Exercise 11.4 - Chapter 11 Probability Distributions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $11.4$
Question 1.
For the random variable $\mathrm{X}$ with the given probability mass function as below, find the mean and variance
(i) $f(x)= \begin{cases}\frac{1}{10} & x=2,5 \\ \frac{1}{5} & x=0,1,3,4\end{cases}$
(ii) $f(x)=\left\{\frac{4-x}{6} \quad x=1,2,3\right.$
(iii) $f(x)= \begin{cases}2(x-1) & 1 (iv) $f(x)= \begin{cases}\frac{1}{2} e^{-\frac{x}{2}} & \text { for } x>0 \\ 0 & \text { otherwise }\end{cases}$
Solution:
(i) Given probability mass function $f(x)= \begin{cases}\frac{1}{10} & x=2,5 \\ \frac{1}{5} & x=0,1,3,4\end{cases}$

Mean $\mathrm{E}(\mathrm{X})=\Sigma x f(x)$
$=0+\frac{1}{5}+\frac{1}{5}+\frac{3}{5}+\frac{4}{5}+\frac{1}{2}$ $=\frac{9}{5}+\frac{1}{2}=\frac{18+5}{10}=\frac{23}{10}=2.3$
$\mathrm{E}\left(\mathrm{X}^2\right)=\Sigma x^2 f(x)$
$=0+\frac{1}{5}+\frac{4}{10}+\frac{9}{5}+\frac{16}{5}+\frac{25}{10}$
$=\frac{1}{5}+\frac{2}{5}+\frac{9}{5}+\frac{16}{5}+\frac{5}{2}=\frac{28}{5}+\frac{5}{2}=\frac{81}{10}$

$\text { Variance } \begin{aligned}
\operatorname{Var}(\mathrm{X}) & =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& =\frac{81}{10}-\frac{529}{100}=\frac{810-529}{100}=\frac{281}{100}=2.81
\end{aligned}$

(ii) Given probability mass function $f(x)=\left\{\frac{4-x}{6}, x=1,2,3\right.$

$
\text { Mean } \begin{aligned}
\mathrm{E}(\mathrm{X}) & =\Sigma x f(x) \\
& =\frac{1}{2}+\frac{2}{3}+\frac{1}{2}=1+\frac{2}{3}=\frac{5}{3}=1.67 \\
\mathrm{E}\left(\mathrm{X}^2\right) & =\Sigma x^2 f(x) \\
& =\frac{1}{2}+\frac{4}{3}+\frac{9}{6}=\frac{3+8+9}{6}=\frac{20}{6}=\frac{10}{3}
\end{aligned}
$
$
\text { Variance } \quad \begin{aligned}
\operatorname{Var}(\mathrm{X}) & =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& =\frac{10}{3}-\left(\frac{5}{3}\right)^2 \\
& =\frac{10}{3}-\frac{25}{9}=\frac{30-25}{9}=\frac{5}{9}=0.56
\end{aligned}
$
(iii) Given probability mass function $f(x)=\left\{\begin{array}{cc}2(x-1), & 1 Here ' $\mathrm{X}$ ' is a continuous random variable
$
\begin{aligned}
\text { Mean } \mathrm{E}(\mathrm{X}) & =\int_{-\infty}^{\infty} x f(x) d x \\
=2 \int_1^2 x(x-1) d x & =2 \int_1^2\left(x^2-x\right) d x \\
& =2\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_1^2=2\left[\left(\frac{8}{3}-2\right)-\left(\frac{1}{3}-\frac{1}{2}\right)\right]
\end{aligned}
$

$\begin{aligned}
& =2\left[\frac{2}{3}+\frac{1}{6}\right]=2\left(\frac{5}{6}\right)=\frac{5}{3} \\
& \mathrm{E}\left(\mathrm{X}^2\right)=\int_{-\infty}^{\infty} x^2 f(x) d x \\
& =2 \int_1^2 x^2(x-1) d x=2 \int_1^2\left(x^3-x^2\right) d x \\
& =2\left[\frac{x^4}{4}-\frac{x^3}{3}\right]_1^2=\frac{17}{6} \\
& \text { Variance } \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& =\frac{17}{6}-\left(\frac{5}{3}\right)^2 \\
& =\frac{17}{6}-\frac{25}{9}=\frac{51-50}{18}=\frac{1}{18} \\
&
\end{aligned}$

