 9 and variance is 6 ?​

Answer 1

Concept:-In Binomial distribution,mean is taken as product of no. of trials(n) and probability of success(p)

Variance is the product of probability of failure(1-p)×mean

Solution:- Mean of distribution=9

n×p=9 -i)

Variance of distribution=6

np×(1-p)=6=>n-np^2=6 --ii)

dividing i) equation by ii) we get

np/{n(1-p)}=9/6

1/1-p=9/6

6=9-9p

p=1/3

putting this value in equation i) we get

n×1/3=9

n=27

this shows that no. of trial is 27

so, binomial distribution=0,1,3,...,27

Answer 2

Concept:-In Binomial distribution,mean is taken as product of no. of trials(n) and probability of success(p)

Variance is the product of probability of failure(1-p)×mean

Solution:- Mean of distribution=9

n×p=9 -i)

Variance of distribution=6

np×(1-p)=6=>n-np^2=6 --ii)

dividing i) equation by ii) we get

np/{n(1-p)}=9/6

1/1-p=9/6

6=9-9p

p=1/3

putting this value in equation i) we get

n×1/3=9

n=27

this shows that no. of trial is 27

so, binomial distribution=0,1,3,...,27