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:

Expected value of X, E(X) = \(\sum_{x} x P_{x}(x)\)

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

= 0 + 0. 1 + 0.8 + 0.9

= 1.8

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 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?

Solution:

The properties of Mathematical expectation are as follows:

- E(a) = a, where ‘a’ is a constant
- Addition theorem: For two r.v’s X and Y, E(X + Y) = E(X) + E(Y)
- Multiplication theorem: E(XY) = E(X) E(Y)
- E(aX) = aE(X), where ‘a’ is a constant
- For constants a and b, E(aX + b) = a E(X) + b

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 a weighted average of the values of a random variable may assume

Question 9.

How do you define variance in terms of Mathematical expectation?

Solution:

Let X be a random variable. Let E(X) denote the expectation of X.

Then the variance is defined in terms of the mathematical expectation as follows.

(a) X is discrete r.v with p.m.f p(x). Then Var(X) = \(\sum[x-\mathrm{E}(\mathrm{X})]^{2} p(x)\)

(b) X is continuous r.v with p.d.f f_{x}(x). Then Var(X) = \(\int_{-\infty}^{\infty}[X-E(X)]^{2} f_{X}(x) d x\)

Question 10.

Define mathematical expectation in tears of a 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 continuous random variables.

Solution:

Let X be a continuous random variable with probability density function f(x). Then the expected value of X is

\(\mathrm{E}(\mathrm{X})=\int_{-\infty}^{\infty} x f(x) d x\)

If the integral exists, E(X) is the mean of the values, otherwise, we say that the mean does not exist.

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 are 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 the variance of X is 5?

Solution:

Given X is a random variable and Y = 2X + 1 and Var(X ) = 5

Var (Y) = Var (2X + 1) = (2)^{2} = 4

Var X = 4(5) = 20