HSC Commerce (English Medium) १२ वीं कक्षा - Maharashtra State Board Important Questions for Mathematics and Statistics

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

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.

The expected value of the sum of two numbers obtained when two fair dice are rolled is ______.

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 ______

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`

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`

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)

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`