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If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e–3 = 0.0497.
P[X = x] = `square`
Since P[X = 2] = P[X = 3]
`square` = `square`
`m^2/2 = m^3/6`
∴ m = `square`
Now, P[X ≥ 2] = 1 – P[x < 2]
= 1 – {P[X = 0] + P[X = 1]
= `1 - {square/(0!) + square/(1!)}`
= 1 – e–3[1 + 3]
= 1 – `square` = `square`
Concept: Poisson Distribution
If X – P(m) with P(X = 1) = P(X = 2), then find the mean and P(X = 2) given that e–2 = 0.1353
Solution: Since P(X = 1) = P(X = 2)
`(e^-mm^1)/(1!) = square`
∴ m = `square`
∴ mean = `square` = `square`
Then P(X = 2) = `square` = `square`
Concept: Poisson Distribution
In a town, 10 accidents take place in the span of 50 days. Assuming that the number of accidents follows Poisson distribution, find the probability that there will be 3 or more accidents on a day.
(Given that e-0.2 = 0.8187)
Solution:
Here, m = `square` and X − P(m) with parameter m.
The p.m.f. X is:
P(X = x) = `(e^(−m).m^x)/(x!), x = 0, 1, 2,...`
P(X ≥ 3) = 1 − P(X < 3)
= 1 − [`square + square + square`]
= `1 − [(e^(− 0.2)(0.2)^0)/(0!) + (e^(−0.2)(0.2)^1)/(1!) + (e^(−0.2)(0.2)^2)/(2!)]`
= 1 − [0.8187(1 + 0.2 + 0.02)]
= 1 − `square`
= `square`
Concept: Poisson Distribution
If X ∼ P(m) with P(X = 1) = P(X = 2), then find the mean and P(X = 2).
Given e–2 = 0.1353
Solution: Since P(X = 1) = P(X = 2)
∴ `("e"^square"m"^1)/(1!) = ("e"^"-m""m"^2)/square`
∴ m = `square`
∴ P(X = 2) = `("e"^-2. "m"^2)/(2!)` = `square`
Concept: Poisson Distribution
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Concept: Nature of Management Principles
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Concept: Concept of Management Principles
