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Probability Distributions (p-2) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Probability Distributions (p-2)3 Marks
Question
Check whether each of the following is $f(x)= \begin{cases}x & \text { for } 0 \leq x \leq 1 \\ 2-x & \text { for } 1<x \leq 2\end{cases}$
✓

Answer

Given function is
$f ( x )= x _1 0 \leq x \leq 1$
Each $f(x) \geq 0$, as $x \geq 0$.
and $\begin{aligned} \int_0^1 f(x) d x & =\int_0^1 x d x \\ &\end{aligned} $
$=\left[\frac{x^2}{2}\right]_0^1 $
$=\frac{1}{2} $
Also, $f(x)=2-x, 1 \leq x \leq 2$
$\Rightarrow$ Each $f(x) \geq 0,$ as $x \leq 2.$
and $\int_1^2 f(x) d x =\int_1^2(2-x) d x$
$ =2 \int_1^2 1 d x-\int_1^2 x d x $
$ =2[x]_1^2-\left(\frac{x^2}{2}\right)_1^2$
$=2(2-1)-\frac{1}{2}(4-1)$
$=1-\frac{3}{2}$
$=\frac{1}{2} $
Now, for the total range of $0 \leq x \leq 2$.
$ \int_0^2 f(x) d x =\int_0^1 f(x) d x+\int_1^2 f(x) d x$
$ =\frac{1}{2}+\frac{1}{2}$
$ =1 \text { } $
$\therefore$ The given function is a $p.d.f.$ of $x$.

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