Advertisements
Advertisements
Question
If X has a Poisson distribution with variance 2, find
Mean of X [Use e-2 = 0.1353]
Advertisements
Solution
Mean = Variance = 2
∴ Mean= 2
APPEARS IN
RELATED QUESTIONS
If X has Poisson distribution with parameter m = 1, find P[X ≤ 1] [Use `e^-1 = 0.367879`]
If X has a Poisson distribution with variance 2, find P (X = 4)
[Use e-2 = 0.1353]
If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e−1 = 0.3678
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has exactly 5 rats inclusive. Given e-5 = 0.0067.
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has more than 5 rats inclusive. Given e-5 = 0.0067.
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has between 5 and 7 rats inclusive. Given e−5 = 0.0067.
If E(X) = m and Var(X) = m then X follows ______.
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.
X : is number obtained on upper most face when a fair die is thrown then E(X) = ______
The expected value of the sum of two numbers obtained when two fair dice are rolled is ______.
Choose the correct alternative:
For the Poisson distribution ______
State whether the following statement is True or False:
If n is very large and p is very small then X follows Poisson distribution with n = mp
If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, using the following activity find the value of m.
Solution: X : Follows Poisson distribution
∴ P(X) = `("e"^-"m" "m"^x)/(x!)`, P(X = 1) = 0.4 and P(X = 2) = 0.2
∴ P(X = 1) = `square` P(X = 2).
`("e"^-"m" "m"^x)/(1!) = square ("e"^-"m" "m"^2)/(2!)`,
`"e"^-"m" = square "e"^-"m" "m"/2`, m ≠ 0
∴ m = `square`
State whether the following statement is true or false:
lf X ∼ P(m) with P(X = 1) = P(X = 2) then m = 1.
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`
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`
