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

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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:
Let X denote the number on the top side of the unbiased die.
The probability mass function is given by the following table.

The expected value for the random variable X is E(X) = \(\sum_{x} x P_{x}(x)\)

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 the probability mass function of the random variable X.

Find the standard deviation of x.
Solution:
The standard deviation of X, σ_x is given by σ_x = √Var[X]
Now Var(X) = E(X²) – [E(X)]²
From the given table,

Hence the standard deviation of X is 2.15

Question 4.
Let X be a continuous random variable with probability density function.

Find the expected value of X.
Solution:
The expected value of the random variable is given by E(X) = \(\int_{-\infty}^{\infty} x f(x) d x\)
According to the problem we have,

Question 5.
Let X be a continuous random variable with probability density function

Find the mean and variance of X.
Solution:
Given that X is a continuous random variable.
The mean of X is the expected value of X.

Question 6.
In investment, a man can make a profit 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:
Let X be the random variable which denotes the gain in the investment. It is given that X takes the value 5000 with probability 0.62 and -8000 with a probability 0.38.
(Note that we take -8000 since it is a loss)
The probability distribution is given by

E(X) = (0.38) (-8000) + (0.62) (5000)
= -3040 + 3100
= 60
Hence the expected gain is ₹ 60

Question 7.
What are the properties of Mathematical expectation?
Solution:
The properties of Mathematical expectation are as follows:
(i) E(a) = a, where ‘a’ is a constant
(ii) Addition theorem: For two r.v’s X and Y, E(X + Y) = E(X) + E(Y)
(iii) Multiplication theorem: E(XY) = E(X) E(Y)
(iv) E(aX) = aE(X), where ‘a’ is a constant
(v) For constants a and b, E(aX + b) = a E(X) + b

Question 8.
What do you understand by Mathematical expectation?
Solution:
The expected value of a random variable gives a measure of the center of the distribution of the variable. In other words, E(X) is the long-term average value of the variable. The expected value is calculated as a weighted average of the values of a random variable in a particular experiment. The weights are the probabilities. The mean of the random variable X is µ_X = E(X).

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 terms 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) = \(\sum_{x} x p(x)\)
In other words, if x₁, x₂, x₃,…… x_n are the different values of X, and p(x₁), p(x₂) …..p(x_n) are the corresponding probabilities, then E(X) = x₁ p(x₁) + x₂ p(x₂) + x₃ p(x₃) +… x_n p(x_n)

Question 11.
State the definition of Mathematical expectation using a continuous random variable.
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 denoting the profit of the business venture.
The probability distribution of X is given as follows

E(X) = (-1000) (0.6) + (2000) (0.4)
= – 600 + 800
= 200
E(X²) = (-1000)² (0.6) + (2000)² (0.4)
= 6,00,000 + 16,00,000
= 22,00,000
V(X) = E(X²) – [E(X)]²
= 22,00,000 – 40000
= 21,60,000
Standard deviation = √Var[X]
= \(\sqrt{2160000}\)
= 1469.69
Thus the expected value of profit is ₹ 200. The variance of profit is ₹ 21,60,000 and the standard deviation of profit is ₹ 1469.69.

Question 13.
The number of miles an automobile tyre 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 tyre would last until it reaches the critical tread wear point.
Solution:
Let the continuous random variable X denote the number of miles (in thousands) till an automobile tyre lasts.
The expected value is E(X) = \(\int_{-\infty}^{\infty} x f(x) d x\)
From the problem we have,
\(E(X)=\int_{0}^{\infty}(x) \frac{1}{30} e^{\frac{-x}{30}} d x\)
We use integration by parts to evaluate the integral

Hence the expected number of miles is 30,000.

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 be the discrete random variable which denotes the gain of the person.
The probability distribution of X is given by

(Here, since a coin is tossed the probability is equal for the outcomes head or tail)

Thus the expectation of his gains is 1 and the variance of his gains is 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