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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 11.3
Question 1.
The probability density function of $\mathrm{X}$ is given by
$
f(x)=\left\{\begin{array}{cl}
k x e^{-2 x} & \text { for } x>0 \\
0 & \text { for } x \leq 0
\end{array} \text { Find the value of } k\right.
$
Solution:
$
f(x)=\left\{\begin{array}{cc}
k x e^{-2 x} & \text { for } x>0 \\
0 & \text { for } x \leq 0
\end{array}\right.
$
Since, $f(x)$ is a Probability density function
$
\text { i.e., } \begin{aligned}
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} 0 d x+k \int_0^{\infty} x e^{-2 x} d x & =1 \\
k \int_0^{\infty} x e^{-2 x} d x & =1
\end{aligned}
$
Using, integration by parts method

Question 2.
The probability density function of $X$ is $f(x)=\left\{\begin{array}{rr}x & 0 Solution:
Given probability density function is $f(x)=\left\{\begin{array}{rr}x, & 0 (i)
$
\begin{aligned}
\mathrm{P}(0.2 \leq \mathrm{X}<0.6) & =\int_{0.2}^{0.6} x d x \\
& =\left[\frac{x^2}{2}\right]_{0.2}^{0.6}=\frac{1}{2}[0.36-0.04]=\frac{0.32}{2}=0.16
\end{aligned}
$
(ii)
$
\begin{aligned}
\mathrm{P}(1.2 \leq \mathrm{X}<1.8) & =\int_{1.2}^{1.8}(2-x) d x \\
& =\left[2 x-\frac{x^2}{2}\right]_{1.2}^{1.8}=(3.6-1.62)-(2.4-0.72) \\
& =1.98-1.68=0.30
\end{aligned}
$
(iii)
$
\begin{aligned}
\mathrm{P}(0.5 \leq \mathrm{X}<1.5) & =\int_{0.5}^1 x d x+\int_1^{1.5}(2-x) d x \\
& =\left[\frac{x^2}{2}\right]_{0.5}^1+\left[2 x-\frac{x^2}{2}\right]_1^{1.5}=0.375+0.375=0.75
\end{aligned}
$

Question 3.
Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function
$
f(x)=\left\{\begin{array}{lc}
k & 200 \leq x \leq 600 \\
0 & \text { otherwise }
\end{array}\right.
$
Find (i) the value of $\mathrm{k}$
(ii) the distribution function
(iii) the probability that daily sales will fall between 300 litres and 500 litres?
Solution:

Given p.d.f. is $f(x)=\left\{\begin{array}{lr}k, & 200 \leq x \leq 600 \\ 0, & \text { otherwise }\end{array}\right.$
(i) Since $f(x)$ is a p.d.f, $\int_{-\infty}^{\infty} f(x) d x=1$
i.e.,
$
\begin{aligned}
k \int_{200}^{600} d x & =1 \\
k[x]_{200}^{600} & =1 \\
k(600-200) & =1 \\
400 k & =1 \Rightarrow k=\frac{1}{400}
\end{aligned}
$
(ii) Now, the probability density function $f(x)=\left\{\begin{array}{cc}1 / 400, & 200 \leq x \leq 600 \\ 0, & \text { otherwise }\end{array}\right.$ Integrating $f(x)$, we get probability density function $\mathrm{F}(x)=\left\{\begin{array}{ccc}0 & \text { for } & x<200 \\ x / 400 & \text { for } & 200 \leq x \leq 600 \\ 0 & \text { for } & x>600\end{array}\right.$
(iii) The probability that daily sales will fall between 300 litres and 500 litres is
$
\begin{aligned}
\int_{300}^{500} f(x) d x & =\frac{1}{400} \int_{300}^{500} d x \\
& =\frac{1}{400}[x]_{300}^{500}=\frac{1}{400}[500-300]=\frac{200}{400}=\frac{1}{2}
\end{aligned}
$

Question 4.
The probability density function of $\mathrm{X}$ is given by $f(x)=\left\{\begin{array}{rr}k e^{-\frac{x}{3}} & \text { for } x>0 \\ 0 & \text { for } x \leq 0\end{array}\right.$
Find ( $i$ ) the value of $k$
(ii) the distribution function
(iii) $\mathbf{P}(\mathbf{X}<3)$
(iv) $\mathrm{P}(5 \leq \mathrm{X})$
(v) $\mathrm{P}(\mathrm{X} \leq 4)$.
Solution:

$\text { Given probability density function } f(x)=\left\{\begin{array}{rr}
k e^{-\frac{x}{3}} & \text { for } x>0 \\
0 & \text { for } x \leq 0
\end{array}\right.$

