Let X sim B(n, p) with mean mu and variance s | vedclass

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(A) We know that for a binomial distribution,mean $\mu = np$ and variance $\sigma^2 = npq$. Given $\mu = 2\sigma^2$,we have $np = 2npq$,which implies

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$\frac{15}{16}$

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(A) We know that for a binomial distribution,mean $\mu = np$ and variance $\sigma^2 = npq$. Given $\mu = 2\sigma^2$,we have $np = 2npq$,which implies $q = \frac{1}{2}$. Since $p + q = 1$,we get $p = 1 - \frac{1}{2} = \frac{1}{2}$. Given $\mu + \sigma^2 = 3$,we have $np + npq = 3$. Substituting $p = \frac{1}{2}$ and $q = \frac{1}{2}$,we get $\frac{n}{2} + \frac{n}{4} = 3$. Multiplying by $4$,we get $2n + n = 12$,so $3n = 12$,which gives $n = 4$. Now,$P(X \leq 3) = 1 - P(X = 4)$. Using the binomial probability formula $P(X=k) = {^nC_k} p^k q^{n-k}$,we have $P(X=4) = {^4C_4} (\frac{1}{2})^4 (\frac{1}{2})^0 = 1 	imes \frac{1}{16} = \frac{1}{16}$. Therefore,$P(X \leq 3) = 1 - \frac{1}{16} = \frac{15}{16}$.

Let $X \sim B(n, p)$ with mean $\mu$ and variance $\sigma^2$. If $\mu=2 \sigma^2$ and $\mu+\sigma^2=3$,then $P(X \leq 3)=$

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