For a Binomial variate X,the mean is 2 and th | vedclass

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(C) Given that for a Binomial distribution,the mean is $np = 2$ and the variance is $npq = 1$. Dividing the variance by the mean,we get $q = \frac{npq

Vedclass · Accepted Answer

$1:15$

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(C) Given that for a Binomial distribution,the mean is $np = 2$ and the variance is $npq = 1$. Dividing the variance by the mean,we get $q = \frac{npq}{np} = \frac{1}{2}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$. Substituting $p = \frac{1}{2}$ into $np = 2$,we get $n(\frac{1}{2}) = 2$,so $n = 4$. The probability $P(X=0)$ is given by the formula $P(X=k) = { }^n C_k p^k q^{n-k}$. For $X=0$,$P(X=0) = { }^4 C_0 (\frac{1}{2})^0 (\frac{1}{2})^4 = 1 	imes 1 	imes \frac{1}{16} = \frac{1}{16}$. The odds in favor of an event $E$ are given by $P(E) : (1 - P(E))$. Therefore,the odds in favor of $X=0$ are $\frac{1}{16} : (1 - \frac{1}{16}) = \frac{1}{16} : \frac{15}{16} = 1 : 15$.

For a Binomial variate $X$,the mean is $2$ and the variance is $1$. Then,the odds in favor of $X=0$ are:

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