Advertisements
Advertisements
प्रश्न
If X has Poisson distribution with parameter m = 1, find P[X ≤ 1] [Use `e^-1 = 0.367879`]
Advertisements
उत्तर
Given X follows Poisson distribution with parameter m = 1
∴ X ∼ P(1)
p.m.f. is given by
P(X = x) = `(e^-m m^x)/(x!)`
= `(e^-1 1^x)/(x!)`
P(X = x) = `e^-1/(x!)`
Now P[X ≤ l] = P (X = 0) + P (X = 1)
= `e^-1/(0!) + e^-1/(1!)`
= `2e^-1`
= 2(0.367879)
= 0.735758
APPEARS IN
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 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~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
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.
If E(X) = m and Var(X) = m then X follows ______.
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 ______
Choose the correct alternative:
A distance random variable X is said to have the Poisson distribution with parameter m if its p.m.f. is given by P(x) = `("e"^(-"m")"m"^"x")/("x"!)` the condition for m is ______
The probability that a bomb will hit the target is 0.8. Using the following activity, find the probability that, out of 5 bombs, exactly 2 will miss the target
Solution: Let p = probability that bomb miss the target
∴ q = `square`, p = `square`, n = 5.
X ~ B`(5, square)`, P(x) = `""^"n""C"_x"P"^x"q"^("n" - x)`
P(X = 2) = `""^5"C"_2 square = 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`
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`
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`
