The p.d.f. of the r.v. x. is given by f_x(x)= l k x , 00, otherwi… - Vidyadip

Question

The $p.d.f.$ of the $r.v. x.$ is given by $ f_x(x)=\left\{\begin{array}{l} \frac{k}{\sqrt{x}}, 00, \text { otherwise } \end{array}\right. $ Determine $k, c.d.f.$ of $X$ and hence find $P(X \leq 2)$ and $P(X \geq 1).$

✓

Answer

We know that
$\int_0^4 \frac{k}{\sqrt{x}} d x=1$
$\therefore k \int_0^4 x^{\frac{-1}{2}} d x=1$
$\therefore k\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_0^4=1$
$\therefore 2 k[\sqrt{4}-\sqrt{0}] =1$
$\therefore 4 k =1$
$\therefore k =\frac{1}{4}$
$\text{c.d.f.}$ of $f(x)$ is
$F(x)=\int_0^x \frac{k}{\sqrt{x}} d x$
$\therefore F(x) =\int_0^x \frac{1}{4} x^{\frac{-1}{2}} d x=\frac{1}{4}\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_0^x=\frac{\sqrt{x}}{2}$
$P(X \leq 2) =F(2)$
$ =\frac{\sqrt{2}}{2}=0.7071$
$P(X \geq 1) =1-P(X<1)=1-F(1)$
$ =1-\frac{1}{2}$
$=\frac{1}{2}$
$=0.5$

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