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Sum

**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.

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#### Solution

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 = m^{2} 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.

Concept: Poisson Distribution

Is there an error in this question or solution?