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 (X^{2})

We know that Var (X) = E(X^{2}) – [E(X)]^{2}

So if E(X) = 0, Var (X) = E(X^{2})

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 x^{2} p(x) = E(X^{2})

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) = a^{2} V(X)

We know V(X) = E(X^{2}) – [E(X)]^{2}

So V(aX) = [E(a^{2} X^{2})] – [E(aX)]^{2}

= a^{2} E(X^{2}) – [aE(X)]^{2}

= a^{2} E(X^{2}) – a^{2} [E(X)]^{2}

= a^{2} {{E(X^{2}) – [E(X)]^{2}}

= a^{2} V(X)

(ii) V(X + b) = V(X)

LHS = V(X + b) = E[(X + b)^{2}] – {E(X + b)}^{2}

= E [X^{2} + 2bX + b^{2}] – [E(X) + b]^{2}

= E(X^{2}) + 2bE(X) + b^{2} – [(E(X))^{2} + b^{2} + 2bE(X)]

= E(X^{2}) + 2bE(X) + b^{2} – [E(X)]^{2} – b^{2} – 2bE(X)

= E(X^{2}) – [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