If the sum of mean and variance of a Binomial | vedclass

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(B) For a Binomial Distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $n$ is the number of trials,$p$ is the probability o

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(B) For a Binomial Distribution,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $n$ is the number of trials,$p$ is the probability of success,and $q = 1-p$ is the probability of failure.Given $n = 10$ and $	ext{mean} + 	ext{variance} = \frac{15}{2}$.Substituting the formulas: $np + npq = \frac{15}{2}$.Since $q = 1-p$,we have $np + np(1-p) = \frac{15}{2}$.$10p + 10p(1-p) = 7.5$.$10p + 10p - 10p^2 = 7.5$.$20p - 10p^2 = 7.5$.Divide by $2.5$: $8p - 4p^2 = 3$.$4p^2 - 8p + 3 = 0$.Solving the quadratic equation: $(2p-1)(2p-3) = 0$.This gives $p = \frac{1}{2}$ or $p = \frac{3}{2}$.Since $0 Then $q = 1 - \frac{1}{2} = \frac{1}{2}$.Therefore,the variance is $\sigma^2 = npq = 10 	imes \frac{1}{2} 	imes \frac{1}{2} = 2.5$.

If the sum of mean and variance of a Binomial Distribution is $\frac{15}{2}$ for $10$ trials,then the variance is (in $.5$)

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