socorraob

2021-12-12

Suppose $f\left(x\right)=0.125x$ for $0. determine the mean and variance of X.

Hector Roberts

Mean $\left[E\left(x\right)\right]$ of $x$
The mean formula is given by,

$=0.125{\left[\frac{{x}^{3}}{3}\right]}_{0}^{4}$ {since
$=0.125\left[\frac{{4}^{3}}{3}-0\right]$
$=0.125\left(\frac{64}{3}\right)$
$E\left(x\right)=\frac{8}{3}$ or $2.667$

Bertha Jordan

The formula for varience of x is given by,
Varience(x)= $E\left[{x}^{2}\right]-{\left[E\left[x\right]\right]}^{2}$
to find $E\left[{x}^{2}\right]$:
$E\left[{x}^{2}\right]={\int }_{x}{x}^{2}f\left(x\right)dx$
$={\int }_{0}^{4}{x}^{2}\cdot 0.125xdx$
$={\int }_{0}^{4}{x}^{3}\cdot 0.125dx$
$=0.125{\int }_{0}^{4}{x}^{3}dx$
$=0.125{\left[\frac{{x}^{4}}{x}\right]}_{0}^{4}\frac{\mathrm{sin}cePSK\int {x}^{n}dx=\left({x}^{n+1}\right)}{n+1}$}
$=0.125\left[\frac{{4}^{4}}{4}-0\right]$
$=0.125\cdot {4}^{3}$
$E\left[{x}^{2}\right]=8$
Thus $=E\left[{x}^{2}\right]-{\left[E\left[x\right]\right]}^{2}$
We know that $E\left[{x}^{2}\right]=8$ and $E\left[x\right]=2.667$
Thus $=8-{\left(2.667\right)}^{2}$
$=8-7.113$
$=0.887$
Varriance of x $0.887$

Do you have a similar question?