If mu and sigma^{2} are the mean and variance | vedclass

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(C) The given probability mass function is of a binomial distribution $B(n, p)$ where $n = 6$ and $p = \frac{1}{3}$.Since $q = 1 - p = 1 - \frac{1}{3}

Vedclass · Accepted Answer

$20$

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(C) The given probability mass function is of a binomial distribution $B(n, p)$ where $n = 6$ and $p = \frac{1}{3}$.Since $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$,the distribution is $P(X=x) = \binom{6}{x} \left(\frac{1}{3}ight)^{x} \left(\frac{2}{3}ight)^{6-x}$.The mean $\mu = np = 6 	imes \frac{1}{3} = 2$.The variance $\sigma^{2} = npq = 6 	imes \frac{1}{3} 	imes \frac{2}{3} = \frac{12}{9} = \frac{4}{3}$.Now,we calculate $2\mu + 12\sigma^{2} = 2(2) + 12\left(\frac{4}{3}ight) = 4 + 16 = 20$.

If $\mu$ and $\sigma^{2}$ are the mean and variance of a random variable $X$ whose probability mass function is given by $P(X=x) = \binom{6}{x} \left(\frac{1}{3}\right)^{x} \left(\frac{2}{3}\right)^{6-x}$ for $x = 0, 1, 2, \ldots, 6$,then the value of $2\mu + 12\sigma^{2}$ is:

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