Question

Let X = amount of time for which a book is taken of a college library by a randomly selected student and suppose X has p.d.f.
$f\left( x \right)=\left\{ \begin{align}
  & 0.5x,\,\,\,\,\,0.5\le x\le 2 \\
 & 0,\,\,\,\,\,\,\,\,\,\,\text{otherwise} \\
\end{align} \right.$
Calculate (a) $P\left( X\le 1 \right)$ (b) $P\left( 0.5\le X\le 1.5 \right)$ (c)$P\left( X>1.5 \right)$.

Answer 1

Hint: The function has a different value in different ranges. To tackle this problem, we need to split the function, we will encounter such a situation in section (c). The integration of the function in the given limits will give the probability of the function.

Complete step-by-step answer:
We have been given a function which is valid in the range of zero to two.
To get the probability of the function we need to integrate the function within the given limit.
We can express that as follows,
$P\left( a\le x\le b \right)=\int_{a}^{b}{f\left( x \right)}...........(i)$
Let's start by solving the (a) part.
From equation (i) we can say that,
$P\left( X\le 1 \right)=\int_{0}^{1}{0.5x\,dx}...........(ii)$
Integrating, the above equation we get that,
$P\left( X\le 1 \right)=\int_{0}^{1}{0.5x\,dx}=0.5\left[ \dfrac{{{x}^{2}}}{2} \right]_{0}^{1}$ .
Solving the equation further we get,
$P\left( X\le 1 \right)=0.5\times \left( \dfrac{1}{2}-0 \right)=0.5\times 0.5=0.25$

Therefore, $P\left( X\le 1 \right)=0.25$ .
Moving on to the (b) part of the problem,
We need to find the value of $P\left( 0.5\le X\le 1.5 \right)$ .
The range of the required probability $0.5\le x\le 1.5$ lies within the range of the function.
Therefore, by integrating the above equation we get,
$P\left( 0.5\le X\le 1.5 \right)=\int_{0.5}^{1.5}{0.5x\,dx}=0.5\left[ \dfrac{{{x}^{2}}}{2} \right]_{0.5}^{1.5}$ .
Solving the equation further we get,
$P\left( 0.5\le X\le 1.5 \right)=0.5\times \left( \dfrac{{{1.5}^{2}}}{2}-\dfrac{{{0.5}^{2}}}{2} \right)=0.5\times \left( \dfrac{2.25-0.25}{2} \right)=0.5\times \dfrac{2}{2}=0.5$ .
Therefore, $P\left( 0.5\le X\le 1.5 \right)=0.5$ .
Let's start with the (c) section.
There is no upper limit of the probability to be found. So we will assume the upper limit to be infinite.
From equation (i) we get,
$P\left( X>1.5 \right)=\int_{1.5}^{\infty }{0.5x\,dx}=\int_{1.5}^{2}{0.5x}dx+\int_{2}^{\infty }{0dx}...............(ii)$
We break the integral into two parts because the function is discontinuous and has a different value at two different integrals.
Solving further we get,
$P\left( X>1.5 \right)=0.5\left[ \dfrac{{{x}^{2}}}{2} \right]_{1.5}^{2}+0=0.5\left( \dfrac{{{2}^{2}}-{{1.5}^{2}}}{2} \right)=0.5\times \left( \dfrac{4-2.25}{2} \right)=\dfrac{7}{16}=0.4375$
Hence, this is the required solution.

Note: we need to keep in mind to check the condition without performing integration because if the required probability crosses the value then we need to break the limits into two parts and solve them individually because the function is no longer continuous at that point. Also, to crosscheck the probability of any function will always be less than equal to 1.