The mean and variance of a binomial distribut | vedclass

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Question

(A) For a binomial distribution,the mean is given by $np = 5$ and the variance is given by $npq = 4$. Substituting $np = 5$ into the variance formula,

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$\frac{4^{24}}{5^{23}}$

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(A) For a binomial distribution,the mean is given by $np = 5$ and the variance is given by $npq = 4$. Substituting $np = 5$ into the variance formula,we get $5q = 4$,which implies $q = \frac{4}{5}$. Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{4}{5} = \frac{1}{5}$. Using $np = 5$ and $p = \frac{1}{5}$,we find $n = \frac{5}{p} = 5 	imes 5 = 25$. The probability mass function for a binomial distribution is $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$. For $X=1$,we have $P(X=1) = {}^{25}C_{1} \left(\frac{1}{5}ight)^{1} \left(\frac{4}{5}ight)^{25-1}$. $P(X=1) = 25 	imes \frac{1}{5} 	imes \left(\frac{4}{5}ight)^{24} = 5 	imes \frac{4^{24}}{5^{24}} = \frac{4^{24}}{5^{23}}$.

The mean and variance of a binomial distribution are $5$ and $4$ respectively. Then what is $P(X=1)$?

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