 # The probability Distribution Function Formula

The Probability Density Function(PDF) is the probability function which is represented for the density of a continuous random variable lying between a certain range of values. The probability Density function is defined by the formula,

 P(a

## Questions on the probability distribution function

### Question 1:

The pdf of a distribution is given as $f(x)= \left\{\begin{matrix}x;\; for\ 0< x< 1 \\ 2-x;\; for \ 1< x< 2 \\ 0;\; for\ x> 2 \end{matrix}\right \}$

Calculate the density within the interval $(0.5< x< 1.5)$

### Solution:

$P(0.5< x< 1.5)=\int_{0.5}^{1.5}f(x)dx$ $=\int_{0.5}^{1}f(x)dx+\int_{1}^{1.5}f(x)dx$ $=\int_{0.5}^{1}xdx+\int_{1}^{1.5}(2-x)dx$ $=\left ( \frac{x^{2}}{2} \right )_{0.5}^{1}+\left ( (2x-\frac{x^{2}}{2}) \right )_{1}^{1.5}$

= 0

Example 2: Let x be a random variable with PDF is given by $f(x)=\left\{\begin{matrix} kx^2; &|x|\le1\\ 0; & otherwise \end{matrix}\right.$. Find the value of k and and P(x ≥ ½).

Solution:
Given,
$f(x)=\left\{\begin{matrix} kx^2; &|x|\le1\\ 0; & otherwise \end{matrix}\right.$ To find the value of k, consider the below expression.
$\int_{-\infty}^{\infty}f(u) du = 1\\ \int_{-1}^{1}cu^2 du=1\\ c[\frac{u^3}{3}]_{-1}^{1}=1\\ c[\frac{1}{3}+\frac{1}{3}]=1\\ \frac{2}{3}c=1\\ c = \frac{3}{2}$

No finding P(x≥ ½),
$P(x\ge\frac{1}{2})=\int_{\frac{1}{2}}^{1}cx^2 dx\\ =\frac{3}{2}[\frac{x^3}{3}]_{\frac{1}{2}}^{1}\\ =\frac{3}{2}[\frac{1}{3}-\frac{1}{24}]\\ =\frac{3}{2}\times \frac{7}{24}\\=\frac{7}{16}$

Therefore, k = 3/2 and P(x≥1/2) = 7/16.

Example 3: Suppose x be a random variable and PDF is give by $f(x)=\left\{\begin{matrix} x^2+1; & x\ge 0\\ 0; &x<0 \end{matrix}\right.$
Find P(1 < x < 3).
Solution:
Given,
$f(x)=\left\{\begin{matrix} x^2+1; & x\ge 0\\ 0; &x<0 \end{matrix}\right.$ $P(1<x<3)=\int_{1}^{3}(x^2+1) dx\\ =[\frac{x^3}{3}+x]_{1}^{3}\\ =[(\frac{27}{3}+3)-(\frac{1}{3}+1)]\\ =[(9+3)-\frac{4}{3}]\\ =\frac{36-4}{3}\\=\frac{32}{3}$

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