The mean and variance of a binomial distribut | vedclass

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(C) For a binomial distribution,the mean is $np = \alpha$ and the variance is $npq = \frac{\alpha}{3}$.Dividing the variance by the mean,we get $q = \

Vedclass · Accepted Answer

$\frac{16}{27}$

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(C) For a binomial distribution,the mean is $np = \alpha$ and the variance is $npq = \frac{\alpha}{3}$.Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{\alpha/3}{\alpha} = \frac{1}{3}$.Since $p + q = 1$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$.Given $P(X=1) = \frac{4}{243}$,we use the formula $P(X=k) = {}^{n}C_{k} p^{k} q^{n-k}$.${}^{n}C_{1} (\frac{2}{3})^{1} (\frac{1}{3})^{n-1} = \frac{4}{243} \implies n \cdot \frac{2}{3} \cdot \frac{1}{3^{n-1}} = \frac{4}{243} \implies \frac{2n}{3^{n}} = \frac{4}{243} \implies \frac{n}{3^{n}} = \frac{2}{243}$.Since $3^{5} = 243$,we have $n=6$.Now,we calculate $P(X=4 	ext{ or } 5) = P(X=4) + P(X=5)$.$P(X=4) = {}^{6}C_{4} (\frac{2}{3})^{4} (\frac{1}{3})^{2} = 15 \cdot \frac{16}{81} \cdot \frac{1}{9} = \frac{240}{729} = \frac{80}{243}$.$P(X=5) = {}^{6}C_{5} (\frac{2}{3})^{5} (\frac{1}{3})^{1} = 6 \cdot \frac{32}{243} \cdot \frac{1}{3} = \frac{192}{729} = \frac{64}{243}$.$P(X=4 	ext{ or } 5) = \frac{80}{243} + \frac{64}{243} = \frac{144}{243} = \frac{16}{27}$.

The mean and variance of a binomial distribution are $\alpha$ and $\frac{\alpha}{3}$ respectively. If $P(X=1)=\frac{4}{243}$,then $P(X=4 \text{ or } 5)$ is equal to.

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