Students can download 12th Business Maths Chapter 6 Random Variable and Mathematical Expectation Additional Problems and Answers, Samacheer Kalvi 12th Business Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

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

One Mark Questions

Question 1.
If a fair coin is tossed three times the probability function p(x) of the number of heads x is _______
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems I Q1
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems I Q1.1
(d) None of these
Answer:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems I Q1.2
Hint:
The sample space is HHH, HHT, HTH, HTT, THH, THT, TTH and TTT
Number of heads is 0, 1, 2, 3 with probability \(\frac{1}{8}, \frac{3}{8}, \frac{3}{8}\) and \(\frac{1}{8}\)

Question 2.
If a discrete random variable has the probability mass function as
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems I Q2
then the value of K is _______
(a) \(\frac{1}{11}\)
(b) \(\frac{1}{12}\)
(c) \(\frac{2}{13}\)
(d) \(\frac{4}{9}\)
Answer:
(b) \(\frac{1}{12}\)
Hint:
k + 3 k + 6k + 2k = 1
⇒ 12k = 1
⇒ k = \(\frac{1}{12}\)

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

Question 3.
If the probability density function of X is f(x) = Cx (2 – x), and 0 < x < 2, then value of C is ______
(a) \(\frac{4}{3}\)
(b) \(\frac{6}{5}\)
(c) \(\frac{3}{4}\)
(d) \(\frac{3}{5}\)
Answer:
(c) \(\frac{3}{4}\)
Hint:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems I Q3

Question 4.
The random variables X and Y are independent if ______
(a) E(XY) = 1
(b) E(XY) = 0
(c) E(XY) = E(X) E(Y)
(d) E(X + Y) = E(X) + E(Y)
Answer:
(c) E(XY) = E(X) E(Y)

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

Question 5.
If a random variable X has the following distribution
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems I Q5
then the expected value of X is ______
(a) \(\frac{3}{2}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{3}\)
Answer:
(b) \(\frac{1}{6}\)
Hint:
E(X) = \(\frac{-1}{3}-\frac{2}{6}+\frac{1}{6}+\frac{2}{3}=\frac{1}{3}-\frac{1}{6}=\frac{1}{6}\)

Question 6.
Var (4X + 7) = _____
(a) 7
(b) 16 Var (X)
(c) 11
(d) None of these
Answer:
(b) 16 Var (X)
Hint:
Var (4X + 7) = (42) Var (X)

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

Question 7.
Match the following:

(a) Random variable (i) Arithmetic mean
(b) Speed of a car (ii) Discrete variable
(c) E(aX + b) (iii) Chance variable
(d) No. of students in a class (iv) Continuous variable
(e) E(X) (v) aE(X) + b

Answer:
(a) – (iii)
(b) – (iv)
(c) – (v)
(d) – (ii)
(e) – (i)

Question 8.
If E(X) = 2 and E(Z) = 4, then E (Z – X) is ______
(a) 2
(b) 6
(c) 0
(d) -2
Answer:
(a) 2
Hint:
E(Z – X) = E(Z) – E(X) = 4 – 2 = 2

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

Question 9.
Fill in the blanks:

  1. The distribution function F (X) is equal to _______
  2. Two types of random variables are ______ and ______
  3. Probability mass function is also called ______
  4. Cumulative distribution function is also called ________
  5. Probability density function is also called ______ and _______
  6. d F(x) is known as ______ of X.
  7. E(X) is denoted by _______
  8. Variance is a measure of ______ or _______ of X.
  9. Standard deviation is defined as _______
  10. Mean is the _______ of a density. Variance is the ________ of a density.

Answers:

  1. P(X ≤ x)
  2. discrete and continuous
  3. discrete probability function
  4. distribution function
  5. continuous probability function, integrating the density function
  6. probability differential
  7. µx
  8. the spread, dispersion of the density
  9. √Var[X]
  10. center of gravity, the moment of inertia.

2 Mark Questions

Question 1.
Verify whether the following function is a probability mass function or not. Hence find c.d.f.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems II Q1
Solution:
Σpi = \(\frac{1}{3}+\frac{2}{3}\) = 1 and pi > 0.
So the given function is a p.m.f. c.d.f is given by
F (x) = P (X ≤ x)
F(1) = P(X ≤ 1) = P(X = 1) = \(\frac{1}{3}\)
F(2) = P (X ≤ 2) = P(X = 1) + P (X = 2) = \(\frac{1}{3}+\frac{2}{3}\) = 1
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems II Q1.1

Question 2.
Consider the following probability distribution of X.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems II Q2
Is p(xi) a p.m.f?
Solution:
p(xi) > 0 for all i
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems II Q2.1
Hence p(xi) is a p.m.f.

