If X is a binomial variate with mean frac{16} | vedclass

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(A) For a binomial distribution,mean $\mu = np = \frac{16}{5}$ and variance $\sigma^2 = npq = \frac{48}{25}$.Dividing variance by mean,we get $q = \fr

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$\frac{3^6(169)}{5^8}$

Vedclass · Answer

(A) For a binomial distribution,mean $\mu = np = \frac{16}{5}$ and variance $\sigma^2 = npq = \frac{48}{25}$.Dividing variance by mean,we get $q = \frac{npq}{np} = \frac{48/25}{16/5} = \frac{48}{25} 	imes \frac{5}{16} = \frac{3}{5}$.Since $p + q = 1$,we have $p = 1 - \frac{3}{5} = \frac{2}{5}$.Substituting $p$ in $np = \frac{16}{5}$,we get $n 	imes \frac{2}{5} = \frac{16}{5}$,so $n = 8$.The probability mass function is $P(X = k) = \binom{n}{k} p^k q^{n-k} = \binom{8}{k} (\frac{2}{5})^k (\frac{3}{5})^{8-k}$.We need $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.$P(X=0) = \binom{8}{0} (\frac{2}{5})^0 (\frac{3}{5})^8 = \frac{3^8}{5^8}$.$P(X=1) = \binom{8}{1} (\frac{2}{5})^1 (\frac{3}{5})^7 = 8 	imes \frac{2}{5} 	imes \frac{3^7}{5^7} = \frac{16 	imes 3^7}{5^8} = \frac{16 	imes 3 	imes 3^6}{5^8} = \frac{48 	imes 3^6}{5^8}$.$P(X=2) = \binom{8}{2} (\frac{2}{5})^2 (\frac{3}{5})^6 = 28 	imes \frac{4}{25} 	imes \frac{3^6}{5^6} = \frac{112 	imes 3^6}{5^8}$.Summing these: $P(X \leq 2) = \frac{3^8 + 48 	imes 3^6 + 112 	imes 3^6}{5^8} = \frac{9 	imes 3^6 + 160 	imes 3^6}{5^8} = \frac{169 	imes 3^6}{5^8}$.

If $X$ is a binomial variate with mean $\frac{16}{5}$ and variance $\frac{48}{25}$,then $P(X \leq 2) = $

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