The mean and variance of a binomial variable | vedclass

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(C) Given,mean $np = 2$ $(i)$ And variance $npq = 1$ $(ii)$ Dividing $(ii)$ by $(i)$,we get $q = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - \fra

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$\frac{11}{16}$

Vedclass · Answer

(C) Given,mean $np = 2$ $(i)$ And variance $npq = 1$ $(ii)$ Dividing $(ii)$ by $(i)$,we get $q = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $(i)$,we get $n 	imes \frac{1}{2} = 2$,so $n = 4$. The binomial distribution is given by $P(X = k) = {}^nC_k p^k q^{n-k} = {}^4C_k (\frac{1}{2})^k (\frac{1}{2})^{4-k} = {}^4C_k (\frac{1}{2})^4$. We need to find $P(X > 1) = P(X = 2) + P(X = 3) + P(X = 4)$. $P(X > 1) = {}^4C_2 (\frac{1}{2})^4 + {}^4C_3 (\frac{1}{2})^4 + {}^4C_4 (\frac{1}{2})^4$. $P(X > 1) = (6 + 4 + 1) 	imes \frac{1}{16} = \frac{11}{16}$.

The mean and variance of a binomial variable $X$ are $2$ and $1$ respectively. The probability that $X$ takes values greater than $1$ is:

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