(iv)
$
f(x)=\left\{\begin{array}{cc}
1 / 2 e^{-x / 2}, & \text { for } x>0 \\
0, & \text { otherwise }
\end{array}\right.
$

Since ' $\mathrm{X}$ ' is a continuous random variable
Mean
$
\begin{aligned}
\mathrm{E}(\mathrm{X}) & =\int_{-\infty}^{\infty} x f(x) d x \\
& =1 / 2 \int_0^{\infty} x e^{-x / 2} d x=1 / 2\left[-2 x e^{-x / 2}-4 e^{-x / 2}\right]_0^{\infty} \\
& =-\left[x e^{-x / 2}+2 e^{-x / 2}\right]_0^{\infty} \\
& =-[0-(0+2)]=2
\end{aligned}
$
(Using integration by parts)
Let $u=x ; d u=d x$
$
\begin{aligned}
\int d v & =\int e^{-x / 2} d x \\
v & =\frac{e^{-x / 2}}{-1 / 2} \\
\int u d v & =u v-\int v d u \\
& =\frac{x \cdot e^{-x / 2}}{-1 / 2}+2 \int e^{-x / 2} d x=-2 x e^{-x / 2}+2 \cdot \frac{e^{-x / 2}}{-1 / 2} \\
& =-2 x e^{-x / 2}-4 e^{-x / 2}
\end{aligned}
$
[ $\because$ Using integration parts twice ]
$
\begin{aligned}
& =-\left[x^2 e^{-x / 2}+4 x e^{-x / 2}+8 e^{-x / 2}\right]_0^{\infty} \\
& =-(0-8)=8 \\
& \text { Variance } \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& =8-4=4 \\
&
\end{aligned}
$

Question 2.
Two balls are drawn in succession without replacement from an urn containing four red balls and
three black balls. Let $\mathrm{X}$ be the possible outcomes drawing red balls. Find the probability mass function and mean for $\mathrm{X}$.
Solution:
Number of Red balls $=4$
Number of Black balls $=3$
Total number of balls $=7$
Given : two balls are drawn in succession without replacement.
Let ' $X$ ' be the number of red balls and ' $X$ ' can take the values 0,1 and 2 .
$
\begin{aligned}
& P(X=0)=P(B B)=\frac{3 C_2}{7 C_2}=\frac{1}{7} \\
& P(X=1)=P(R B)+P(B R)=\left(\frac{4 C_1}{7 C_1} \times \frac{3 C_1}{6 C_1}\right)+\left(\frac{3 C_1}{7 C_1} \times \frac{4 C_1}{6 C_1}\right)=\frac{4}{7} \\
& P(X=2)=P(R R)=\frac{4 C_2}{7 C_2}=\frac{2}{7}
\end{aligned}
$
$\therefore$ Probability mass function

$
E(X)=\Sigma x f(x)=0+\frac{4}{7}+\frac{4}{7}=\frac{8}{7}
$

Question 3.
If $\mu$ and $\sigma^2$ are the mean and variance of the discrete random variable $X$, and $E(X+3)=10$ and $\mathrm{E}(\mathrm{X}+3)^2=116$, find $\mu$ and $\sigma^2$.
Solution:
Mean $=\mu$,
Variance $=\sigma^2$

$\begin{aligned}
& \text { Variance } \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& 65-49=16=\sigma^2 \\
& \therefore \mu=7 \text { and } \sigma^2=16
\end{aligned}$

Question 4.
Four fair coins are tossed once. Find the probability mass function, mean and variance for number of heads occurred.
Solution:
Let ' $\mathrm{X}$ ' be the number of heads occurred when four coins are tossed once. Hence ' $\mathrm{X}$ ' can take the values $0,1,2,3$ and 4 .
When 4 coins are tossed, the sample space is,
$\mathrm{S}=\{$ HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT $\}$
$
P(X=0)=\frac{1}{16}, P(X=1)=\frac{4}{16}, P(X=2)=\frac{6}{16}, P(X=3)=\frac{4}{16}, P(X=4)=\frac{1}{16} \text {. }
$
Probability mass function