(i) Since $f(x)$ is a p.d.f, $\int_{-\infty}^{\infty} f(x) d x=1$
i.e.,
$
\begin{aligned}
k \int_0^{\infty} e^{-x / 3} d x & =1 \\
k\left[\frac{e^{-x / 3}}{-1 / 3}\right]_0^{\infty} & =1 \\
-3 k\left[e^{-\infty}-e^0\right] & =1 \\
-3 k(0-1) & =1 \Rightarrow 3 k=1 \Rightarrow k=1 / 3
\end{aligned}
$
(ii) Probability density function $f(x)=\left\{\begin{array}{ccc}1 / 3 e^{-x / 3} & \text { for } & x>0 \\ 0 & \text { for } & x \leq 0\end{array}\right.$ Integrating $f(x)$, we get
Probability density function $\mathrm{F}(x)=\left\{\begin{array}{ccc}\frac{1}{3} \cdot \frac{e^{-x / 3}}{-1 / 3} & \text { for } & x>0 \\ 0 & \text { for } & x \leq 0\end{array}\right.$
$
\therefore \mathrm{F}(x)=\left\{\begin{array}{ccc}
-e^{-x / 3} & \text { for } & x>0 \\
0 & \text { for } & x \leq 0
\end{array}\right.
$
(iii)
$
\begin{aligned}
\mathrm{P}(\mathrm{X}<3) & =\int_0^3 f(x) d x \\
& =1 / 3 \int_0^3 e^{-x / 3} d x=-\left[e^{-x / 3}\right]_0^3=-\left[e^{-1}-e^0\right]=1-e^{-1}
\end{aligned}
$

(iv)
$
\begin{aligned}
\mathrm{P}(5 \leq \mathrm{X}) & =1 / 3 \int_5^x e^{-x / 3} d x \\
& =-\left[e^{-x / 3}\right]_5^x=-\left[e^{-x / 3}-e^{-5 / 3}\right] \\
& =e^{-5 / 3}-e^{-x / 3}
\end{aligned}
$
(v)
$
\begin{aligned}
\mathrm{P}(\mathrm{X} \leq 4) & =1 / 3 \int_0^4 e^{-x / 3} d x \\
& =-\left[e^{-x / 3}\right]_0^4=-\left[e^{-4 / 3}-1\right]=1-e^{-4 / 3}
\end{aligned}
$

Question 5.
If $\mathrm{X}$ is a random variable with probability density function $\mathrm{f}(\mathrm{x})$ given by,
$
f(x)=\left\{\begin{array}{rr}
x+1, & -1 \leq x<0 \\
-x+1, & 0 \leq x<1 \\
0 & \text { otherwise }
\end{array}\right.
$
Solution:

Given probability density function is $f(x)=\left\{\begin{array}{rr}x+1, & -1 \leq x<0 \\ -x+1, & 0 \leq x<1 \\ 0, & \text { otherwise }\end{array}\right.$
(i) Integrating $f(x)$ and separating with respect to the intervals we get, Probability distribution function $\mathrm{F}(x)=\left\{\begin{array}{ccc}0 & \text { for } & x \leq-1 \\ \frac{x^2}{2}+x+\frac{1}{2} & \text { for } & -1 \leq x<0 \\ -\frac{x^2}{2}+x & \text { for } & 0 \leq x<1 \\ x-1 & \text { for } & 1 \leq x\end{array}\right.$
$f(x)$ is a Probability density function.
$
\begin{aligned}
\int_{-1}^0(x+1) d x & =1 \\
{\left[\frac{x^2}{2}+x\right]_{-1}^0+c } & =1 \\
1 / 2+c & =1 \Rightarrow c=1-1 / 2=1 / 2
\end{aligned}
$
$
\text { (ii) } \begin{aligned}
\quad \mathrm{P}(-0.5 \leq \mathrm{X} \leq 0.5) & =\int_{-0.5}^0(x+1) d x+\int_0^{0.5}(-x+1) d x \\
= & {\left[\frac{x^2}{2}+x\right]_{-0.5}^0+\left[-\frac{x^2}{2}+x\right]_0^{0.5} } \\
= & {[0-(0.125-0.5)]+[-0.125+0.5] } \\
= & 1-0.25=0.75
\end{aligned}
$

Question 6.
If $\mathrm{X}$ is the random variable with probability density function $\mathrm{F}(\mathrm{x})$ given by,
$
F(x)= \begin{cases}0, & x<0 \\ \frac{1}{2}\left(x^2+x\right) & 0 \leq x<1 \\ 1, & x \geq 1\end{cases}
$
then find (i) the distribution function $f(x)$
(ii) $\mathrm{P}(0.3 \leq \mathrm{X} \leq 0.6)$
Solution:
Given probability distribution function $\mathrm{F}(x)= \begin{cases}0, & x<0 \\ \frac{1}{2}\left(x^2+x\right) & 0 \leq x<1 \\ 1, & x \geq 1\end{cases}$
(i) Differentiating $\mathrm{F}(x)$, we get $f(x)$
i.e., Probability density function $f(x)=\mathrm{F}^{\prime}(x)= \begin{cases}0, & x<0 \\ \frac{1}{2}(2 x+1), & 0 \leq x<1 \\ 0, & x \leq 1\end{cases}$
(ii)
$
\begin{aligned}
\mathrm{P}(0.3 \leq \mathrm{X} \leq 0.6) & =1 / 2 \int_{0.3}^{0.6}(2 x+1) d x=1 / 2\left[x^2+x\right]_{0.3}^{0.6} \\
& =1 / 2[(0.36+0.6)-(0.09+0.3)] \\
& =1 / 2(0.57)=0.285
\end{aligned}
$