Question

In a binomial distribution $B(n, p)$, the sum and the product of the mean and the variance are 5 and 6 respectively, then $6(n+p-q)$ is equal to

• A. 50
• B. 53
• C. 52
• D. 51

Answer 1

Answer: (C)

52

Answer 2

np+npq=5,np⋅npq=6
np(1+q)=5,n2p2q=6
n2p2(1+q)2=25,n2p2q=6
q6​(1+q)2=25
6q2+12q+6=25q
6q2−13q+6=0
6q2−9q−4q+6=0
(3q−2)(2q−3)=0
q=32​,23​,q=32​ is accepted
p=31​⇒n⋅31​+n⋅31​⋅32​=5
93n+2n​=5
n=9
So 6(n+p−q)=6(9+31​−32​)=52