A random variable X sim B(n, p). If the value | vedclass

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(B) For a binomial distribution $X \sim B(n, p)$,the mean is given by $E(X) = np = 18$ and the variance is given by $Var(X) = npq = 12$.Dividing the v

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$55$

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(B) For a binomial distribution $X \sim B(n, p)$,the mean is given by $E(X) = np = 18$ and the variance is given by $Var(X) = npq = 12$.Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{12}{18}$.This simplifies to $q = \frac{2}{3}$.Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.Substituting $p = \frac{1}{3}$ into the mean equation $np = 18$,we get $n 	imes \frac{1}{3} = 18$,which implies $n = 54$.The random variable $X$ can take values from $0, 1, 2, \dots, n$.Therefore,the possible values of $X$ are $0, 1, 2, \dots, 54$.The total number of such values is $n + 1 = 54 + 1 = 55$.

$A$ random variable $X \sim B(n, p)$. If the values of the mean and variance of $X$ are $18$ and $12$ respectively,then the total number of possible values of $X$ is:

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