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प्रश्न
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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उत्तर
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.
संबंधित प्रश्न
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If X has a Poisson distribution with variance 2, find
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