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

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(D) For a binomial distribution $X \sim B(n, p)$,the mean $\mu = np$ and variance $\sigma^2 = npq$,where $q = 1-p$.Given $\mu = 2\sigma^2$,we have $np

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$\frac{15}{16}$

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(D) For a binomial distribution $X \sim B(n, p)$,the mean $\mu = np$ and variance $\sigma^2 = npq$,where $q = 1-p$.Given $\mu = 2\sigma^2$,we have $np = 2npq$,which implies $1 = 2q$,so $q = \frac{1}{2}$ and $p = 1 - q = \frac{1}{2}$.Given $\mu + \sigma^2 = 3$,we substitute $\mu = 2\sigma^2$ to get $3\sigma^2 = 3$,so $\sigma^2 = 1$.Since $\sigma^2 = npq = n(\frac{1}{2})(\frac{1}{2}) = \frac{n}{4} = 1$,we find $n = 4$.Thus,$X \sim B(4, \frac{1}{2})$.We need to calculate $P(X \leq 3) = 1 - P(X = 4)$.$P(X = 4) = \binom{4}{4} p^4 q^0 = 1 	imes (\frac{1}{2})^4 	imes 1 = \frac{1}{16}$.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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