Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 1.

Find the expected value for the random variable of an unbiased die

Solution:

When a un based die is thrown , any one of the number 1, 2, 3, 4, 5, 6, can turn up that x denote the random variable taking the values from 1 to 6

Question 2.

Let X be a random variable defining number of students getting A grade. Find the expected value of X from the given table

Solution:

Let X be a random variable taking values 0, 1, 2, 3,

Expected value E(x) = Σp_{1}x_{1}

= (0.2 × 0) + (0.1 × 1) + (0.4 × 2) + (0.3 × 3)

= 0 + 0.1 + 0.8 + 0.9

∴ E(x) = 0.18

Question 3.

The following table is describing about the probability mass function of the random variable X

Find the standard deviation of x.

Solution:

Let x be the random variable taking the values 3, 4, 5

E(x) = Σpixi

= (0.1 × 3) + (0.1 × 4) + (0.2 × 5)

0.3 + 0.4 + 1.0

E(x) = 1.7

E(x²) = Σpixi²

= (0.1 × 3²)+ (0.1 × 4²) + (0.2 × 5²)

= (0.1 × 9) + (0.1 × 16) + (0.2 × 25)

E(x²) = 7.5

Var(x) = E(x²) – (E(x)]²

= 7.5 – (1.7)²

= 7.5 – 2.89

Var(x) = 4.61

Standard deviation(S.D) = \(\sqrt { var (x)}\)

= \(\sqrt { 4.61}\)

σ = 2.15

Question 4.

Let X be a continuous random variable with probability density function

f_{x} (x) = \(\left\{\begin{array}{l}

2 x, 0 \leq x \leq 1 \\

0, \text { otherwise }

\end{array}\right.\)

Find the expected value of X.

Solution:

Let x be a continuous random variable. In the probability density function, Expected

Question 5.

Let X be a continuous random variable with probability density function

f_{x} (x) = \(\left\{\begin{array}{l}

2 x, 0 \leq x \leq 1 \\

0, \text { otherwise }

\end{array}\right.\)

Find the mean and variance of X.

Solution:

Let x be a continuous random variable. In the probability density function,

Question 6.

In an investment, a man can make a profit I of ₹ 5,000 with a probability of 0.62 or a loss of ₹ 8,000 with a probability of 0.38 . Find the expected gain.

Solution:

X | 5000 | -8000 |

P(x = x) | 0.62 | 0.38 |

Let x be the random variable of getting gain in an Investment

E(x) be the random variable of getting gain in an Investment

E(x) = ΣPixi

= (0.62 × 5000) + [0.38 × (-8000)]

= 3100 – 3040

E(x) = 60

∴ Expected gain = ₹ 60

Question 7.

What are the properties of mathematical expectation?

Properties of Mathematical expectation

Solution:

i. E(a) = a, where ‘a’ is a constant.

ii. E(aX) = aE(X)

iii. E(aX + b) = aE(X) + b, where ‘a’ and ‘b’ are constants.

iv. If X ≥ 0, then E(X) ≥ 0

v. V(a) = 0

vi. If X is random variable, then V(aX + b) = a²V(X)

Question 8.

What do you understand by mathematical expectation?

Solution:

The average value of a random Phenomenon is termed as mathematical expectation or expected value.

The expected value is weighted average of the values of a random variable may assume

Question 9.

How do you define variance in terms of Mathematical expectation?

Solution:

The variance of X is defined by

Var(X) = Σ[x – E(X)]² p(x)

If X is discrete random variable with probability mass function p(x).

Var(X) = \(\int_{ -∞ }^{∞}\) [x- E(X)]² f_{x} (x) dx

If X is continuous random variable with probability density function f_{x} (x).

Question 10.

Define mathematical expectation in tears of discrete random variable?

Solution:

Let X be a discrete random variable with probability mass function (p.m.f.) p(x). Then, its expected value is defined by

E(X) = \(\sum_{ x }\) x p(x) ……(1)

Question 11.

State the definition of mathematical expectation using continous random variable.

Solution:

If X is a continuous random variable and fix) is the value of its probability density function at x, the expected value of X is

E(X) = \(\int_{ -∞ }^{∞}\) x f(x) dx …….(2)

Question 12.

In a business venture a man can make a profit of ₹ 2,000 with a probability of 0.4 or have a loss of ₹ 1,000 with a probability of 0.6. What is his expected, variance and standard deviation of profit?

Solution:

Let X be the random variable of getting profit in a business

X | 2000 | -1000 |

P(x = x) | 0.4 | 0.6 |

E(x) = Σ_{x}xp_{x}(x)

= (0.4 × 2000) +[0.6 × (-1000)]

= 800 – 600

E(X) = 200

∴ Expected value of profit = ₹ 200

E(X²) = Σx² P_{x}(x)

= [(2000)² × 0.4] + [(-1000)² × 0.6]

= (4000000 × 0.4) + (1000000 × 0.6)

E(X²) = 2200000

Var(X) = E(X²) – [E(X)]²

= 22000000 – (200)²

= 2200000 – 40000

Var(X) = 21,60,000

Variance of his profit = ₹ 21,60,000

Standard deviation(S.D) = \(\sqrt { var (x)}\)

σ = \(\sqrt { 2160000}\)

σ = 1469.69

Standard deviation of his profit is ₹ 1,469.69

Question 13.

The number of miles an automobile tire lasts before it reaches a critical point in tread wear can be represented by a p.d.f.

Find the expected number of miles (in thousands) a tire would last until it reaches the critical tread wear point.

Solution:

We know that,

Question 14.

A person tosses a coin and is to receive ₹ 4 for a head and is to pay ₹ 2 for a tail. Find the expectation and variance of his gains.

Solution:

Let X denote the amount the person receives in a game

Then X takes values 4,-2 and

So P(X = 4) = P (of getting a head)

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

P(X = – 2) = P (of getting a tail)

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

Hence the Probability distribution is

X | 4 | -2 |

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

E(x²) = 10

Var(x) = E(x²) – E(x²)

= 10 – (1)²

Var(x) = 9

∴ His expected gain = ₹ 1

His variance of gain = ₹ 9

Question 15.

Let X be a random variable and Y = 2X + 1. What is the variance of Y if variance of X is 5?

Solution:

Var(X) = 5

Var(Y) = var (2x + 1) {∴ v(ax + b) = a²v(x)}

= (2)² var(X)

= 4(5)

Var() = 20