For a binomial variate X sim B(n, p),the diff | vedclass

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(B) Given that the difference between the mean and variance is $1$: $np - npq = 1 \Rightarrow np(1-q) = 1 \Rightarrow np^2 = 1$ $\qquad (i)$ Also,the

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$7 : 10$

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(B) Given that the difference between the mean and variance is $1$: $np - npq = 1 \Rightarrow np(1-q) = 1 \Rightarrow np^2 = 1$ $\qquad (i)$ Also,the difference between their squares is $11$: $(np)^2 - (npq)^2 = 11$ $\qquad (ii)$ $n^2p^2 - n^2p^2q^2 = 11 \Rightarrow n^2p^2(1-q^2) = 11$ Dividing $(ii)$ by $(i)$,we get: $\frac{n^2p^2(1-q^2)}{np^2} = 11 \Rightarrow n(1-q^2) = 11$ Since $n=36$,$36(1-q^2) = 11 \Rightarrow 1-q^2 = \frac{11}{36} \Rightarrow q^2 = 1 - \frac{11}{36} = \frac{25}{36} \Rightarrow q = \frac{5}{6}$ Thus,$p = 1 - q = 1 - \frac{5}{6} = \frac{1}{6}$ Now,$P(X=2) = {}^{36}C_2 \left(\frac{1}{6}ight)^2 \left(\frac{5}{6}ight)^{34} = m\left(\frac{5}{6}ight)^{36}$ $\frac{36 	imes 35}{2} 	imes \frac{1}{36} 	imes \left(\frac{5}{6}ight)^{34} = m\left(\frac{5}{6}ight)^{36}$ $\frac{35}{2} 	imes \left(\frac{5}{6}ight)^{34} = m 	imes \left(\frac{5}{6}ight)^{34} 	imes \left(\frac{5}{6}ight)^2$ $m = \frac{35}{2} 	imes \left(\frac{6}{5}ight)^2 = \frac{35}{2} 	imes \frac{36}{25} = \frac{7 	imes 18}{5} = \frac{126}{5} = 25.2$ Wait,re-evaluating $m$: $m = \frac{35}{2} 	imes \frac{36}{25} = \frac{7 	imes 18}{5} = 25.2$. Given $m : n = 25.2 : 36 = 252 : 360 = 7 : 10$.

For a binomial variate $X \sim B(n, p)$,the difference between the mean and variance is $1$ and the difference between their squares is $11$. If the probability $P(X=2) = m\left(\frac{5}{6}\right)^n$ and $n=36$,then $m : n =$

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