The random variable X follows a binomial dist | vedclass

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(C) For a binomial distribution $B(n, p)$,the mean is $np$ and the variance is $npq$,where $q = 1-p$.Given $np - npq = 1$,we have $np(1-q) = 1$,which

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$11$

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(C) For a binomial distribution $B(n, p)$,the mean is $np$ and the variance is $npq$,where $q = 1-p$.Given $np - npq = 1$,we have $np(1-q) = 1$,which implies $np^2 = 1$.Given $2 P(X=2) = 3 P(X=1)$,we substitute the binomial probability formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$:$2 \binom{n}{2} p^2 q^{n-2} = 3 \binom{n}{1} p^1 q^{n-1}$$2 \cdot \frac{n(n-1)}{2} p^2 q^{n-2} = 3n p q^{n-1}$$(n-1) p = 3q$Since $q = 1-p$,we have $(n-1)p = 3(1-p) \Rightarrow np - p = 3 - 3p \Rightarrow np + 2p = 3$.From $np^2 = 1$,we have $n = \frac{1}{p^2}$. Substituting this into $np + 2p = 3$:$\frac{1}{p^2} \cdot p + 2p = 3 \Rightarrow \frac{1}{p} + 2p = 3 \Rightarrow 2p^2 - 3p + 1 = 0$.Solving the quadratic equation: $(2p-1)(p-1) = 0$. Since $p Then $n = \frac{1}{(1/2)^2} = 4$.We need to find $n^2 P(X > 1) = 16(1 - (P(X=0) + P(X=1)))$.$P(X=0) = \binom{4}{0} (1/2)^4 = 1/16$.$P(X=1) = \binom{4}{1} (1/2)^1 (1/2)^3 = 4/16 = 1/4$.$P(X > 1) = 1 - (1/16 + 4/16) = 1 - 5/16 = 11/16$.Thus,$n^2 P(X > 1) = 16 	imes \frac{11}{16} = 11$.

The random variable $X$ follows a binomial distribution $B(n, p)$ for which the difference between the mean and the variance is $1$. If $2 P(X=2) = 3 P(X=1)$,then $n^2 P(X > 1)$ is equal to $......$.

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