A random variable X has a probability mass fu | vedclass

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(B) The given probability mass function is $P(X=x) = \frac{{}^4C_x}{2^4} = {}^4C_x \left(\frac{1}{2}\right)^x \left(\frac{1}{2}\right)^{4-x}$.This is

Vedclass · Accepted Answer

$\mu = 2, \sigma^2 = 1$

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(B) The given probability mass function is $P(X=x) = \frac{{}^4C_x}{2^4} = {}^4C_x \left(\frac{1}{2}ight)^x \left(\frac{1}{2}ight)^{4-x}$.This is a binomial distribution $B(n, p)$ with $n = 4$ and $p = \frac{1}{2}$.The mean $\mu$ of a binomial distribution is given by $\mu = np = 4 	imes \frac{1}{2} = 2$.The variance $\sigma^2$ of a binomial distribution is given by $\sigma^2 = npq$,where $q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$.Thus,$\sigma^2 = 4 	imes \frac{1}{2} 	imes \frac{1}{2} = 1$.Therefore,$\mu = 2$ and $\sigma^2 = 1$.

$A$ random variable $X$ has a probability mass function $P(X=x) = \frac{{}^4C_x}{2^4}$ for $x = 0, 1, 2, 3, 4$. If $\mu$ and $\sigma^2$ are the mean and variance of the random variable $X$ respectively,then:

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