The sum and product of the mean and variance | vedclass

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Question

(D) Let the mean be $m = np$ and the variance be $v = npq$,where $p + q = 1$.Given,the sum $m + v = 82.5 = \frac{165}{2}$ and the product $mv = 1350$.

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(D) Let the mean be $m = np$ and the variance be $v = npq$,where $p + q = 1$.Given,the sum $m + v = 82.5 = \frac{165}{2}$ and the product $mv = 1350$.Since $m$ and $v$ are roots of the quadratic equation $x^2 - (m+v)x + mv = 0$,we have:$x^2 - \frac{165}{2}x + 1350 = 0$$2x^2 - 165x + 2700 = 0$Solving this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{165 \pm \sqrt{165^2 - 4(2)(2700)}}{2(2)} = \frac{165 \pm \sqrt{27225 - 21600}}{4} = \frac{165 \pm \sqrt{5625}}{4} = \frac{165 \pm 75}{4}$Thus,$x_1 = \frac{240}{4} = 60$ and $x_2 = \frac{90}{4} = 22.5$.Since $m > v$ for a binomial distribution (as $q Now,$v = mq \implies 22.5 = 60q \implies q = \frac{22.5}{60} = \frac{225}{600} = \frac{3}{8}$.Since $p = 1 - q$,we have $p = 1 - \frac{3}{8} = \frac{5}{8}$.Finally,$m = np \implies 60 = n 	imes \frac{5}{8} \implies n = 60 	imes \frac{8}{5} = 12 	imes 8 = 96$.

The sum and product of the mean and variance of a binomial distribution are $82.5$ and $1350$ respectively. The number of trials in the binomial distribution is:

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