statistics

Variance of binomial distribution proof

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The_Don. answered · 15/04/2021

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Variance of Binomial Distribution Proof

E(X) is the expected value of Binomial Distribution

$V a r (x) = E (X^{2}) - {[E (X)]}^{2}$

$M_{X}^{I I} (t) = n P e^{t} (n - 1) {(P e^{t} + q)}^{n - 2} \times P e^{t} + {(P e^{t} + q)}^{n - 1} \times n P e^{t}$

Replace t =0

$M_{X}^{I I} (0) = n P e^{0} (n - 1) {(P e^{0} + q)}^{n - 2} \times P e^{0} + {(P e^{0} + q)}^{n - 1} \times n P e^{0}$

$L e t n P e^{t} = U a n d {(P e^{t} + q)}^{n - 1} = V$

$n p^{2} (n - 1) + n p = n^{2} p^{2} - n p^{2} + n p$

$V a r (x) = E (X^{2}) - {[E (X)]}^{2}$

$= n^{2} p^{2} - n p^{2} + n p - {(n p)}^{2}$

$= n^{2} p^{2} - n^{2} p^{2} + n p - {(n p)}^{2}$

$= n p - {n p}^{2}$

$= n p (1 - p)$

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