logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

A random variable $x$ has a probability density function $f(x) = \left\{ \begin{array}{l l} k & \quad \text{0<x<2n}\\ 0 & \quad \text{elsewhere} \end{array} \right.$ Find $p(0<x<\large\frac{\pi}{2})$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • The probability density function (continuous probability function $f(x)$ satisfies the following properties :
  • (i) $P(a\leq x\leq b)=\int_a^b f(x) dx$
  • (ii) $f(x)$ is non-negative for all real $x$
  • (iii) $\int_{-\infty}^\infty f(x) dx=1$
  • Also $P(x=a)=0$
  • $P(a\leq x\leq b)=P(a\leq x\leq b)$=P(a < x < b)
Step 1:
$\int_{-\infty}^{\infty} f(x) dx=1$
$\int_0^{2\pi} kdx=1$
Since $f(x)=0$ elsewhere
$\therefore kx\bigg]_0^{2\pi}=1$
$2\pi k=1$
$k=\large\frac{1}{2\pi}$
Step 2:
$P(0 < x <\large\frac{\pi}{2})=\int_0^{\Large\frac{\pi}{2}}$$f(x)dx$
$\qquad\qquad\;\;\;\;\;\;=\int_0^{\Large\frac{\pi}{2}}\large\frac{dx}{2\pi}$
$\qquad\qquad\;\;\;\;\;\;=\large\frac{x}{2\pi}\bigg]_0^{\Large\frac{\pi}{2}}$
$\qquad\qquad\;\;\;\;\;\;=\large\frac{1}{4}$
answered Sep 17, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...