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
fx (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
fx (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:

1. E(a) = a, where ‘a’ is a constant
2. Addition theorem: For two r.v’s X and Y, E(X + Y) = E(X) + E(Y)
3. Multiplication theorem: E(XY) = E(X) E(Y)
4. E(aX) = aE(X), where ‘a’ is a constant
5. 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 fx(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) = Σxxpx(x)
= (0.4 × 2000) +[0.6 × (-1000)]
= 800 – 600
E(X) = 200
∴ Expected value of profit = ₹ 200
E(X²) = Σx² Px(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