HSC Commerce (English Medium) इयत्ता १२ वी - 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.

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)

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`

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`