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 = -2) + P(x = 0)

= \(\frac { 1 }{4}\) + \(\frac { 1 }{4}\) = \(\frac { 2 }{4}\) = \(\frac { 1 }{2}\)

(ii) P(x < 0) = P(x = -2) = \(\frac { 1 }{4}\)

(iii) P(|x| ≤ 2) = P(-2 ≤ X ≤ 2)

= P(X = -2) + P(X = 0) = \(\frac { 1 }{4}\) + \(\frac { 1 }{4}\)

= \(\frac { 2 }{4}\) = \(\frac { 1 }{2}\)

(iv) p(0 ≤ X ≤ 10) = P (X = 0) + P(X = 10)

= \(\frac { 1 }{4}\) + \(\frac { 1 }{2}\) = \(\frac { 1+2 }{4}\) = \(\frac { 3 }{4}\)

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 continuous variable of density function

Question 6.

Prove that if E(X) = 0, then V(X) = E(X²)

Solution:

V(X) = E(X²) – [E(X)]²

= E(X²) – 0 {Given that E(X) = 0}

Var(X) = E(X²)

Hence proved.

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:

LHS = V(ax)

= E(ax)² – [E(ax)]²

= a² E(x²) – [aE(x)]²

= a²E(x²) – a²E(x)]²

= a²E(x²) – E(x)²]

= a²v(x)

= RHS

Hence proved

(ii) 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

LHS = RHS Hence proved.

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