Question

Derive the mean and variance of binomial distribution

Answer 1

Derivation of the Mean and Variance of Binomial distribution:

The mean of the binomial distribution

`"E"("X") =sum_(x = 0)^"n" (("n"),(x))"p"^x "q"^("n" - x)`

= `"p" sum_(x = 0)^"n" x*(("n")/(x)) (("n"- 1),(x - 1)) "p"^(x - 1)"q"^("n"- x)`

= np(q + p)n – 1 ......[since p + q = 1]

= np

E(X) = np

The mean of the binomial distribution is np.

Var(X) = E(X²) – E(X²)

Here `"E"("X"^2) = sum_(x = 0)^"n" x^2 (("n"),(x))"p"^x"q"^("n" - x)`

`sum_(x - 0)^"n" {x(x - 1) + x} (("n"),(x))"p"^x"q"^("n" - x)`

`sum_(x - 0)^"n" {x(x - 1) + x} (("n"),(x))"p"^x"q"^("n" - x) + sum x (("n"),(x))"p"^x"q"^("n" - x)`

`sum_(x = 0)^"n" {x(x - 1)} (("n"("n" - 1))/(x(x - 1)))(("n" - 2),(x - 2))"p"^(x - 2)"q"^(n - x) + sum x (("n"),(x))"p"^x"q"^("n" - x)`

= `"n"("n" - 1)"p"^2 {sum(("n" - 2),(x - ))"p"^(x - 2)"q"^("n" - x)} + "np"`

= n(n – 1)p²(q + p)(n – 2) + np

n(n – 1 )p² + np

Variance = E(X²) – [E(X)]²

= n²p² – np² + np – n²p²

= np(1 – p) = npq

Hence, mean of the BD is np and the Variance is npq.