Advertisements
Advertisements
प्रश्न
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.
उत्तर
Let X denote the number of rats per bungalow.
Given, m = 5 and e–5 = 0.0067
∴ X ~ P(m) ≡ X ~ P(5)
The p.m.f. of X is given by
P(X = x) = `("e"^-"m" "m"^x)/(x!)`
∴ P(X = x) = `("e"^-5*(5)^x)/(x!), x` = 0, 1, ..., 5
P(exactly five rats)
= P(X = 5)
= `("e"^-5*(5)^5)/(5!)`
= `(0.0067 xx 5^5)/(5 xx 4 xx 3 xx 2 xx1)`
= `(0.0067 xx 625)/(24)`
= `(4.1875)/(24)`
= 0.1745
संबंधित प्रश्न
The number of complaints which a bank manager receives per day is a Poisson random variable with parameter m = 4. Find the probability that the manager will receive -
(a) only two complaints on any given day.
(b) at most two complaints on any given day
[Use e-4 =0.0183]
If a random variable X follows Poisson distribution such that P(X = l) =P(X = 2), then find P(X ≥ 1). [Use e-2 = 0.1353]
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 a Poisson distribution with variance 2, find
Mean of X [Use e-2 = 0.1353]
If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e−1 = 0.3678
If X~P(0.5), then find P(X = 3) given e−0.5 = 0.6065.
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
The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives only two complaints on a given day
The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives a) only two complaints on a given day, b) at most two complaints on a given day. Use e−4 = 0.0183.
A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows Poisson distribution with mean 1.5. Find the probability that (i) no car is used on a given day, (ii) some demand is refused on a given day, given e−1.5 = 0.2231.
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.
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.
Solve the following problem :
If X follows Poisson distribution with parameter m such that
`("P"("X" = x + 1))/("P"("X" = x)) = (2)/(x + 1)`
Find mean and variance of X.
X : is number obtained on upper most face when a fair die is thrown then E(X) = ______
Choose the correct alternative:
For the Poisson distribution ______
State whether the following statement is True or False:
X is the number obtained on upper most face when a die is thrown, then E(x) = 3.5
State whether the following statement is True or False:
A discrete random variable X is said to follow the Poisson distribution with parameter m ≥ 0 if its p.m.f. is given by P(X = x) = `("e"^(-"m")"m"^"x")/"x"`, x = 0, 1, 2, .....
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.
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)
