7 coins are tossed simultaneously and the num | vedclass

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Question

(C) For a binomial distribution with $n=7$ and $p=q=\frac{1}{2}$:$\mu = np = 7 	imes \frac{1}{2} = \frac{7}{2}$$\sigma^2 = npq = 7 	imes \frac{1}{2}

Vedclass · Accepted Answer

$\frac{112}{5}$

Vedclass · Answer

(C) For a binomial distribution with $n=7$ and $p=q=\frac{1}{2}$:$\mu = np = 7 	imes \frac{1}{2} = \frac{7}{2}$$\sigma^2 = npq = 7 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{7}{4}$$P(X=3) = {}^{7}C_{3} 	imes (\frac{1}{2})^{3} 	imes (\frac{1}{2})^{4} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} 	imes (\frac{1}{2})^{7} = 35 	imes \frac{1}{128} = \frac{35}{128}$$\frac{\mu \sigma^2}{P(X=3)} = \frac{(\frac{7}{2}) 	imes (\frac{7}{4})}{\frac{35}{128}} = \frac{49}{8} 	imes \frac{128}{35} = \frac{7}{1} 	imes \frac{16}{5} = \frac{112}{5}$

$7$ coins are tossed simultaneously and the number of heads turned up is denoted by the random variable $X$. If $\mu$ is the mean and $\sigma^2$ is the variance of $X$,then $\frac{\mu \sigma^2}{P(X=3)}=$

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