The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively. The P(X=1) is

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The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively. The P(X=1) is
A. 1//4
B. 1//32
C. 1//16
D. 1//8

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Let n and p be the parameters of the binomial distribution.
We have,
Mean =4 and Variance =2
rArr np=4 and npq=2
rArr p=1//2 =q and n=8
therefore Required Probability =P(X-1)=.^8C_(1)((1)/(2))^1((1)/(2))^7=(1)/(32)