Question

Solve the following problem :

If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, find variance of X.

Answer 1

Given, P[X = 1] = 0.4, P[X = 2] = 0.2,

e^–1 = 0.3678

For Poisson distribution,

X ~ P(m)

The p.m.f. of X is given by

P[X = x] = `("e"^(-"m")"m"^x)/(x!)`

Now,
P[X = 1] = `("e"^(-"m")"m"^1)/(1!)` = me^-m

∴ 0.4 = me^-m               ...(i)

P[X = 2] = `("e"^(-"m")"m"^2)/(2!)` = `("m"^2"e"^(-"m"))/(2)`

∴ 0.2 = `("m"^2"e"^(-"m"))/(2)`

∴ 0.4 = m² e^–m          ...(ii)

∴ `(0.4)/(0.4) = ("m"^2"e"^(-"m"))/("me"^(-"m"))`     ...[From (i) and (ii)]

∴ m = 1

∴ X ~ P(1)

∴ Var (X) = m = 1.