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,
Questions on the probability distribution function
Question 1:
The pdf of a distribution is given as
\(\begin{array}{l}f(x)= \left\{\begin{matrix}x;\; for\ 0< x< 1 \\ 2-x;\; for \ 1< x< 2 \\ 0;\; for\ x> 2 \end{matrix}\right \}\end{array} \)
Calculate the density within the interval
\(\begin{array}{l}(0.5< x< 1.5)\end{array} \)
Solution:
\(\begin{array}{l}P(0.5< x< 1.5)=\int_{0.5}^{1.5}f(x)dx\end{array} \)
\(\begin{array}{l}=\int_{0.5}^{1}f(x)dx+\int_{1}^{1.5}f(x)dx\end{array} \)
\(\begin{array}{l}=\int_{0.5}^{1}xdx+\int_{1}^{1.5}(2-x)dx\end{array} \)
\(\begin{array}{l}=\left ( \frac{x^{2}}{2} \right )_{0.5}^{1}+\left ( (2x-\frac{x^{2}}{2}) \right )_{1}^{1.5}\end{array} \)
= 0
Example 2: Let x be a random variable with PDF is given by \(\begin{array}{l}f(x)=\left\{\begin{matrix} kx^2; &|x|\le1\\ 0; & otherwise \end{matrix}\right.\end{array} \)
. Find the value of k and and P(x ≥ ½).
Solution:
Given,
\(\begin{array}{l}f(x)=\left\{\begin{matrix} kx^2; &|x|\le1\\ 0; & otherwise \end{matrix}\right.\end{array} \)
To find the value of k, consider the below expression.
\(\begin{array}{l}\int_{-\infty}^{\infty}f(u) du = 1\\ \int_{-1}^{1}cu^2 du=1\\ c[\frac{u^3}{3}]_{-1}^{1}=1\\ c[\frac{1}{3}+\frac{1}{3}]=1\\ \frac{2}{3}c=1\\ c = \frac{3}{2}\end{array} \)
No finding P(x≥ ½),
\(\begin{array}{l}P(x\ge\frac{1}{2})=\int_{\frac{1}{2}}^{1}cx^2 dx\\ =\frac{3}{2}[\frac{x^3}{3}]_{\frac{1}{2}}^{1}\\ =\frac{3}{2}[\frac{1}{3}-\frac{1}{24}]\\ =\frac{3}{2}\times \frac{7}{24}\\=\frac{7}{16}\end{array} \)
Therefore, k = 3/2 and P(x≥1/2) = 7/16.
Example 3: Suppose x be a random variable and PDF is give by \(\begin{array}{l}f(x)=\left\{\begin{matrix} x^2+1; & x\ge 0\\ 0; &x<0 \end{matrix}\right.\end{array} \)
Find P(1 < x < 3).
Solution:
Given,
\(\begin{array}{l}f(x)=\left\{\begin{matrix} x^2+1; & x\ge 0\\ 0; &x<0 \end{matrix}\right.\end{array} \)
\(\begin{array}{l}P(1<x<3)=\int_{1}^{3}(x^2+1) dx\\ =[\frac{x^3}{3}+x]_{1}^{3}\\ =[(\frac{27}{3}+3)-(\frac{1}{3}+1)]\\ =[(9+3)-\frac{4}{3}]\\ =\frac{36-4}{3}\\=\frac{32}{3}\end{array} \)
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