For a binomial variate X with parameters n=5 | vedclass

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(C) We know that the binomial distribution is given by $P(X=r) = { }^n C_r p^r q^{n-r}$ where $p+q=1$. Given $n=5, p=\frac{3}{4}, q=\frac{1}{4}$. Firs

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(C) We know that the binomial distribution is given by $P(X=r) = { }^n C_r p^r q^{n-r}$ where $p+q=1$. Given $n=5, p=\frac{3}{4}, q=\frac{1}{4}$. First,calculate $P(X \geq 3) = P(X=3) + P(X=4) + P(X=5)$. $P(X=3) = { }^5 C_3 (\frac{3}{4})^3 (\frac{1}{4})^2 = 10 	imes \frac{27}{64} 	imes \frac{1}{16} = \frac{270}{1024}$. $P(X=4) = { }^5 C_4 (\frac{3}{4})^4 (\frac{1}{4})^1 = 5 	imes \frac{81}{256} 	imes \frac{1}{4} = \frac{405}{1024}$. $P(X=5) = { }^5 C_5 (\frac{3}{4})^5 (\frac{1}{4})^0 = 1 	imes \frac{243}{1024} 	imes 1 = \frac{243}{1024}$. $P(X \geq 3) = \frac{270+405+243}{1024} = \frac{918}{1024}$. Then $\alpha = \frac{1}{9} 	imes \frac{918}{1024} = \frac{102}{1024}$. Next,calculate $\beta = P(X \leq 2) = 1 - P(X \geq 3) = 1 - \frac{918}{1024} = \frac{106}{1024}$. Finally,$256(\beta - \alpha) = 256(\frac{106}{1024} - \frac{102}{1024}) = 256(\frac{4}{1024}) = 256(\frac{1}{256}) = 1$.

For a binomial variate $X$ with parameters $n=5$ and $p=\frac{3}{4}$. If $\alpha=\frac{1}{9} P(X \geq 3)$ and $\beta=P(X \leq 2)$,then $256(\beta-\alpha)=$

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