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

socorraob 2021-12-12 Answered
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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Expert Answer

Hector Roberts
Answered 2021-12-13 Author has 3535 answers
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
Answered 2021-12-14 Author has 1632 answers
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 \(\displaystyle{0.887}\)
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