Exercise 6.1 - Chapter 6 Random Variable and Mathematical Expectation 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Random Variable and Mathematical Expectation
Ex $6.1$
Question 1.
Construct cumulative distribution function for the given probability distribution.

Solution:

Thus the cumulative distribution function is given by,
Question 2.
Let $\mathrm{X}$ be a discrete random variable with the following p.m.f.
$
p(x)= \begin{cases}0.3 & \text { for } x=3 \\ 0.2 & \text { for } x=5 \\ 0.3 & \text { for } x=8 \\ 0.2 & \text { for } x=10 \\ 0 & \text { otherwise }\end{cases}
$

Find and plot the c.d.f. of $X$.
Solution:
The probability distribution function can be written as follows:

We obtain
$
\begin{aligned}
& \mathrm{F}(3)=\mathrm{P}(\mathrm{x} \leq 3)=\mathrm{P}(3)=0.3 \\
& \mathrm{~F}(5)=\mathrm{P}(\mathrm{x} \leq 5)=\mathrm{P}(3)+\mathrm{P}(5)=0.3+0.2=0.5 \\
& \mathrm{~F}(8)=\mathrm{P}(\mathrm{x} \leq 8)=\mathrm{P}(3)+\mathrm{P}(5)+\mathrm{P}(8)=0.5+0.3=0.8 \\
& \mathrm{~F}(10)=\mathrm{P}(\mathrm{x} \leq 10)=\mathrm{P}(3)+\mathrm{P}(5)+\mathrm{P}(8)+\mathrm{P}(10)=0.8+0.2=1
\end{aligned}
$
$
\mathrm{F}_{\mathrm{X}}(x)=\left\{\begin{array}{clc}
0 & \text { for } \quad x<3 \\
0.3 & \text { for } 3 \leq x<5 \\
0.5 & \text { for } 5 \leq x<8 \\
0.8 & \text { for } 8 \leq x<10 \\
1 & \text { for } \quad x \leq 10
\end{array}\right.
$

Question 3.
The discrete random variable $\mathrm{X}$ has the following probability function.

$
\mathrm{P}(\mathrm{X}=x)= \begin{cases}k x & x=2,4,6 \\ k(x-2) & x=8 \\ 0 & \text { otherwise }\end{cases}
$
where $\mathrm{k}$ is a constant. Show that $\mathrm{k}=\frac{1}{18}$
Solution:
The given probability function can be written as follows:

Since the condition of probability mass function is $\sum_{i=1}^{\infty} p\left(x_i\right)=1$, we have
$
\begin{aligned}
& 2 \mathrm{k}+4 \mathrm{k}+6 \mathrm{k}+6 \mathrm{k}=1 \\
& \Rightarrow 18 \mathrm{k}=1 \\
& \Rightarrow \mathrm{k}=\frac{1}{18}
\end{aligned}
$
Question 4.
The discrete random variable $\mathrm{X}$ has the probability function.

Show that $\mathrm{k}=0.1$
Solution:
$
\begin{aligned}
& \sum_{i=1}^{\infty} p\left(x_i\right)=1 \\
& \text { gives } \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+4 \mathrm{k}=1 \\
& \Rightarrow 10 \mathrm{k}=1 \\
& \Rightarrow \mathrm{k}=\frac{1}{10}=0.1
\end{aligned}
$
Question 5.
Two coins are tossed simultaneously. Getting a head is termed as a success. Find the probability distribution of the number of successes.
Solution:
Let $\mathrm{X}$ be the number of observed heads. The sample space $\mathrm{S}=\{(\mathrm{H} \mathrm{H}),(\mathrm{H} \mathrm{T}),(\mathrm{T} \mathrm{H}),(\mathrm{T} \mathrm{T})\}$ $\mathrm{X}$ takes the values $2,1,1,0$. Hence the probability distribution of the number of successes is given by

Question 6.
A random variable $\mathrm{X}$ has the following probability function.

