In a Binomial distribution,if n is the number | vedclass

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(A) Let $X$ be the binomial variate for which mean $= 4$ and variance $= 3$. Then $np = 4$ and $npq = 3$. Dividing the variance by the mean,we get $q

Vedclass · Accepted Answer

${}^{16}C_8(3^8)$

Vedclass · Answer

(A) Let $X$ be the binomial variate for which mean $= 4$ and variance $= 3$. Then $np = 4$ and $npq = 3$. Dividing the variance by the mean,we get $q = \frac{3}{4}$. Since $p = 1 - q$,we have $p = 1 - \frac{3}{4} = \frac{1}{4}$. Substituting $p$ into $np = 4$,we get $n 	imes \frac{1}{4} = 4$,so $n = 16$. The probability mass function is $P(X=k) = {}^{n}C_k p^k q^{n-k}$. We need to calculate $2^{32} P\left(X=\frac{16}{2}ight) = 2^{32} P(X=8)$. $P(X=8) = {}^{16}C_8 \left(\frac{1}{4}ight)^8 \left(\frac{3}{4}ight)^{16-8} = {}^{16}C_8 \left(\frac{1}{4}ight)^8 \left(\frac{3}{4}ight)^8 = {}^{16}C_8 \frac{3^8}{4^{16}}$. Since $4^{16} = (2^2)^{16} = 2^{32}$,we have $P(X=8) = {}^{16}C_8 \frac{3^8}{2^{32}}$. Therefore,$2^{32} P(X=8) = 2^{32} 	imes {}^{16}C_8 \frac{3^8}{2^{32}} = {}^{16}C_8 (3^8)$.

In a Binomial distribution,if $n$ is the number of trials and the mean and variance are $4$ and $3$ respectively,then $2^{32} P\left(X=\frac{n}{2}\right)=$

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