Advertisements
Advertisements
Question
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`
Advertisements
Solution
Since P(X = 1) = P(X = 2)
`(e^-mm^1)/(1!) = bb((e^-mm^2)/(2!))`
∴ m = 2
∴ mean = m = 2
Then P(X = 2) = `bb((e^-mm^2)/(2!)) = bb((e^-2 2^2)/(2!))`
= `bb((0.1353 xx 4)/2)`
= 0.2706
APPEARS IN
RELATED QUESTIONS
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 has Poisson distribution with m = 1, then find P(X ≤ 1) given e−1 = 0.3678
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 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 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) = ______
The expected value of the sum of two numbers obtained when two fair dice are rolled is ______.
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, .....
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`
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)