(i) Find $k$
(ii) Evaluate $P(X<6), P(X \geq 6)$ and $P(0 (iii) If $P(X \leq x)>\frac{1}{2}$, then find the minimum value of $x$.
Solution:
(i) Since the condition of probability mass function
$
\begin{aligned}
& \sum_{i=1}^{\infty} p\left(x_i\right)=1 \\
& \sum_{i=0}^7 p\left(x_i\right)=1 \\
& 0+\mathrm{k}+2 \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}+\mathrm{k}^2+2 \mathrm{k}^2+7 \mathrm{k}^2+\mathrm{k}=1 \\
& \Rightarrow 10 \mathrm{k}^2+9 \mathrm{k}-1=0 \\
& \Rightarrow(10 \mathrm{k}-1)(\mathrm{k}+1)=0 \\
& \Rightarrow \mathrm{k}=\frac{1}{10} \text { and } \mathrm{k}=-1
\end{aligned}
$
Since $p(x)$ cannot be negative, $k=-1$ is not applicable. Hence $k=\frac{1}{10}$
Method 2:
$
\begin{aligned}
& \mathrm{P}(\mathrm{X}<6)=1-\mathrm{P}(\mathrm{X} \geq 6) \\
& =1-[\mathrm{P}(\mathrm{X}=6)+\mathrm{P}(\mathrm{X}=7)] \\
& =1-\left[2 \mathrm{k}^2+7 \mathrm{k}^2+\mathrm{k}\right] \\
& =1-9 \mathrm{k}^2-\mathrm{k} \\
& =1-9\left(\frac{1}{10}\right)^2-\left(\frac{1}{10}\right) \\
& =1-\frac{9}{100}-\frac{1}{10}=\frac{81}{100} \\
& P(X \geq 6)=1-P(X<6) \\
& =1-\frac{81}{100}=\frac{100-81}{100}=\frac{19}{100} \\
&
\end{aligned}
$

$
\begin{aligned}
& \mathrm{P}(0<\mathrm{X}<5)=\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=4) \\
& =\mathrm{k}+2 \mathrm{k}+2 \mathrm{k}+3 \mathrm{k}=8 \mathrm{k} \\
& =\frac{8}{10}
\end{aligned}
$
(iii) $\mathrm{P}(\mathrm{X} \leq \mathrm{x})>\frac{1}{2}$
To find the minimum value of $\mathrm{x}$, let us construct the cumulative distribution function of $\mathrm{X}$.

From the table we see that
$
\begin{aligned}
& \mathrm{P}(\mathrm{X} \leq 0)=0 \\
& \mathrm{P}(\mathrm{X} \leq 1)=\frac{1}{10}=0.1 \\
& \mathrm{P}(\mathrm{X} \leq 2)=\frac{3}{10}=0.3 \\
& \mathrm{P}(\mathrm{X} \leq 3)=\frac{5}{10}=0.5 \\
& \mathrm{P}(\mathrm{X} \leq 4)=\frac{8}{10}=0.8 \\
& \mathrm{So} \mathrm{X}=4
\end{aligned}
$
Question 7.
The distribution of a continuous random variable $\mathrm{X}$ in range $(-3,3)$ is given by p.d.f.
$f(x)= \begin{cases}\frac{1}{16}(3+x)^2, & -3 \leq x \leq-1 \\ \frac{1}{16}\left(6-2 x^2\right), & -1 \leq x \leq 1 \\ \frac{1}{16}(3-x)^2, & 1 \leq x \leq 3\end{cases}$
Verify that the area under the curve is unity.
Solution:
We know that area under a curve $\mathrm{f}(\mathrm{x})$ between $\mathrm{x}=\mathrm{a}$ and $\mathrm{x}=\mathrm{b}$ is $\int_a^b f(x) d x$
Here the area is given by $\int_{-3}^3 f(x) d x$