$
\text { Mean } \begin{aligned}
\mathrm{E}(\mathrm{X}) & =\Sigma x f(x) \\
& =0+\frac{1}{4}+\frac{3}{4}+\frac{3}{4}+\frac{1}{4}=\frac{8}{4}=2
\end{aligned}
$
$
\begin{aligned}
& E\left(\mathrm{X}^2\right)=\Sigma x^2 f(x) \\
& =0+\frac{1}{4}+\frac{3}{2}+\frac{9}{4}+1=\frac{1+6+9+4}{4}=\frac{20}{4}=5 \\
& \text { Variance } \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\
& =5-4=1 \\
&
\end{aligned}
$

Question 5.
A commuter train arrives punctually at a station every half hour. Each morning, a student leaves his house to the train station. Let $\mathrm{X}$ denote the amount of time, in minutes, that the student waits for the train from the time he reaches the train station. It is known that the p.d.f. of $\mathrm{X}$ is.
$
f(x)= \begin{cases}1 / 30, & 0 $
Obtain and interpret the expected value of the random variable $\mathrm{X}$.
Solution:
Given, p.d.f. is $f(x)= \begin{cases}1 / 30, & 0 ' $\mathrm{X}$ ' is a continuous random variable.
$
\begin{aligned}
\therefore \text { Expected value of } \mathrm{X}=\mathrm{E}(\mathrm{X}) & =\int_0^{30} x f(x) d x \\
& =1 / 30 \int_0^{30} x d x=1 / 30\left[\frac{x^2}{2}\right]_0^{30} \\
& =\frac{1}{30}\left(\frac{30 \times 30}{2}\right)=15 \\
& =15 \text { minutes }
\end{aligned}
$

Question 6.
The time to failure in thousands of hours of an electronic equipment used in a manufactured $f(x)=\left\{\begin{array}{ll}3 e^{-3 x} & x>0 \\ 0 & \text { elsewhere }\end{array}\right.$ Find the expected life of this
computer has the density function. electronic equipment.
Solution:
Given p.d.f. is $f(x)= \begin{cases}3 e^{-3 x}, & x>0 \\ 0, & \text { elsewhere }\end{cases}$
Expected life
$
\begin{aligned}
\mathrm{E}(\mathrm{X}) & =\int_0^{\infty} x f(x) d x \\
& =3 \int_0^{\infty} x e^{-3 x} d x
\end{aligned}
$
(Using integration by parts method)
Let $u=x \Rightarrow d u=d x$
$
\int d v=\int e^{-3 x} d x \Rightarrow v=\frac{e^{-3 x}}{-3}
$

$
\begin{aligned}
\int u d v & =u v-\int v d u \\
\int x e^{-3 x} d x & =\frac{x e^{-3 x}}{-3}+\frac{1}{3} \int e^{-3 x} d x=\frac{x e^{-3 x}}{-3}+\frac{1}{3} \cdot \frac{e^{-3 x}}{-3} \\
& =-1 / 3\left[x e^{-3 x}+\frac{e^{-3 x}}{3}\right] \\
\therefore \mathrm{E}(\mathrm{X}) & =3(-1 / 3)\left[x e^{-3 x}+\frac{e^{-3 x}}{3}\right]_0^{\infty} \\
& =-\left[0-\left(0+\frac{1}{3}\right)\right]=\frac{1}{3}
\end{aligned}
$
Only $E(X)$ is required.

Question 7.
The probability density function of the random variable $\mathrm{X}$ is given by
$
f(x)=\left\{\begin{array}{ll}
16 x e^{-4 x} & \text { for } x>0 \\
0 & \text { for } x \leq 0
\end{array} \text { Find the mean and variance of } X .\right.
$
Solution:
Given p.d.f. is $f(x)= \begin{cases}16 x e^{-4 x} & \text { for } x>0 \\ 0 & \text { for } x \leq 0\end{cases}$
$
\text { Mean } \quad \mathrm{E}(\mathrm{X})=\int_{-\infty}^{\infty} x f(x) d x=16 \int_0^{\infty} x^2 e^{-4 x} d x
$
(Using integration by parts method twice)
Let $u=x^2 \Rightarrow d u=2 x d x$
and $\int d v=\int e^{-4 x} ; v=\frac{e^{-4 x}}{-4}$
[ $\because$ Integration by parts method]
$
u=x \Rightarrow d u=d x
$
and $\int d v=\int e^{-4 x} d x ; v=\frac{e^{-4 x}}{-4}$
$
\begin{aligned}
\int u d v & =u v-\int v d u \\
\int x e^{-4 x} d x & =\frac{-x e^{-4 x}}{4}+\frac{1}{4} \int e^{-4 x} d x=\frac{-x e^{-4 x}}{4}-\frac{1}{16} e^{-4 x}
\end{aligned}
$

