The mean and variance of a binomial variable | vedclass

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(B) For a binomial distribution,the mean is $np = 2$ and the variance is $npq = 1$.Dividing the variance by the mean,we get $q = \frac{1}{2}$.Since $p

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$\frac{11}{16}$

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(B) For a binomial distribution,the mean is $np = 2$ and the variance is $npq = 1$.Dividing the variance by the mean,we get $q = \frac{1}{2}$.Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n(\frac{1}{2}) = 2$,so $n = 4$.The probability mass function is $P(X = k) = \binom{n}{k} p^k q^{n-k} = \binom{4}{k} (\frac{1}{2})^k (\frac{1}{2})^{4-k} = \binom{4}{k} (\frac{1}{2})^4 = \binom{4}{k} \frac{1}{16}$.We need to find $P(X > 1) = P(X = 2) + P(X = 3) + P(X = 4)$.$P(X = 2) = \binom{4}{2} \frac{1}{16} = 6 	imes \frac{1}{16} = \frac{6}{16}$.$P(X = 3) = \binom{4}{3} \frac{1}{16} = 4 	imes \frac{1}{16} = \frac{4}{16}$.$P(X = 4) = \binom{4}{4} \frac{1}{16} = 1 	imes \frac{1}{16} = \frac{1}{16}$.Therefore,$P(X > 1) = \frac{6}{16} + \frac{4}{16} + \frac{1}{16} = \frac{11}{16}$.

The mean and variance of a binomial variable $X$ are $2$ and $1$ respectively. Then,the probability that $X$ takes values greater than $1$ is

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