The mean and variance of a binomial variate X | vedclass

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(A) For a binomial distribution,the mean is $np = 16/5$ and the variance is $npq = 48/25$.Dividing the variance by the mean,we get $q = (48/25) / (16/

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$19$

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(A) For a binomial distribution,the mean is $np = 16/5$ and the variance is $npq = 48/25$.Dividing the variance by the mean,we get $q = (48/25) / (16/5) = (48/25) 	imes (5/16) = 3/5$.Since $p + q = 1$,we have $p = 1 - 3/5 = 2/5$.Substituting $p$ into the mean equation: $n(2/5) = 16/5 \Rightarrow n = 8$.We need to find $P(X > 1) = 1 - P(X = 0) - P(X = 1)$.$P(X = 0) = ^8C_0 (2/5)^0 (3/5)^8 = (3/5)^8$.$P(X = 1) = ^8C_1 (2/5)^1 (3/5)^7 = 8 	imes (2/5) 	imes (3/5)^7 = (16/5) 	imes (3/5)^7$.Thus,$P(X > 1) = 1 - (3/5)(3/5)^7 - (16/5)(3/5)^7 = 1 - (3/5 + 16/5)(3/5)^7 = 1 - (19/5)(3/5)^7$.Comparing this with $1 - K(3/5)^7$,we get $K = 19/5$.Therefore,$5K = 5 	imes (19/5) = 19$.

The mean and variance of a binomial variate $X$ are $16/5$ and $48/25$ respectively. If $P(X > 1) = 1 - K (3/5)^7$,then $5 K =$

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