Miscellaneous Problems - Chapter 6 - Random Variable and Mathematical Expectation - 12th Maths Guide Guide Samacheer Kalvi Solutions

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Miscellaneous Problems
Question 1.
The probability function of a random variable $\mathrm{X}$ is given by
$
p(x)=\left\{\begin{array}{l}
\frac{1}{4}, \text { for } x=-2 \\
\frac{1}{4}, \text { for } x=0 \\
\frac{1}{2}, \text { for } x=10 \\
0, \text { elsewhere }
\end{array}\right.
$
Evaluate the following probabilities.
(i) $\mathrm{P}(\mathrm{X} \leq 0)$
(ii) $\mathrm{P}(\mathrm{X}<0)$
(iii) $\mathrm{P}(|\mathrm{X}| \leq 2)$
(iv) $\mathrm{P}(0 \leq \mathrm{X} \leq 10)$
Solution:
$
\begin{aligned}
& \text { (i) } \mathrm{P}(\mathrm{X} \leq 0)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=-2) \\
& =\frac{1}{4}+\frac{1}{4}=\frac{1}{2}
\end{aligned}
$
(ii) $\mathrm{P}(\mathrm{X}<0)=\mathrm{P}(\mathrm{X}=-2)=\frac{1}{4}$$\begin{aligned}
& \text { (iii) } \mathrm{P}(|\mathrm{X}| \leq 2)=\mathrm{P}(-2 \leq \mathrm{X} \leq 2) \\
& =\mathrm{P}(\mathrm{X}=-2)+\mathrm{P}(\mathrm{X}=-1)+\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2) \\
& =\frac{1}{4}+0+\frac{1}{4}+0+0 \\
& =\frac{1}{2}
\end{aligned}$

$\begin{aligned}
& \text { (iv) } \mathrm{P}(0 \leq \mathrm{X} \leq 10)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=10)+0 \\
& =\frac{1}{4}+\frac{1}{2}=\frac{3}{4}
\end{aligned}$

Question 2.
Let $\mathrm{X}$ be a random variable with cumulative distribution function.
$
F(x)=\left\{\begin{array}{lr}
0, & \text { if } x<0 \\
\frac{x}{8} & \text {, if } 0 \leq x<1 \\
\frac{1}{4}+\frac{x}{8}, & \text { if } 1 \leq x<2 \\
\frac{3}{4}+\frac{x}{12}, & \text { if } 2 \leq x<3 \\
1, & \text { for } 3 \leq x
\end{array}\right.
$
(a) Compute: (i) $\mathrm{P}(1 \leq \mathrm{X} \leq 2)$ and (ii) $\mathrm{P}(\mathrm{X}=3)$.
(b) Is $\mathrm{X}$ a discrete random variable? Justify your answer.
Solution:
(a) (i) $\mathrm{P}(1 \leq \mathrm{X} \leq 2)=\mathrm{F}(2)-\mathrm{F}(1)$
$
\begin{aligned}
& =\left(\frac{3}{4}+\frac{2}{12}\right)-\left(\frac{1}{4}+\frac{1}{8}\right) \\
& =\frac{11}{12}-\frac{3}{8}=\frac{13}{24} \\
\mathrm{~F}(2) & =\frac{3}{4}+\frac{x}{12} \text { at } x=2 \\
\mathrm{~F}(1) & =\frac{1}{4}+\frac{x}{8} \text { at } x=1
\end{aligned}
$
(a) (ii) $\mathrm{P}(\mathrm{X}=3)=0$. The given random variable is continuous r.v. Hence the probability for a particular value of $\mathrm{X}$ is zero.
(b) $\mathrm{X}$ is not discrete since the cumulative distribution function is a continuous function. It is not a step function.
Question 3.
The p.d.f. of $\mathrm{X}$ is defined as
$
f(x)=\left\{\begin{array}{l}
k, \text { for } 0 0, \quad \text { otherwise }
\end{array}\right.
$

Find the value of $\mathrm{k}$ and also find $\mathrm{P}(2 \leq \mathrm{X} \leq 4)$.
Solution:
Since $f(x)$ is a p.d.f $\int_{-\infty}^{\infty} f(x) d x=1$
From the problem, $\int_0^4 f(x) d x=1$
$
\begin{aligned}
& \Rightarrow \quad \int_0^4 k d x=1 \Rightarrow(k x)_0^4=1 \\
& 4 k=1 \\
& k=\frac{1}{4} \\
&
\end{aligned}
$
We know $\mathrm{P}(a \leq \mathrm{X} \leq b)=\int_a^b f(x) d x$
So $P(2 \leq X \leq 4)=\int_2^4 \frac{1}{4} d x=\left(\frac{1}{4} x\right)_2^4=\left(\frac{1}{4}\right)(2)=\frac{1}{2}$
Question 4.
The probability distribution function of a discrete random variable $\mathrm{X}$ is $f(x)=\left\{\begin{array}{lc}2 k, & x=1 \\ 3 k, & x=3 \\ 4 k, & x=5 \\ 0, & \text { otherwise }\end{array}\right.$
where $\mathrm{k}$ is some constant.
Find (a) $k$ and (b) $\mathrm{P}(\mathrm{X}>2)$.
Solution:
(a) Given $\mathrm{X}$ is a discrete random variable.

