If the mean and variance of a binomial distri | vedclass

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(B) For a binomial distribution,the mean is given by $np = \frac{4}{3}$ and the variance is given by $npq = \frac{10}{9}$.Dividing the variance by the

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$\frac{741}{6^8}$

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(B) For a binomial distribution,the mean is given by $np = \frac{4}{3}$ and the variance is given by $npq = \frac{10}{9}$.Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{10/9}{4/3} = \frac{10}{9} 	imes \frac{3}{4} = \frac{30}{36} = \frac{5}{6}$.Since $p + q = 1$,we have $p = 1 - \frac{5}{6} = \frac{1}{6}$.Substituting $p = \frac{1}{6}$ into $np = \frac{4}{3}$,we get $n(\frac{1}{6}) = \frac{4}{3}$,which implies $n = \frac{4}{3} 	imes 6 = 8$.We need to find $P(X \geq 6) = P(X=6) + P(X=7) + P(X=8)$.The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k} = \binom{8}{k} (\frac{1}{6})^k (\frac{5}{6})^{8-k}$.$P(X=6) = \binom{8}{6} (\frac{1}{6})^6 (\frac{5}{6})^2 = 28 	imes \frac{25}{6^8} = \frac{700}{6^8}$.$P(X=7) = \binom{8}{7} (\frac{1}{6})^7 (\frac{5}{6})^1 = 8 	imes \frac{5}{6^8} = \frac{40}{6^8}$.$P(X=8) = \binom{8}{8} (\frac{1}{6})^8 (\frac{5}{6})^0 = 1 	imes \frac{1}{6^8} = \frac{1}{6^8}$.Summing these,$P(X \geq 6) = \frac{700 + 40 + 1}{6^8} = \frac{741}{6^8}$.

If the mean and variance of a binomial distribution are $\frac{4}{3}$ and $\frac{10}{9}$ respectively,then $P(X \geq 6)=$

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