If the mean and variance of a binomial variab | vedclass

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(D) For a binomial distribution,Mean $= np = 2$ and Variance $= npq = 1$.Since $q = 1 - p$,we have $np(1 - p) = 1$.Substituting $np = 2$,we get $2(1 -

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$\frac{11}{16}$

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(D) For a binomial distribution,Mean $= np = 2$ and Variance $= npq = 1$.Since $q = 1 - p$,we have $np(1 - p) = 1$.Substituting $np = 2$,we get $2(1 - p) = 1$,which implies $1 - p = 1/2$,so $p = 1/2$.Then $n(1/2) = 2$,so $n = 4$.We need to find $P(X > 1) = 1 - P(X \leq 1) = 1 - [P(X = 0) + P(X = 1)]$.Using the formula $P(X = k) = ^nC_k p^k q^{n-k}$:$P(X = 0) = ^4C_0 (1/2)^0 (1/2)^4 = 1 	imes 1 	imes 1/16 = 1/16$.$P(X = 1) = ^4C_1 (1/2)^1 (1/2)^3 = 4 	imes 1/2 	imes 1/8 = 4/16$.Therefore,$P(X > 1) = 1 - (1/16 + 4/16) = 1 - 5/16 = 11/16$.

If the mean and variance of a binomial variable $X$ are $2$ and $1$ respectively,then $P(X>1)=$

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