So
$
\begin{aligned}
\int_{-3}^3 f(x) d x & =\int_{-3}^{-1} f(x) d x+\int_{-1}^1 f(x) d x+\int_1^3 f(x) d x \\
& =\int_{-3}^{-1} \frac{1}{16}(3+x)^2 d x+\int_{-1}^1 \frac{1}{16}\left(6-2 x^2\right) d x+\int_1^3 \frac{1}{16}(3-x)^2 d x \\
& =\frac{1}{16}\left[\frac{(3+x)^3}{3}\right]_{-3}^{-1}+\frac{1}{16}\left[6 x-\frac{2}{3} x^3\right]_{-1}^1+\frac{1}{16}\left[\frac{(3-x)^3}{-3}\right]_1^3 \\
& =\frac{1}{16}\left[\frac{(3-1)^3}{3}-\frac{(3-3)^3}{3}\right]+\frac{1}{16}\left[6-\frac{2}{3}+6-\frac{2}{3}\right]+\frac{1}{16}\left[\frac{(3-3)^3}{-3}-\frac{(3-1)^3}{-3}\right] \\
& =\frac{1}{16}\left(\frac{8}{3}\right)+\frac{1}{16}\left(12-\frac{4}{3}\right)+\frac{1}{16}\left(\frac{8}{3}\right) \\
& =\frac{1}{6}+\frac{1}{16}\left(\frac{32}{3}\right)+\frac{1}{6}=\frac{1}{6}+\frac{2}{3}+\frac{1}{6}=1
\end{aligned}
$
Thus the area under the curve is unity.

Question 8.
A continuous random variable $\mathrm{X}$ has the following distribution function:
$
F(x)= \begin{cases}0 & , \text { if } x \leq 1 \\ k(x-1)^4, & \text { if } 13\end{cases}
$
Find (i) $\mathrm{k}$ and (ii) the probability density function.
Solution:
We have $\frac{d}{d x} F(x)=f(x) \geq 0$, where $F(x)$ is the distribution function and $f(x)$ is the probability density function.
Here $F(x)=0$ for $x \leq 1$
$\mathrm{f}(\mathrm{x})=0$ for $\mathrm{x} \leq 1$
Again $F(x)=1$ for $x>3$
$\mathrm{f}(\mathrm{x})=\frac{d}{d x}(1)=0$ for $\mathrm{x}>3$
In $1<\mathrm{x} \leq 3, \mathrm{~F}(\mathrm{x})=\mathrm{k}(\mathrm{x}-1) 4$
$\mathrm{f}(\mathrm{x})=\frac{d}{d x}(\mathrm{k}(\mathrm{x}-1) 4)=4 \mathrm{k}(\mathrm{x}-1) 3$
(i) We know that $\int_{-\infty}^{\infty} f(x) d x=1$
This gives $\int_1^3 4 k(x-1)^3 d x=1$
$
\begin{aligned}
& {\left[k(x-1)^4\right]_1^3=1} \\
& \mathrm{k}[16-0]=1 \\
& \mathrm{k}=\frac{1}{16}
\end{aligned}
$
(ii) The probability density function is
$\mathrm{f}(\mathrm{x})=\frac{1}{4}(x-1)^3, 1<\mathrm{x} \leq 3$
$=0$, otherwise

Question 9.
The length of time (in minutes) that a certain person speaks on the telephone is found to be a random phenomenon, with a probability function specified by the probability density function $f(x)$ as
$
f(x)= \begin{cases}A e^{-x / 5}, \text { for } x \geq 0 \\ 0, & \text { otherwise }\end{cases}
$
(a) Find the value of $A$ that makes $f(x)$ a p.d.f.
(b) What is the probability that the number of minutes that person will talk over the phone is (i) more than 10 minutes, (ii) less than 5 minutes and (iii) between 5 and 10 minutes.
Solution:
(a) For $\mathrm{f}(\mathrm{x})$ to be a p.d.f we must have $\int_{-\infty}^{\infty} f(x) d x=1$
According to the problem,
$
\begin{aligned}
\int_0^{\infty} f(x) d x & =1 \\
\int_0^{\infty} A e^{\frac{-x}{5}} d x & =1 \\
{\left[\frac{A e^{\frac{-x}{5}}}{\left(\frac{-1}{5}\right)}\right]_0^{\infty} } & =1 \\
-5 \mathrm{~A}(0-1) & =1 \\
\mathrm{~A} & =\frac{1}{5}
\end{aligned}
$
(b) (i) The probability that the number of minutes that person will talk is more than 10 minutes is given by

