Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 1.
The probability function of a random variable X is given by

Evaluate the following probabilities.
(i) P(X ≤ 0)
(ii) P(X ≤ 0)
(iii) P( X ≤ 2) and
(iv) P(0 ≤ X ≤ 10)
Solution:
(i) From the data

x	-2	0	10
P(x = x)	1/4	1/4	1/2

(i) P(X ≤ 0) = P (X = 0) + P (X = -2)
\(=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
(ii) P(X < 0) = P (X = – 2) = \(\frac{1}{4}\)
(iii) P(|X| ≤ 2) = P(-2 ≤ X ≤ 2)
= P(X = -2) + P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac{1}{4}\) + 0 + \(\frac{1}{4}\) + 0 + 0
= \(\frac{1}{2}\)
(iv) P(0 ≤ X ≤ 10) = P(X = 0) + P(X = 10) + 0
\(=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\)

Question 2.
The probability function of a random variable X is given by

(a) Compute: (i) P(1 ≤ X ≤ 2) and (ii) P(X = 3).
(b) Is X a discrete random variable? Justify your answer.
Solution:
W.K.T Probability density Function
f(x) = \(\frac { d[F(x)] }{dx}\)

(ii) P(X = 3) = 0
(b) X is not discrete since f is not a step function.

Question 3.
The p.d.f. of X is defined as
f(x) = \(\left\{\begin{array}{l}
\mathrm{k}, \text { for } 0<x \leq 4 \\
0, \text { otherwise }
\end{array}\right.\)
Solution:
Let X and a random variable if a Probability density function

Question 4.
The probability distribution function of a discrete random variable X is

where k is some constant. Find (a) k and (b) P(X > 2).
Solution:
(a) Let X be the random variable of a probability distribution function
W.K.T Σpi = 1
P(x = 1) + P(x = 3) + P(x = 5) = 1
2k + 3k + 4k = 1
9k – 1 ⇒ k = 1/9

(b) P(x > 1) = P(x = 3) + P(x = 5)
= 3k + 4k = 7k
= 7(1/9) = 7/9

Question 5.
The probability distribution function of a discrete random variable X is
f(x) = \(\left\{\begin{array}{l}
\mathrm{a}+\mathrm{b} x^{2}, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
where a and b are some constants. Find (i) a and b if E(X) = 3/5 (ii) Var(X).
Solution:
Let X be due to continuous variable of the density function

Question 6.
Prove that if E(X) = 0, then V(X) = E(X²)
Solution:
Given E(X) = 0. To show V(X) = E (X²)
We know that Var (X) = E(X²) – [E(X)]²
So if E(X) = 0, Var (X) = E(X²)
From the definition of the variance of X also we can see the result.
Var(X) = Σ[x – E(x)]² p(x)
If E (X) = 0, then V(X) = Σ x² p(x) = E(X²)

Question 7.
What is the expected value of a game that works as follows: I flip a coin and if tails pay you ₹ 2; if heads pay you ₹ 1. In either case, I also pay you ₹ 50.
Solution:
Let x be the remain variable denoting the amount paying for a game of flip coin then x and takes 2 and 1
P(X = 2) = \(\frac { 1 }{2}\) (getting a head)
p(X = 1) = \(\frac { 1 }{2}\) (getting a tail)
Hence the probability of X is

x	2	1
P(x = x)	1/2	1/2

Expected value E(X) = \(\sum_{ x }\)x P(x)
= (2)(\(\frac { 1 }{2}\)) + 1 (\(\frac { 1 }{2}\))
= 1 + \(\frac { 1 }{2}\) = \(\frac { 3 }{2}\)
Since I pay you ₹ 50 in either case
E(X) = 50 × 3/2 = ₹ 75

Question 8.
Prove that, (i) V(aX) = a²V(X) , and (ii) V(X + b) = V(X).
Solution:
(i) To show V(aX) = a² V(X)
We know V(X) = E(X²) – [E(X)]²
So V(aX) = [E(a² X²)] – [E(aX)]²
= a² E(X²) – [aE(X)]²
= a² E(X²) – a² [E(X)]²
= a² {{E(X²) – [E(X)]²}
= a² V(X)

(ii) V(X + b) = V(X)
LHS = V(X + b) = E[(X + b)²] – {E(X + b)}²
= E [X² + 2bX + b²] – [E(X) + b]²
= E(X²) + 2bE(X) + b² – [(E(X))² + b² + 2bE(X)]
= E(X²) + 2bE(X) + b² – [E(X)]² – b² – 2bE(X)
= E(X²) – [E(X)]²
= V(X)
= RHS

Question 9.
Consider a random variable X with p.d.f.
f(x) = \(\left\{\begin{array}{l}
3 x^{2}, \text { if } 00 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected life of this piece of equipment.
Solution:
Let X be the random variable

Question 10.
The time to failure in thousands of hours of an important piece of electronic equipment used in a manufactured DVD player has the density function
f(x) = \(\left\{\begin{array}{l}
2 e^{-2 x}, x>0 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected life of this piece of equipment.
Solution:
Let X be the random variable