Substituting in (1)
$
\begin{aligned}
\int x^2 e^{-4 x} d x & =\frac{x^2 e^{-4 x}}{-4}+\frac{1}{2}\left[\frac{-x e^{-4 x}}{4}-\frac{1}{16} e^{-4 x}\right] \\
\mathrm{E}(\mathrm{X}) & =16\left[\frac{x^2 e^{-4 x}}{-4}-\frac{x e^{-4 x}}{8}-\frac{e^{-4 x}}{32}\right]_0^{\infty} \\
& =16[0-(-1 / 32)]=16(1 / 32)=1 / 2 \\
\mathrm{E}\left(\mathrm{X}^2\right) & =\int_{-\infty}^{\infty} x^2 f(x) d x=16 \int_0^{\infty} x^3 e^{-4 x} d x
\end{aligned}
$
(Using integration by parts method)
Let, $u=x^3 \Rightarrow d u=3 x^2 d x$
and $\int d v=\int e^{-4 x} d x \Rightarrow v=\frac{e^{-4 x}}{-4}$
$
\begin{aligned}
\int u d v & =u v-\int v d u \\
\int x^3 e^{-4 x} d x & =-\frac{x^3 e^{-4 x}}{4}+\frac{3}{4} \int e^{-4 x} \cdot x^2 d x \\
& =-\frac{x^3 e^{-4 x}}{4}+\frac{3}{4}\left[\frac{x^2 e^{-4 x}}{-4}-\frac{x e^{-4 x}}{8}-\frac{e^{-4 x}}{32}\right]
\end{aligned}
$
$[\because$ Using $E(X)$ integration]
$
=-\frac{x^3 e^{-4 x}}{4}-\frac{3}{16} x^2 e^{-4 x}-\frac{3}{32} x e^{-4 x}-\frac{3}{128} e^{-4 x}
$
$
\begin{aligned}
\mathrm{E}\left(\mathrm{X}^2\right) & =16\left[-\frac{x^3 e^{-4 x}}{4}-\frac{3}{16} x^2 e^{-4 x}-\frac{3}{32} x e^{-4 x}-\frac{3}{128} e^{-4 x}\right]_0^{\infty} \\
& =16[0-(-3 / 128)]=16\left(\frac{3}{128}\right)=\frac{3}{8}
\end{aligned}
$
Variance $\quad \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2$

$
=\frac{3}{8}-\frac{1}{4}=\frac{3-2}{8}=\frac{1}{8}
$
Question 8.
A lottery with 600 tickets gives one prize of ₹ 200 , four prizes of ₹ 100 , and six prizes of ₹ 50 . If the
ticket costs is ₹ 2 , find the expected winning amount of a ticket.
Solution:
Given, total number of tickets $=600$
Prizes to be given : One prize of Rs. 200
Four prizes of Rs. 100
Six prizes of Rs. 50
Let ' $\mathrm{X}$ ' be the random variable denotes the winning amount and it can take the values 200,100 and 50.
Probability of winning Rs. $200=\frac{1}{600}$
Probability of winning Rs. $100=\frac{4}{600}$
Probability of winning Rs. $50=\frac{6}{600}$
$\therefore$ Probability mass function is

$
\begin{aligned}
\therefore \mathrm{E}(\mathrm{X}) & =200 \Sigma x f(x) \\
& =\frac{200}{600}+\frac{400}{600}+\frac{300}{600}=\frac{900}{600}=1.5
\end{aligned}
$
Expected winning amount $=$ Amount won $-$ Cost of lottery
$
=1.50-2.00=-0.50
$
i.e., Loss of Rs. $0.50$