$
\begin{aligned}
\mathrm{P}(\mathrm{X} \geq 10) & =\int_{10}^{\infty} f(x) d x \\
& =\int_{10}^{\infty} \frac{1}{5} e^{\frac{-x}{5}} d x=\frac{1}{5}\left(\frac{e^{\frac{-x}{5}}}{\frac{-1}{5}}\right)_{10}^{\infty} \\
& =-\left(e^{\frac{-x}{5}}\right)_{10}^{\infty}=0+e^{-2}=\frac{1}{e^2}
\end{aligned}
$
Hence $P(X \geq 10)=\frac{1}{e^2}$
(b) (ii) The probability that the number of minutes is less than 5 is given by 5
$
\begin{aligned}
\mathrm{P}(0 \leq \mathrm{X} \leq 5) & =\int_0^5 f(x) d x \\
& =\int_0^5 \frac{1}{5} e^{\frac{-x}{5}} d x=\frac{1}{5}\left[\frac{e^{\frac{-x}{5}}}{\left(\frac{-1}{5}\right)}\right]_0^5 \\
& =-\left(e^{\frac{-x}{5}}\right)_0^5=-e^{-1}+e^0=\frac{-1}{e}+1=\frac{e-1}{e}
\end{aligned}
$
Hence $\mathrm{P}(0 \leq \mathrm{X} \leq 5)=\frac{e-1}{e}$
(b) (iii) The probability that the number of minutes that person will talk is between 5 and 10

minutes is given by
$
\begin{aligned}
\mathrm{P}(5 \leq \mathrm{X} \leq 10) & =\int_5^{10} \frac{1}{5} e^{\frac{-x}{5}} d x \\
& =-\left[e^{\frac{-x}{5}}\right]_5^{10}=-e^{-2}+e^{-1}=\frac{-1}{e^2}+\frac{1}{e} \\
& =\frac{e-1}{e^2}
\end{aligned}
$
Hence $\mathrm{P}(5 \leq \mathrm{X} \leq 10)=\frac{e-1}{e^2}$
Question 10.
Suppose that the time in minutes that a person has to wait at a certain station for a train is found to be a random phenomenon with a probability function specified by the distribution function.
$
F(x)=\left\{\begin{array}{lr}
0, & \text { for } \quad x \leq 0 \\
\frac{x}{2}, & \text { for } 0 \leq x<1 \\
\frac{1}{2}, & \text { for } 1 \leq x<2 \\
\frac{x}{4}, & \text { for } 2 \leq x<4 \\
1, & \text { for } \quad x \geq 4
\end{array}\right.
$
(a) Is the distribution function continuous? If so, give its probability density function?
(b) What is the probability that a person will have to wait (i) more than 3 minutes, (ii) less than 3 minutes and (iii) between 1 and 3 minutes?
Solution:
(a) Yes. The distribution function is continuous. It is defined for all values of $-\infty

For $x<0, \quad \mathrm{~F}(x)=0 \Rightarrow f(x)=0$
For $0 \leq x<1, \quad \mathrm{~F}(x)=\frac{x}{2} \Rightarrow f(x)=\frac{d}{d x}\left(\frac{x}{2}\right)=\frac{1}{2}$.
For $1 \leq x<2, \quad \mathrm{~F}(x)=\frac{1}{2} \Rightarrow f(x)=\frac{d}{d x}\left(\frac{1}{2}\right)=0$
For $2 \leq x<4, \quad \mathrm{~F}(x)=\frac{x}{4} \Rightarrow f(x)=\frac{d}{d x}\left(\frac{x}{4}\right)=\frac{1}{4}$
For $x \geq 4, \quad \mathrm{~F}(x)=1 \Rightarrow f(x)=0$
Thus the probability density function is given by

$\mathrm{F}(x)=\left\{\begin{array}{l}
0, \text { for } \quad x<0 \\
\frac{1}{2}, \text { for } 0 \leq x<1 \\
0, \text { for } 1 \leq x<2 \\
\frac{1}{4}, \text { for } 2 \leq x<4 \\
0, \text { for } \quad x \geq 4
\end{array}\right.$
(b) (i) The probability that a person will have to wait for more than 3 minutes is given by
$
\begin{aligned}
\mathrm{P}(3 \leq \mathrm{X}<4) & =\int_3^4 f(x) d x \\
& =\int_3^4 \frac{1}{4} d x=\left(\frac{1}{4} x\right)_3^4=\frac{1}{4}(4-3)=\frac{1}{4}
\end{aligned}
$
Method 2:
$\mathrm{P}(\mathrm{X} \geq 3)=1-\mathrm{P}(\mathrm{X} \leq 3)=1-\mathrm{F}(3)$
From the question $F(3)=\frac{3}{4}$
So $\mathrm{P}(\mathrm{X} \geq 3)=1-\frac{3}{4}=\frac{1}{4}$
(b) (ii) The probability that a person will have to Wait for less than 3 minutes is $\mathrm{P}(\mathrm{X} \leq 3)=\mathrm{F}(3)=\frac{3}{4}$

