Exercise 6.2 - Chapter 6 Random Variable and Mathematical Expectation 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $6.2$
Question 1.
Find the expected value for the random variable of an unbiased die.
Solution:
Let $\mathrm{X}$ 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 $\mathrm{X}$ is $\mathrm{E}(\mathrm{X})=\sum_x x P_x(x)$
$
\begin{aligned}
& =\left(1 \times \frac{1}{6}\right)+\left(2 \times \frac{1}{6}\right)+\left(3 \times \frac{1}{6}\right)+\left(4 \times \frac{1}{6}\right)+\left(5 \times \frac{1}{6}\right)+\left(6 \times \frac{1}{6}\right) \\
& =\frac{1}{6}(1+2+3+4+5+6) \\
& =\frac{21}{6}=\frac{7}{2}=3.5
\end{aligned}
$
Question 2.
Let $\mathrm{X}$ be a random variable defining number of students getting A grade. Find the expected value of $\mathrm{X}$ from the given table.

Solution:
Expected value of $\mathrm{X}, \mathrm{E}(\mathrm{X})=\sum_x x P_x(x)$
$
\mathrm{E}(\mathrm{X})=(0 \times 0.2)+(1 \times 0.1)+(2 \times 0.4)+(3 \times 0.3)
$
$
\begin{aligned}
& =0+0.1+0.8+0.9 \\
& =1.8
\end{aligned}
$

Question 3.
The following table is describing the probability mass function of the random variable $\mathrm{X}$

Find the standard deviation of $\mathrm{x}$.
Solution:
The standard deviation of $\mathrm{X}, \sigma_{\mathrm{x}}$ is given by $\sigma_{\mathrm{x}}=\sqrt{ } \operatorname{Var}[\mathrm{X}]$
Now $\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2$
From the given table,
$
\begin{aligned}
& \mathrm{E}(\mathrm{X})=\sum_x x \mathrm{P}_{\mathrm{X}}(x)=(3 \times 0.1)+(4 \times 0.1)+(5 \times 0.2) \\
&=0.3+0.4+1 \\
&=1.7 \\
& \mathrm{E}\left(\mathrm{X}^2\right)=\sum_x x^2 \mathrm{P}_{\mathrm{X}}(x) \\
&=\left(3^2 \times 0.1\right)+\left(4^2 \times 0.1\right)+\left(5^2 \times 0.2\right) \\
&=9 \times 0.1+16 \times 0.1+25 \times 0.2 \\
&=0.9+1.6+5 \\
&=7.5 \\
& \Rightarrow \operatorname{Var}(\mathrm{X})=7.5-(1.7)^2=7.5-2.89=4.61 \\
& \sigma_x=\sqrt{4.61}=2.15
\end{aligned}
$
Hence the standard deviation of $\mathrm{X}$ is $2.15$
Question 4.
Let $\mathrm{X}$ be a continuous random variable with probability density function.
$
f_{\mathrm{X}}(x)=\left\{\begin{array}{c}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.
$

Find the expected value of $\mathrm{X}$.
Solution:
The expected value of the random variable is given by $\mathrm{E}(\mathrm{X})=\int_{-\infty}^{\infty} x f(x) d x$ According to the problem we have,
$
\mathrm{E}(\mathrm{X})=\int_0^1 x(2 x) d x=\int_0^1 2 x^2 d x=\left(\frac{2}{3} x^3\right)_0^1=\frac{2}{3}
$
Question 5.
Let $\mathrm{X}$ be a continuous random variable with probability density function
$
f(x)=\left\{\begin{array}{l}
\frac{3}{x^4}, \quad x \geq 1 \\
0, \text { otherwise }
\end{array}\right.
$
Find the mean and variance of $\mathrm{X}$.
Solution:
Given that $\mathrm{X}$ is a continuous random variable.
The mean of $\mathrm{X}$ is the expected value of $\mathrm{X}$.
$
\begin{aligned}
\mathrm{E}(\mathrm{X}) & =\int_1^{\infty} x\left(\frac{3}{x^4}\right) d x=\int_1^{\infty} 3 x^{-3} d x=\left(\frac{3 x^{-2}}{-2}\right)_1^{\infty} \\
& =\left(-\frac{3}{2 x^2}\right)_1^{\infty}=0+\frac{3}{2}=\frac{3}{2} \\
\mathrm{E}\left(\mathrm{X}^2\right) & =\int_{-\infty}^{\infty} x^2 f(x) d x=\int_1^{\infty}\left(x^2\right)\left(\frac{3}{x^4}\right) d x=\int_1^{\infty} \frac{3}{x^2} d x \\
& =\left(\frac{-3}{x}\right)_1^{\infty}=0+3=3 \\
\operatorname{Var}(\mathrm{X}) & =\mathrm{E}(\mathrm{X})^2-[\mathrm{E}(\mathrm{X})]^2=3-\left(\frac{3}{2}\right)^2=3-\frac{9}{4}=\frac{3}{4}
\end{aligned}
$
Hence the mean of $X$ is $\frac{3}{2}$ and variance of $X$ is $\frac{3}{4}$

