If the mean and variance of a Binomial variab | vedclass

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(C) Given that the mean $\mu = np = \frac{5}{2}$ and the variance $\sigma^2 = npq = \frac{5}{4}$.Dividing the variance by the mean,we get $q = \frac{n

Vedclass · Accepted Answer

$\frac{13}{16}$

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(C) Given that the mean $\mu = np = \frac{5}{2}$ and the variance $\sigma^2 = npq = \frac{5}{4}$.Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{5/4}{5/2} = \frac{1}{2}$.Since $p = 1 - q$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.Substituting $p = \frac{1}{2}$ into $np = \frac{5}{2}$,we get $n 	imes \frac{1}{2} = \frac{5}{2}$,which implies $n = 5$.We need to find $P(X > 1) = 1 - \{P(X = 0) + P(X = 1)\}$.Using the Binomial probability formula $P(X = k) = {}^nC_k p^k q^{n-k}$:$P(X = 0) = {}^5C_0 (\frac{1}{2})^0 (\frac{1}{2})^5 = 1 	imes 1 	imes \frac{1}{32} = \frac{1}{32}$.$P(X = 1) = {}^5C_1 (\frac{1}{2})^1 (\frac{1}{2})^4 = 5 	imes \frac{1}{2} 	imes \frac{1}{16} = \frac{5}{32}$.Therefore,$P(X > 1) = 1 - (\frac{1}{32} + \frac{5}{32}) = 1 - \frac{6}{32} = 1 - \frac{3}{16} = \frac{13}{16}$.

If the mean and variance of a Binomial variable $X$ are $\frac{5}{2}$ and $\frac{5}{4}$ respectively,then $P(X>1)=$

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