Exercise 11.3-Additional Problems - Chapter 11 Probability Distributions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
For the probability density function $f(x)=\left\{\begin{array}{cc}2 e^{-2 x}, & x>0 \\ 0, & x \leq 0\end{array}\right.$, find $F(2)$
Solution:
$
\begin{aligned}
\mathrm{F}(2)=\mathrm{P}(\mathrm{X} \leq 2) & =\int_{-\infty}^2 f(x) d x \\
& =\int_0^2 2 e^{-2 x} d x=2 \cdot\left[\frac{e^{-2 x}}{-2}\right]_0^2=-\left[e^{-4}-1\right]=1-e^{-4}=\frac{e^4-1}{e^4}
\end{aligned}
$
Question 2.
For the probability density function $f(x)=\left\{\begin{array}{lc}c x(1-x)^3, & 0<x<1 \\ 0, & \text { elsewhere }\end{array}\right.$ (i) find the constant $c$; (ii) $\mathrm{P}(x<1 / 2)-$
Solution:

(i) Since $f(x)$ is a probability density function $\int_{-\infty}^{\infty} f(x) d x=1$
$
\begin{aligned}
& \text { i.e., } \quad \int_0^1 c x(1-x)^3 d x=1 \text {; i.e., } \quad c \int_0^1(1-x)[1-(1-x)]^3 d x=1 \\
& c \int_0^1(1-x)\left(x^3\right) d x=1 \\
& c \int_0^1\left(x^3-x^4\right) d x=1 \text {; i.e., } c\left[\frac{x^4}{4}-\frac{x^5}{5}\right]_0^1=1 \\
& c\left[\frac{1}{4}-\frac{5}{5}\right]=1 \text {; i.e., } c\left[\frac{5-4}{20}\right]=1 \Rightarrow c=20 \\
&
\end{aligned}
$

(ii)
$
\begin{aligned}
& P(x<1 / 2)=\int_{-\infty}^{1 / 2} f(x) d x=\int_0^{1 / 2} c x(1-x)^3 d x=20 \int_0^{1 / 2} x(1-x)^3 d x \\
& \text { Put } 1-x=t ; x=1-t \\
& -d x=d t \Rightarrow d x=-d t \\
& 1=20 \int(1-t) t^3(-d t)=-20 \int\left(t^3-t^4\right) d t \\
& =-20\left[\frac{t^4}{4}-\frac{t^5}{5}\right]=-20\left[\frac{1}{4}(1-x)^4-\frac{1}{5}(1-x)^5\right]_0^{1 / 2} \\
& =-20\left\{\left[\frac{1}{4}\left(\frac{1}{2}\right)^4-\frac{1}{5}\left(\frac{1}{2}\right)^5\right]-\left[\frac{1}{4}-\frac{1}{5}\right]\right\} \\
& =-20\left[\frac{1}{4} \times \frac{1}{16}-\frac{1}{5} \times \frac{1}{32}-\frac{1}{4}+\frac{1}{5}\right]=-20\left[\frac{1}{64}-\frac{1}{160}-\frac{1}{4}+\frac{1}{5}\right] \\
& =\frac{-5}{6}+\frac{1}{8}+5-4=\frac{-5+2+80-64}{16}=\frac{13}{16} \\
&
\end{aligned}
$

Question 3.
$
f(x)=\left\{\begin{array}{cc}
k x^{\alpha-1} e^{-\beta x^\alpha}, & x, \alpha, \beta>0 \\
0, & \text { elsewhere }
\end{array}\right. \text { Find }
$
The probability density function of a random variable $\mathrm{x}$ is
(i) $\mathrm{k}$;
(ii) $\mathrm{P}(\mathrm{X}>10)$
Solution:

(i) Since $f(x)$ is a probability density function $\int_{-\infty}^{\infty} f(x) d x=1$
$
\begin{aligned}
& \text { i.e., } \quad \int_0^{\infty} k x^{\alpha-1} e^{-\beta x^\alpha} d x=1 \\
& \text { Put } e^{-\beta x^\alpha}=t ; d t=e^{-\beta x^\alpha}\left\{-\beta \alpha x^{\alpha-1} d x\right\} \\
& \text { i.e., } \left.d t=t\left(-\alpha \beta x^{\alpha-1} d x\right) \text {; So, } x^{\alpha-1} d x=\frac{-d t}{t \alpha \beta}\right] \\
& \therefore \text { (1) becomes } \quad k \int t \frac{(-d t)}{t \alpha \beta}=\frac{-k}{\alpha \beta} \int d t=-k(t) \\
& =\frac{-k}{\alpha \beta}\left[e^{-\beta x^\alpha}\right]_0^{\infty}=1 \quad \text { (given) } \\
& =\frac{-k}{\alpha \beta}\{0-1\}=1 \\
& \text { i.e., } \frac{k}{\alpha \beta}=1 \Rightarrow k=\alpha \beta \\
&
\end{aligned}
$

(ii)
$
\begin{aligned}
P(X>10) & =\int_{10}^{\infty} f(x) d x=\frac{-\alpha \beta}{\alpha \beta}\left\{e^{-\beta x^\alpha}\right\}_{10}^{\infty} \\
& =-1\left\{0-e^{-\beta 10^\alpha}\right\}=e^{-\beta\left(10^\alpha\right)}
\end{aligned}
$

Question 4.
A continuous random variable $\mathrm{x}$ has the p.d.f. defined by
$
f(x)=\left\{\begin{array}{cc}
C e^{-\alpha x}, & 0<x<\infty \\
0 & \text { elsewhere }
\end{array}\right.
$
Find the value of $\mathrm{C}$ if $\mathrm{a}>0$.
Solution:
Since $f(x)$ is a probability density function $\int_{-\infty}^{\infty} f(x) d x=1$ i.e., $\int_0^{\infty} \mathrm{C} e^{-a x} d x=1$
$
\mathrm{C}\left[\frac{e^{-a x}}{-a}\right]_0^{\infty}=1 \text {; i.e., } \frac{-\mathrm{C}}{a}\{0-1\}=1 \Rightarrow \frac{\mathrm{C}}{a}=1 \therefore \mathrm{C}=a
$