If the sum and product of the mean and varian | vedclass

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(A) For a binomial distribution with $n$ trials and probability of success $p$,let $q = 1 - p$. Mean $\mu = np$ and Variance $\sigma^2 = npq$. Given:

Vedclass · Accepted Answer

$27$

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(A) For a binomial distribution with $n$ trials and probability of success $p$,let $q = 1 - p$. Mean $\mu = np$ and Variance $\sigma^2 = npq$. Given: $np + npq = 15$ $np(npq) = 54$ Let $X = np$ and $Y = npq$. Then $X + Y = 15$ and $XY = 54$. These are roots of the quadratic equation $t^2 - 15t + 54 = 0$. $(t - 9)(t - 6) = 0$,so the roots are $9$ and $6$. Since $np > npq$ (because $q Dividing the two equations: $\frac{npq}{np} = \frac{6}{9} \implies q = \frac{2}{3}$. Then $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$. Substituting $p$ into $np = 9$: $n(\frac{1}{3}) = 9 \implies n = 27$. Thus,the number of trials is $27$.

If the sum and product of the mean and variance of a binomial distribution are $15$ and $54$ respectively,then the number of trials in it is

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