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Question
Derive the mean and variance of binomial distribution
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Solution
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.
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