(b) (iii) Probability that the person will have to wait between 1 and 3 minutes is
$
\begin{aligned}
\mathrm{P}(1 \leq \mathrm{X} \leq 3) & =\int_1^3 f(x) d x=\int_{-\infty}^3 f(x) d x-\int_{-\infty}^1 f(x) d x \\
& =\mathrm{P}(\mathrm{X} \leq 3)-\mathrm{P}(\mathrm{X} \leq 1) \\
& =\mathrm{F}(3)-\mathrm{F}(1) \\
& =\frac{3}{4}-\frac{1}{2}=\frac{1}{4}
\end{aligned}
$
Question 11.
Define the random variable.
Solution:
A Random variable is a set of possible values from a random experiment. The set of possible values is called the sample space. A random variable (r.v) is denoted by a capital letter such as $\mathrm{X}$, $\mathrm{Y}$ and $\mathrm{Z}$ etc. If $\mathrm{X}$ and $\mathrm{Y}$ are r.v's then $\mathrm{X}+\mathrm{Y}$ is also an r.v.
Question 12.
Explain what are the types of a random variable?
Solution:
Random variables are classified into two types namely discrete and continuous random variables. Discrete random variable: A discrete random variable has a finite number of possible values or an infinite sequence of countable real numbers.
Examples:
1. X denotes the number of hits when trying 20 free throws.
2. $\mathrm{X}$ denotes the number of customers who arrive at the bank from 8.30-9.30 AM Mon - Fri.
3. $\mathrm{X}$ denotes the number of balls sold during a week in a particular shop.
Continuous random variable:
A continuous random variable takes all values in an interval of real numbers.
Examples:
1. $\mathrm{X}$ denotes the time it takes for a bulb to bum out.
2. $\mathrm{X}$ denotes the weight of a truck in a truck - weighing station.
3. $\mathrm{X}$ denotes the amount of water in a bottle.

Question 13.
Define discrete random variable.
Solution:
A variable which can assume a finite number of possible values or an infinite sequence of countable real numbers is called a discrete random variable.
Examples:
1. Marks obtained in an exam.
2. Number of chocolates in a box.
3. Number of phone calls during a day.
4. Number of TV sets sold during a month by a dealer.
Question 14.
What do you understand by continuous random variable?
Solution:
A random variable which can take on any value (integral as well as fraction) in the interval is called a continuous random variable.
Examples:
1. The speed of a train.
2. Electricity consumption in kilowatt-hours.
3. Height of people in a population
4. Weight of students in a class.
Question 15.
Describe what is meant by a random variable.
Solution:
A Random variable is a set of possible values from a random experiment. The set of possible values is called the sample space. A random variable (r.v) is denoted by a capital letter such as $\mathrm{X}$, $Y$ and $\mathrm{Z}$ etc. If $X$ and $Y$ are r.v's then $X+Y$ is also an r.v.
Question 16.
Distinguish between discrete and continuous random variable.
Solution:

distributions come under this category.
Question 17.
Explain the distribution function of a random variable.
Solution:
The discrete cumulative distribution function or distribution function of a real-valued discrete random variable $X$, which takes the countable number of points $\mathrm{x}_1, \mathrm{x}_2, \ldots$. with corresponding probabilities $\mathrm{p}\left(\mathrm{x}_1\right), \mathrm{p}\left(\mathrm{x}_2\right), \ldots$ is defined by
$
\mathrm{F}_{\mathrm{X}}(x)=\mathrm{P}(\mathrm{X} \leq x) \text { for all } x \in \mathrm{R}
$
i.e., $\mathrm{F}_{\mathrm{X}}(x)=\sum_{x_i \leq x} p\left(x_i\right)$
If $X$ is a continuous random variable with probability density function $f_X(x)$, then the function $\mathrm{F}_{\mathrm{X}}(\mathrm{x})$ is defined by
$
\mathrm{F}_{\mathrm{X}}(x)=\mathrm{P}[\mathrm{X} \leq x]=\int_{-\infty}^x f(t) d t,-\infty $
is called the distribution function of the continuous random variable.

