The mean and variance of a random variable X | vedclass

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(B) For a binomial distribution,the mean is given by $np = 6$ and the variance is given by $npq = 3$. Dividing the variance by the mean,we get $q = \f

Vedclass · Accepted Answer

$\frac{13}{4096}$

Vedclass · Answer

(B) For a binomial distribution,the mean is given by $np = 6$ and the variance is given by $npq = 3$. Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{3}{6} = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $np = 6$,we get $n(\frac{1}{2}) = 6$,which implies $n = 12$. The probability mass function is $P(X = k) = \binom{n}{k} p^k q^{n-k} = \binom{12}{k} (\frac{1}{2})^k (\frac{1}{2})^{12-k} = \binom{12}{k} (\frac{1}{2})^{12}$. We need to find $P(X $P(X = 0) = \binom{12}{0} (\frac{1}{2})^{12} = 1 	imes \frac{1}{4096} = \frac{1}{4096}$. $P(X = 1) = \binom{12}{1} (\frac{1}{2})^{12} = 12 	imes \frac{1}{4096} = \frac{12}{4096}$. Therefore,$P(X < 2) = \frac{1}{4096} + \frac{12}{4096} = \frac{13}{4096}$.

The mean and variance of a random variable $X$ having a binomial distribution are $6$ and $3$ respectively. The probability of variable $X$ less than $2$ is . . . . . . .

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