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(B) For a binomial distribution $B(n, p)$,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1-p$.Given that the sum of mean and v

Vedclass · Accepted Answer

$52$

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(B) For a binomial distribution $B(n, p)$,the mean is $\mu = np$ and the variance is $\sigma^2 = npq$,where $q = 1-p$.Given that the sum of mean and variance is $5$: $np + npq = 5 \Rightarrow np(1+q) = 5$.Given that the product of mean and variance is $6$: $np \cdot npq = 6 \Rightarrow n^2p^2q = 6$.From the first equation,$np = \frac{5}{1+q}$. Substituting this into the second equation:$(\frac{5}{1+q})^2 \cdot q = 6 \Rightarrow 25q = 6(1+q)^2$.$25q = 6(1 + 2q + q^2) \Rightarrow 6q^2 + 12q + 6 = 25q$.$6q^2 - 13q + 6 = 0$.Solving the quadratic equation: $6q^2 - 9q - 4q + 6 = 0 \Rightarrow 3q(2q-3) - 2(2q-3) = 0$.$(3q-2)(2q-3) = 0$. Since $q Then $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.Using $np(1+q) = 5$: $n(\frac{1}{3})(1 + \frac{2}{3}) = 5 \Rightarrow n(\frac{1}{3})(\frac{5}{3}) = 5 \Rightarrow n(\frac{5}{9}) = 5 \Rightarrow n = 9$.Finally,$6(n+p-q) = 6(9 + \frac{1}{3} - \frac{2}{3}) = 6(9 - \frac{1}{3}) = 54 - 2 = 52$.

In a binomial distribution $B(n, p)$,the sum and product of the mean and variance are $5$ and $6$ respectively. Find the value of $6(n+p-q)$.

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