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

Question 3.
The probability that a man fishing at a particular place will catch 1, 2, 3 and 4 fish are 0.4, 0.3, 0.2 and 0.1. What is the expected number of fish caught?
Solution:
Let X denote the no.of fish caught by the man.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems II Q3
E(X) = 1(0.4) + 2 (0.3) + 3 (0.2) + 4 (0.1)
= 0.4 + 0.6 + 0.6 + 0.4
= 2

Question 4.
A random variable X has the following probability function.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems II Q4
Find the value of K.
Solution:
We know that Σp(xi) = 1
⇒ 0.1 + K + 0.2 + 2K + 0.3 + K = 1
⇒ 4K + 0.6 = 1
⇒ 4K = 0.4
⇒ K = 0.1

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

Question 5.
A person receives a sum of rupees equal to the square of the number that appears on the face when a die is tossed. How much money can he expect to receive?
Solution:
Let the random variable X denote the square of the number that can appear on the face of a die. Then the distribution is given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems II Q5

3 and 5 Marks Questions

Question 1.
For the following distribution of X.
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q1
(i) P(X ≤ 1)
(ii) P(X ≤ 2 )
(iii) P(0 < X < 2)
Solution:
(i) P(X ≤ 1) = P (X = 1) + P (X = 0)
\(=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\)
(ii) P(X ≤ 2) = P(X = 2) + P(X = 1) + P(X = 0)
\(=\frac{3}{10}+\frac{1}{2}+\frac{1}{6}=\frac{29}{30}\)
(iii) P(0 < X < 2) = P(X = 1) = \(\frac{1}{2}\)

Question 2.
Given that p.d.f of a random variable X as follows
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q2
Find K and c.d.f.
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q2.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q2.2

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

Question 3.
Suppose that the life in hours of a certain part of radio tube is an r.v X with p.d.f given by
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q3
(i) What is the probability that all of three Such tubes in a given radio set will have to be replaced in the first 150 hours?
(ii) What is the probability that none of the three tubes will be replaced?
Solution:
(i) A tube in the radio set will have to be replaced during the first 150 hours if its life is less than 150 hours. Hence the required probability is
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q3.1
The probability that all three of the original tubes will have to be replaced during the first 150 hours is \(\left(\frac{1}{3}\right)^{3}=\frac{1}{27}\)
(ii) The probability that a tube is not replaced is given by P(X > 150)
= 1 – P(X ≤ 150)
= 1 – \(\frac{1}{3}\)
= \(\frac{2}{3}\)
Hence the probability that none of the three tubes will be replaced during the 150 hours of operation is \(\left(\frac{2}{3}\right)^{3}=\frac{8}{27}\)

Question 4.
Let X be a continuous random variable with p.d.f:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q4
(i) Find ‘a’
(ii) compute P(X ≤ 1.5)
Solution:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q4.1
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q4.2

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

Question 5.
A random variable X has the probability function as follows:
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q5
Find E(3X + 1), E(X2) and Var(X).
Solution:
E(X) = (-1) (0.2) + 0(0.3) + 1(0.5) = -0.2 + 0.5 = 0.3
So E(3X + 1) = 3 E(X) + 1 = 3(0.3) + 1 = 1.9
E(X2) = (-1)2 (0.2) + 02 (0.3) + 12 (0.5) = 0.2 + 0.5 = 0.7
Var(X) = E(X2) – [E(X)]2 = 0.7 – (0.3)2 = 0.61

Question 6.
A player tossed two coins. If two heads show he wins Rs.4. If one head shows he wins Rs.2, but if two tails show he must pay Rs.3 as a penalty. Calculate the expected value of ‘ the sum won by him.
Solution:
Let X be the discrete random variable denoting the sum won by the player. We know that,
Probability of getting 2 heads = \(\frac{1}{4}\)
Probability of getting 1 head is \(\frac{1}{2}\)
Probability of getting 2 tail is \(\frac{1}{4}\)
So the player wins Rs.4 with probability \(\frac{1}{4}\), he wins Rs. 2 with probability \(\frac{1}{2}\) and loses Rs.3 with probability \(\frac{1}{4}\)
Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Additional Problems III Q6
E(X) = 4(\(\frac{1}{4}\)) + 2(\(\frac{1}{2}\)) – 3(\(\frac{1}{4}\))
= 1 + 1 – \(\frac{3}{4}\)
= \(\frac{5}{4}\)
Thus the expected value of the sum won by him is Rs.1.25.

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