For a binomial distribution B(n, p),if the me | vedclass

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(C) In a binomial distribution $B(n, p)$,the mean is given by $\mu = np$ and the standard deviation is given by $\sigma = \sqrt{npq}$.Given,$np = 200$

Vedclass · Accepted Answer

$160008$

Vedclass · Answer

(C) In a binomial distribution $B(n, p)$,the mean is given by $\mu = np$ and the standard deviation is given by $\sigma = \sqrt{npq}$.Given,$np = 200$ and $\sqrt{npq} = 10$.Squaring the standard deviation equation,we get $npq = 100$.Substituting $np = 200$ into $npq = 100$,we get $200q = 100$,which implies $q = \frac{1}{2}$.Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.Now,$np = 200 \implies n(\frac{1}{2}) = 200 \implies n = 400$.We need to calculate $n^2 + \frac{1}{p^2} + \frac{1}{q^2}$.Substituting the values: $(400)^2 + \frac{1}{(1/2)^2} + \frac{1}{(1/2)^2} = 160000 + 4 + 4 = 160008$.

For a binomial distribution $B(n, p)$,if the mean $= 200$ and the standard deviation $= 10$,then the value of $n^2 + \frac{1}{p^2} + \frac{1}{q^2}$ is equal to:

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