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 $\mathrm{X}$ be the random variable which denotes the gain in the investment. It is given that $\mathrm{X}$ 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

$
\mathrm{E}(\mathrm{X})=(0.38)(-8000)+(0.62)(5000)
$
$
\begin{aligned}
& =-3040+3100 \\
& =60
\end{aligned}
$
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) $\mathrm{E}(\mathrm{a})=\mathrm{a}$, where ' $\mathrm{a}$ ' is a constant
(ii) Addition theorem: For two r.v's $\mathrm{X}$ and $\mathrm{Y}, \mathrm{E}(\mathrm{X}+\mathrm{Y})=\mathrm{E}(\mathrm{X})+\mathrm{E}(\mathrm{Y})$
(iii) Multiplication theorem: $\mathrm{E}(\mathrm{XY})=\mathrm{E}(\mathrm{X}) \mathrm{E}(\mathrm{Y})$
(iv) $\mathrm{E}(\mathrm{aX})=\mathrm{aE}(\mathrm{X})$, where ' $\mathrm{a}$ ' is a constant
(v) For constants a and $\mathrm{b}, \mathrm{E}(\mathrm{aX}+\mathrm{b})=\mathrm{aE}(\mathrm{X})+\mathrm{b}$
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, $\mathrm{E}(\mathrm{X})$ 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 $\mathrm{X}$ is $\mu_{\mathrm{X}}=\mathrm{E}(\mathrm{X})$.

Question 9.
How do you define variance in terms of Mathematical expectation?
Solution:
Let $\mathrm{X}$ be a random variable. Let $\mathrm{E}(\mathrm{X})$ denote the expectation of $\mathrm{X}$.
Then the variance is defined in terms of the mathematical expectation as follows.
(a) $\mathrm{X}$ is discrete r.v with p.m.f $\mathrm{p}(\mathrm{x})$. Then $\operatorname{Var}(\mathrm{X})=\sum[x-\mathrm{E}(\mathrm{X})]^2 p(x)$
(b) $\mathrm{X}$ is continuous r.v with p.d.f $\mathrm{f}_{\mathrm{x}}(\mathrm{x})$. Then $\operatorname{Var}(\mathrm{X})=\int_{-\infty}^{\infty}[X-E(X)]^2 f_X(x) d x$
Question 10.
Define Mathematical expectation in terms of a discrete random variable.
Solution:
Let $\mathrm{X}$ be a discrete random variable with probability mass function (p.m.f) $\mathrm{P}(\mathrm{x})$. Then, its expected value is defined by $\mathrm{E}(\mathrm{X})=\sum_x x p(x)$
In other words, if $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \ldots \ldots \mathrm{x}_{\mathrm{n}}$ are the different values of $\mathrm{X}$, and $\mathrm{p}\left(\mathrm{x}_1\right), \mathrm{p}\left(\mathrm{x}_2\right)$
$\ldots . . \mathrm{p}\left(\mathrm{x}_{\mathrm{n}}\right)$ are the corresponding probabilities, then $\mathrm{E}(\mathrm{X})=\mathrm{x}_1 \mathrm{p}\left(\mathrm{x}_1\right)+\mathrm{x}_2 \mathrm{p}\left(\mathrm{x}_2\right)+\mathrm{x}_3 \mathrm{p}\left(\mathrm{x}_3\right)$ $+\ldots \mathrm{x}_{\mathrm{n}} \mathrm{p}\left(\mathrm{x}_{\mathrm{n}}\right)$
Question 11.
State the definition of Mathematical expectation using a continuous random variable.
Solution:
Let $\mathrm{X}$ be a continuous random variable with probability density function $f(x)$. Then the expected value of $\mathrm{X}$ is
$\mathrm{E}(\mathrm{X})=\int_{-\infty}^{\infty} x f(x) d x$
If the integral exists, $\mathrm{E}(\mathrm{X})$ 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 $\mathrm{X}$ be the random variable denoting the profit of the business venture.
The probability distribution of $\mathrm{X}$ is given as follows

