# Suppose f(x) = 0.125x for 0 < x < 4.

Suppose $$\displaystyle{f{{\left({x}\right)}}}={0.125}{x}$$ for $$\displaystyle{0}{ < }{x}{ < }{4}$$. determine the mean and variance of X.

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Hector Roberts
Mean $$\displaystyle{\left[{E}{\left({x}\right)}\right]}$$ of $$\displaystyle{x}$$
The formula for mean is given by,
$$\displaystyle{E}{\left({x}\right)}=\int_{{x}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{4}}^{{0}}}{x}\cdot{0.125}{x}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{4}}^{{0}}}{x}^{{2}}\cdot{0.125}{x}{\left.{d}{x}\right.}$$
$$\displaystyle={0.125}={\int_{{4}}^{{0}}}{x}^{{2}}{\left.{d}{x}\right.}$$
$$\displaystyle={0.125}{\left[\frac{{{x}^{{3}}}}{{{3}}}\right]}^{{4}}_{0}$$ {since $$\displaystyle\int{x}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}$$}
$$\displaystyle={0.125}{\left[\frac{{{4}^{{3}}}}{{{3}}}-{0}\right]}$$
$$\displaystyle={0.125}{\left(\frac{{{64}}}{{3}}\right)}$$
$$\displaystyle{E}{\left({x}\right)}=\frac{{8}}{{3}}$$ or $$\displaystyle{2.667}$$
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Bertha Jordan
The formula for varience of x is given by,
Varience(x)= $$\displaystyle{E}{\left[{x}^{{2}}\right]}-{\left[{E}{\left[{x}\right]}\right]}^{{2}}$$
to find $$\displaystyle{E}{\left[{x}^{{2}}\right]}$$:
$$\displaystyle{E}{\left[{x}^{{2}}\right]}=\int_{{x}}{x}^{{2}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\int^{{4}}_{0}{x}^{{2}}\cdot{0.125}{x}{\left.{d}{x}\right.}$$
$$\displaystyle=\int^{{4}}_{0}{x}^{{3}}\cdot{0.125}{\left.{d}{x}\right.}$$
$$\displaystyle={0.125}\int^{{4}}_{0}{x}^{{3}}{\left.{d}{x}\right.}$$
$$\displaystyle={0.125}{\left[{\frac{{{x}^{{4}}}}{{{x}}}}\right]}^{{4}}_{0}\frac{{{\sin{{c}}}{e}{P}{S}{K}\int{x}^{{n}}{\left.{d}{x}\right.}={\left({x}^{{{n}+{1}}}\right)}}}{{{n}+{1}}}$$}
$$\displaystyle={0.125}{\left[\frac{{{4}^{{4}}}}{{4}}-{0}\right]}$$
$$\displaystyle={0.125}\cdot{4}^{{3}}$$
$$\displaystyle{E}{\left[{x}^{{2}}\right]}={8}$$
Thus $$\displaystyle={E}{\left[{x}^{{2}}\right]}-{\left[{E}{\left[{x}\right]}\right]}^{{2}}$$
We know that $$\displaystyle{E}{\left[{x}^{{2}}\right]}={8}$$ and $$\displaystyle{E}{\left[{x}\right]}={2.667}$$
Thus $$\displaystyle={8}-{\left({2.667}\right)}^{{2}}$$
$$\displaystyle={8}-{7.113}$$
$$\displaystyle={0.887}$$
Varriance of x $$\displaystyle{0.887}$$