Question: Find the value of c such that

$ f(x)= \begin{cases} \frac {x} {6} + c & 0\le x \le 3 \\ 0 & \mbox{elsewhere} \end{cases} $

is a p.d.f. Also find $ P(1\le x \le 2) $

Solution:

$ f(x)\le 0 if c\ge 0 $; Also we must have

$ \int\limits_{-infty}^{infty}f(x)dx=1 $

i.e., $ \int\limits_{0}^{3} ( \frac {x} {6} + c)dx=1 $

On integrating this we get,

$ \frac {3} {4} + 3c=1 $

Therefore,

$ c=\frac {1} {12} $

Now,

$ P(1\le x\le 2)=\int\limits_{1}^{2} f(x) dx $

$ =\int\limits_{1}^{2}(\frac {x} {6} + \frac {1} {12}) dx $

Which on further integration leads to,

$ \frac {1} {12} [(4+2)-(1+1)]=\frac {1} {3} $

Thus,

$ P(1\le x \le 2)= \frac {1} {3} $

Back to ECE302 Fall 2008 Prof. Sanghavi