The probability distribution can be written as

We know that $\Sigma p(x)=1$
$
\begin{aligned}
& \Rightarrow 2 \mathrm{k}+3 \mathrm{k}+4 \mathrm{k}=1 \\
& \Rightarrow 9 \mathrm{k}=1 \\
& \Rightarrow \mathrm{k}=1 / 9
\end{aligned}
$

$\begin{aligned}
& \text { (b) } \mathrm{P}(\mathrm{X}>2)=\mathrm{P}(\mathrm{X}=3)+\mathrm{P}(\mathrm{X}=5) \\
& =3 \mathrm{k}+4 \mathrm{k} \\
& =7 \mathrm{k} \\
& =\frac{7}{9}
\end{aligned}$
Question 5.
The probability density function of a continuous random variable $\mathrm{X}$ is
$
f(x)= \begin{cases}a+b x^2, 0 \leq x \leq 1 \\ 0, & \text { otherwise }\end{cases}
$
where $a$ and $b$ are some constants.
Find (i) $a$ and $\mathrm{b}$ if $\mathrm{E}(\mathrm{X})=\frac{3}{5}$
(ii) $\operatorname{Var}(\mathrm{X})$
Solution:
Given that $X$ is a continuous random variable and $f(x)$ is density function.
$
\begin{aligned}
& \text { So } \int_{-\infty}^{\infty} f(x) d x=1 \\
& \Rightarrow \int_0^1\left(a+b x^2\right) d x=1 \\
&\left(a x+\frac{b x^3}{3}\right)_0^1=1 \\
& a+\frac{b}{3}=1
\end{aligned}
$

Again it is given that $E(X)=\frac{3}{5}$
Now
$
\begin{aligned}
\Rightarrow \int_0^1 x\left(a+b x^2\right) d x & =\frac{3}{5} \\
\int_0^1\left(a x+b x^3\right) d x & =\frac{3}{5} \\
\left(\frac{a x^2}{2}+\frac{b x^4}{4}\right)_0^1 & =\frac{3}{5} \\
\frac{a}{2}+\frac{b}{4} & =\frac{3}{5}
\end{aligned}
$
We solve (1) and (2) to find the values of $a$ and $b$.
$
\begin{aligned}
& a+\frac{b}{3}=1 \\
& a+\frac{b}{2}=\frac{6}{5}
\end{aligned}
$
Subtract $b\left(\frac{1}{3}-\frac{1}{2}\right)=1-\frac{6}{5}$
$
\begin{aligned}
\frac{-1}{6} b & =\frac{-1}{5} \\
\Rightarrow \quad b & =\frac{6}{5} \\
a & =1-\frac{b}{3}=1-\frac{1}{3}\left(\frac{6}{5}\right)=1-\frac{2}{5}=\frac{3}{5} \\
a & =\frac{3}{5}, b=\frac{6}{5}
\end{aligned}
$
Thus
(ii) $\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2$

$
\begin{aligned}
\mathrm{E}\left(\mathrm{X}^2\right) & =\int_{-\infty}^{\infty} x^2 f(x) d x=\int_0^1 x^2\left(\frac{3}{5}+\frac{6}{5} x^2\right) d x \\
& =\int_0^1\left(\frac{3}{5} x^2+\frac{6}{5} x^4\right) d x \\
& =\left[\frac{x^3}{5}+\frac{6 x^5}{25}\right]_0^1=\frac{1}{5}+\frac{6}{25}=\frac{11}{25} \\
\Rightarrow \operatorname{Var}(\mathrm{X}) & =\frac{11}{25}-\left(\frac{3}{5}\right)^2=\frac{11}{25}-\frac{9}{25}=\frac{2}{25}
\end{aligned}
$
Question 6.
Prove that if $\mathrm{E}(\mathrm{X})=0$, then $\mathrm{V}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)$.
Solution:
Given $\mathrm{E}(\mathrm{X})=0$. To show $\mathrm{V}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)$
We know that $\operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2$
So if $\mathrm{E}(\mathrm{X})=0, \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)$
From the definition of the variance of $\mathrm{X}$ also we can see the result.
$\operatorname{Var}(\mathrm{X})=\Sigma[\mathrm{X}-\mathrm{E}(\mathrm{x})]^2 \mathrm{p}(\mathrm{x})$
If $\mathrm{E}(\mathrm{X})=0$, then $\mathrm{V}(\mathrm{X})=\Sigma \mathrm{x}^2 \mathrm{p}(\mathrm{x})=\mathrm{E}\left(\mathrm{X}^2\right)$
Question 7.
What is the expected value of a game that works as follows: I flip a coin and if tails pay you ₹ 2 ; if heads pay you ₹ 1 . In either case, I also pay you ₹ 50 .
Solution:
Let $\mathrm{X}$ be the expected value of the game.