Question 18.
Explain the terms
(i) Probability mass function
(ii) Probability density function and
(iii) Probability distribution function.
Solution:
(i) Probability mass function: A probability mass function (p.m.f) is a function that gives the probability that a discrete random variable is exactly equal to some value. This is the means of defining a discrete probability distribution. Let $\mathrm{X}$ be a random variable with values $\mathrm{x}_1, \mathrm{x}_2, \ldots$ $\mathrm{x}_{\mathrm{n}} \ldots$. Then the p.m.f is defined by

The p.m.f satisfies the conditions, $\begin{array}{ll}p\left(x_i\right) \geq 0 \forall i & \text { (ii) } \sum_{i=1}^{\infty} p\left(x_i\right)=1\end{array}$
(ii) Probability density function: A probability density function (p.d.f) or density of a continuous random variable, is a function whose value at any given sample (or point) in the sample space gives the probability of the r.v. falling within the range of values. This probability is given by the area under the density function. The probability that an r.v. $\mathrm{X}$ takes a value in the interval $[\mathrm{a}, \mathrm{b}]$ is given by

$
\mathrm{P}(\mathrm{a} \leq \mathrm{X} \leq \mathrm{b})=\int_a^b f_{\mathrm{X}}(x) d x
$
The p.d.f $f_{\mathrm{x}}(x)$ satisfied the conditions,
(i) $f(x) \geq 0 \forall x$
(ii) $\int_{-\infty}^{\infty} f_{\mathbf{X}}(x) d x=1$
(iii) Probability distribution function: A probability distribution is a list of all of the possible outcomes of a random variable along with their corresponding probability values. When the r.v is discrete, the distribution function is called a probability mass function (p.m.f), and when the r.v is continuous, the distribution function is called a probability density function (p.d.f) Examples:
(i) $\mathrm{P}(\mathrm{X}=i)=\frac{1}{2^i}$ for $i=1,2,3, \ldots \ldots \ldots$.
This is a p.m.f.
(ii) $f(x)=\frac{1}{\sqrt{2 \pi}} e^{\frac{-x^2}{2}}$ for $-\infty

Question 19.
What are the properties of
(i) discrete random variable and
(ii) continuous random variable?
Solution:
Discrete Random Variable:
- A variable which can take only certain values.
- The value of the variables can increase incomplete numbers
- Binomial, Poisson, Hypergeometric probability distributions come under this category
- Example: Number of students who opt for commerce in class 11 , say $30,35,40,45$, and 50.
Continuous random variable:
- A variable which can take any value in a particular limit.
- Its value increases infractions but not in jumps.
- Normal, student's t and chi-square distribution come under this category.
- Example: Height, Weight and age of family members: $50.5 \mathrm{~kg}, 30 \mathrm{~kg}, 42.8 \mathrm{~kg}$ and $18.6 \mathrm{~kg}$.
Question 20.
State the properties of the distribution function.
Solution:
- Property 1: The distribution function $F$ is increasing, (i.e) if $x \leq y$, then $F(x) \leq F(y)$
- Property 2: $\mathrm{F}(\mathrm{x})$ is continuous from right, (i.e) for each $\mathrm{x} \in \mathrm{R}, \mathrm{F}\left(\mathrm{x}^{+}\right)=\mathrm{F}(\mathrm{x})$
- Property 3: $\mathrm{F}(\infty)=1$
- Property 4: $\mathrm{F}(-\infty)=0$
- Property $5: F^{\prime}(x)=f(x)$
- Property 6: $\mathrm{P}(\mathrm{a} \leq \mathrm{X} \leq \mathrm{b})=\mathrm{F}(\mathrm{b})-\mathrm{F}(\mathrm{a})$

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