Balbharati Maharashtra State Board12th Commerce Maths Solution Book PdfChapter 8 Probability Distributions Ex 8.2 Questions and Answers.
Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2
Question 1
Maharashtra Board Solution
Check whether each of the following is p.d.f. (i)
Solution & Step-by-Step Answer:
Given function is f(x) = x, 0 ≤ x ≤ 1 Each f(x) ≥ 0, as x ≥ 0. ∴ The given function is a p.d.f. of x.
(ii) f(x) = 2 for 0 < x < 1
Solution:
Given function is
f(x) = 2 for 0 < x < 1 Each f(x) > 0,
∴ The given function is not a p.d.f.
Question 2
Maharashtra Board Solution
The following is the p.d.f. of a r.v. X. Find (i) P(X < 1.5), (ii) P(1 < X < 2), (iii) P(X > 2)
Solution & Step-by-Step Answer:
Question 3
Maharashtra Board Solution
It is felt that error in measurement of reaction temperature (in Celsius) in an experiment is a continuous r.v. with p.d.f. (i) Verify whether f(x) is a p.d.f. (ii) Find P(0 < X ≤ 1). (iii) Find the probability that X is between 1 and 3.
Solution & Step-by-Step Answer:
(i) f(x) is p.d.f. of r.v. X if (a) f(x) ≥ 0, ∀ x ∈ R (b) = 1
Question 4
Maharashtra Board Solution
Find k, if the following function represents the p.d.f. of a r.v. X. (i) Also find P[ < X < ]
Solution & Step-by-Step Answer:
(ii)
Also find (a) P[ < X < ], (b) P[X < ]
Solution:
We know that
Question 5
Maharashtra Board Solution
Let X be the amount of time for which a book is taken out of the library by a randomly selected student and suppose that X has p.d.f. Calculate (i) P(X ≤ 1), (ii) P(0.5 ≤ X ≤ 1.5), (iii) P(X ≥ 1.5).
Solution & Step-by-Step Answer:
Given p.d.f. of X is f(x) = 0.5x for 0 ≤ x ≤ 2 ∴ Its c.d.f. F(x) is given by
(i) P(X < 1) = F(1)
= 0.25(1)2
= 0.25
(ii) P(0.5 < X < 1.5) = F(1.5) – F(0.5)
= 0.25(1.5)2– 0.25(0.5)2
= 0.25[2.25 – 0.25]
= 0.25(2)
= 0.5
(iii) P(X ≥ 1.5) = 1 – P(X ≤ 1.5)
= 1 – F(1.5)
= 1 – 0.25(1.5)2
= 1 – 0.25(2.25)
= 1 – 0.5625
= 0.4375
Question 6
Maharashtra Board Solution
Suppose X is the waiting time (in minutes) for a bus and its p.d.f. is given by Find the probability that (i) waiting time is between 1 and 3 minutes, (ii) waiting time is more than 4 minutes.
Solution & Step-by-Step Answer:
p.d.f. of r.v. X is given by f(x) = for 0 ≤ x ≤ 5 This is a constant function. (i) Probability that waiting time X is between 1 and 3 minutes (ii) Probability that waiting time X is more than 4 minutes
Question 7
Maharashtra Board Solution
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f. Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution & Step-by-Step Answer:
Since given f(x) is a p.d.f. of r.v. X Since -2 ≤ x ≤ 2 ∴ x2 ≤ 4 ∴ 4 – x2 ≥ 0 ∴ k(4 – x2) ≥ 0 ∴ k ≥ 0 [∵ f(x) ≥ 0]
Question 8
Maharashtra Board Solution
Following is the p.d.f. of a continuous r.v. X. (i) Find an expression for the c.d.f. of X. (ii) Find F(x) at x = 0.5, 1.7, and 5.
Solution & Step-by-Step Answer:
The p.d.f. of a continuous r.v. X is (i) c.d.f. of continuous r.v. X is given by
(ii) F(0.5) = = 0.015
F(1.7) = = 0.18
For any of x greater than or equal to 4, F(x) = 1
∴ F(5) = 1
Question 9
Maharashtra Board Solution
The p.d.f. of a continuous r.v. X is Determine the c.d.f. of X and hence find (i) P(X < 1), (ii) P(X < -2), (iii) P(X > 0), (iv) P(1 < X < 2).
Solution & Step-by-Step Answer:
The p.d.f. of a continuous r.v. X is
Question 10
Maharashtra Board Solution
If a r.v. X has p.d.f. Find c, E(X) and V(X). Also find f(x).
Solution & Step-by-Step Answer:
The p.d.f. of r.v. X is f(x) = , 1 < x < 3, c > 0 For p.d.f. of X, we have