$
\begin{aligned}
& \mathrm{E}(\mathrm{X})=(-1000)(0.6)+(2000)(0.4) \\
& =-600+800 \\
& =200 \\
& \mathrm{E}\left(\mathrm{X}^2\right)=(-1000)^2(0.6)+(2000)^2(0.4) \\
& =6,00,000+16,00,000 \\
& =22,00,000 \\
& \mathrm{~V}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& =22,00,000-40000 \\
& =21,60,000 \\
& \text { Standard deviation }=\sqrt{\operatorname{Var}}[\mathrm{X}] \\
& =\sqrt{2160000} \\
& =1469.69 \\
&
\end{aligned}
$
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.
$
f(x)=\left\{\begin{array}{cc}
\frac{1}{30} e^{-\frac{x}{30}}, \text { for } x>0 \\
0, & \text { for } x \leq 0
\end{array}\right.
$
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 $\mathrm{X}$ denote the number of miles (in thousands) till an automobile tyre lasts.
The expected value is $\mathrm{E}(\mathrm{X})=\int_{-\infty}^{\infty} x f(x) d x$
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

$
\begin{aligned}
\mathrm{E}(\mathrm{X}) & =\frac{1}{30}\left[\frac{x e^{\frac{-x}{30}}}{\frac{-1}{30}}\right]_0^x-\frac{1}{30} \int_0^{\infty} \frac{e^{\frac{-x}{30}}}{\left(\frac{-1}{30}\right)} d x \\
& =\left(-x e^{\frac{-x}{30}}\right)_0^{\infty}+\int_0^{\infty} e^{\frac{-x}{30}} d x \\
& =0+\left(\frac{e^{\frac{-x}{30}}}{-\frac{1}{30}}\right)_0^{\infty}=-30(0-1)=30
\end{aligned}
$
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 $\mathrm{X}$ be the discrete random variable which denotes the gain of the person. The probability distribution of $\mathrm{X}$ is given by

(Here, since a coin is tossed the probability is equal for the outcomes head or tail)
$
\begin{aligned}
\mathrm{E}(\mathrm{X}) & =4\left(\frac{1}{2}\right)+(-2)\left(\frac{1}{2}\right) \\
& =2-1=1 \\
\mathrm{E}\left(\mathrm{X}^2\right) & =(4)^2\left(\frac{1}{2}\right)+(-2)^2\left(\frac{1}{2}\right) \\
& =16\left(\frac{1}{2}\right)+4\left(\frac{1}{2}\right)=8+2=10
\end{aligned}
$

$
\begin{aligned}
\operatorname{Var}(\mathrm{X}) & =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& =10-1=9
\end{aligned}
$
Thus the expectation of his gains is 1 and the variance of his gains is 9 .
Question 15 .
Let $\mathrm{X}$ be a random variable and $\mathrm{Y}=2 \mathrm{X}+1$. What is the variance of $\mathrm{Y}$ if the variance of $\mathrm{X}$ is 5 ?
Solution:
Given $\mathrm{X}$ is a random variable and $\mathrm{Y}=2 \mathrm{X}+1$ and $\operatorname{Var}(\mathrm{X})=5$
$\operatorname{Var}(\mathrm{Y})=\operatorname{Var}(2 \mathrm{X}+1)=(2)^2=4$
$\operatorname{Var} \mathrm{X}=4(5)=20$