The probability distribution is given by,

$
E(X)=(52)\left(\frac{1}{2}\right)+51\left(\frac{1}{2}\right)=\frac{1}{2}(52+51)=\frac{103}{2}=51.5
$
Question 8.
Prove that
(i) $\mathrm{V}(\mathrm{aX})=\mathrm{a}^2 \mathrm{~V}(\mathrm{X})$
(ii) $V(X+b)=V(X)$
Solution:
(i) To show $\mathrm{V}(\mathrm{aX})=\mathrm{a}^2 \mathrm{~V}(\mathrm{X})$
We know $\mathrm{V}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2$
So $\mathrm{V}(\mathrm{aX})=\left[\mathrm{E}\left(\mathrm{a}^2 \mathrm{X}^2\right)\right]-[\mathrm{E}(\mathrm{aX})]^2$
$
\begin{aligned}
& =\mathrm{a}^2 \mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{aE}(\mathrm{X})]^2 \\
& =\mathrm{a}^2 \mathrm{E}\left(\mathrm{X}^2\right)-\mathrm{a}^2[\mathrm{E}(\mathrm{X})]^2 \\
& =\mathrm{a}^2\left\{\left\{\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2\right\}\right. \\
& =\mathrm{a}^2 \mathrm{~V}(\mathrm{X})
\end{aligned}
$
(ii) $\mathrm{V}(\mathrm{X}+\mathrm{b})=\mathrm{V}(\mathrm{X})$
$
\begin{aligned}
& \text { LHS }=\mathrm{V}(\mathrm{X}+\mathrm{b})=\mathrm{E}\left[(\mathrm{X}+\mathrm{b})^2\right]-\{\mathrm{E}(\mathrm{X}+\mathrm{b})\}^2 \\
& =\mathrm{E}\left[\mathrm{X}^2+2 \mathrm{bX}+\mathrm{b}^2\right]-[\mathrm{E}(\mathrm{X})+\mathrm{b}]^2 \\
& =\mathrm{E}\left(\mathrm{X}^2\right)+2 \mathrm{bE}(\mathrm{X})+\mathrm{b}^2-\left[(\mathrm{E}(\mathrm{X}))^2+\mathrm{b}^2+2 \mathrm{bE}(\mathrm{X})\right] \\
& =\mathrm{E}\left(\mathrm{X}^2\right)+2 \mathrm{bE}(\mathrm{X})+\mathrm{b}^2-[\mathrm{E}(\mathrm{X})]^2-\mathrm{b}^2-2 \mathrm{bE}(\mathrm{X}) \\
& =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& =\mathrm{V}(\mathrm{X}) \\
& =\text { RHS }
\end{aligned}
$
Question 9.
Consider a random variable $\mathrm{X}$ with p.d.f

$
f(x)=\left\{\begin{aligned}
3 x^2, & \text { if } 0 0, & \text { otherwise }
\end{aligned}\right.
$
Find $\mathrm{E}(\mathrm{X})$ and $\mathrm{V}(3 \mathrm{X}-2)$.
Solution:
$
\begin{aligned}
\mathrm{E}(\mathrm{X}) & =\int_0^1 x\left(3 x^2\right) d x=\int_0^1 3 x^3 d x=\frac{3}{4} \\
\mathrm{~V}(3 \mathrm{X}-2) & =(3)^2 \mathrm{~V}(\mathrm{X})=9 \mathrm{~V}(\mathrm{X}) \\
\text { Now } \mathrm{E}\left(\mathrm{X}^2\right) & =\int_0^1 x^2\left(3 x^2\right) d x=\int_0^1 3 x^4 d x=\frac{3}{5} \\
\mathrm{~V}(\mathrm{X}) & =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2=\frac{3}{5}-\left(\frac{3}{4}\right)^2=\frac{3}{5}-\frac{9}{16}=\frac{3}{80}
\end{aligned}
$
Thus $\mathrm{V}(3 \mathrm{X}-2)=9\left(\frac{3}{80}\right)=\frac{27}{80}$
Question 10.
The time to failure in thousands of hours of an important piece of electronic equipment used in a manufactured DVD player has the density function.
$
f(x)=\left\{\begin{array}{cc}
2 e^{-2 x}, & x>0 \\
0, & \text { otherwise }
\end{array}\right.
$
Find the expected life of this piece of equipment.
Solution:
Let $\mathrm{X}$ be the random variable denoting the life of the piece of equipment.

$
\mathrm{E}(\mathrm{X})=\int_{-\infty}^{\infty} x f(x) d x
$
From the problem we have,
$
\begin{aligned}
\mathrm{E}(\mathrm{X}) & =\int_0^{\infty} x\left(2 e^{-2 x}\right) d x=2 \int_0^{\infty} x e^{-2 x} d x \\
& =2\left[\frac{x e^{-2 x}}{-2}\right]_0^{\infty}-2 \int_0^{\infty} \frac{e^{-2 x}}{-2} d x \\
& =-\left[x e^{-2 x}\right]_0^{\infty}+\left(\frac{e^{-2 x}}{-2}\right)_0^{\infty} \\
& =0+0+\frac{1}{2}=\frac{1}{2}
\end{aligned}
$
Thus the expected life of the piece of equipment is $\frac{1}{2}$ hrs (in thousands).