Probability Question 71
Question: The mean and variance of a binomial distribution are 6 and 4. The parameter n is
[MP PET 2000]
Options:
A) 18
B) 12
C) 10
D) 9
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ np=6,npq=4 $ \ $ \frac{npq}{np}=\frac{4}{6}\Rightarrow q=\frac{2}{3} $ and $ p=\frac{1}{3} $
$ \because $ $ np=6\Rightarrow n\times \frac{1}{3}=6 $
$ \Rightarrow n=18 $ .
BETA
