MATH103 April 2005
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Question 02

A certain random variable X takes values in the interval $0\leq x\leq 2$, with probability density
 $\displaystyle p(x)=1x/2$
Find the mean and variance of X.

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Hint 2

For a continuous probability density p(x) with mean $\mu$ the variance is given by
 $V=\int _{a}^{b}(x\mu )^{2}p(x)dx$
Alternatively, the variance can also be calculated by
 $V=\int _{a}^{b}x^{2}p(x)dx\mu ^{2}$

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Solution

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Given the probability density p(x)=1x/2 over the interval $0\leq x\leq 2$ we plug this into the formula for the mean to get
 ${\begin{aligned}\mu &=\int _{0}^{2}x\left(1{\frac {x}{2}}\right)dx\\&=\int _{0}^{2}\left(x{\frac {x^{2}}{2}}\right)dx\\&=\left[{\frac {x^{2}}{2}}{\frac {x^{3}}{6}}\right]_{0}^{2}\\&=\left({\frac {2^{2}}{2}}{\frac {2^{3}}{6}}\right)\left({\frac {0^{2}}{2}}{\frac {0^{3}}{6}}\right)\\&=\mathbf {\frac {2}{3}} \end{aligned}}$
Using the first formula for the variance with $\mu ={\frac {2}{3}}$,
 ${\begin{aligned}V&=\int _{0}^{2}\left(x{\frac {2}{3}}\right)^{2}\left(1{\frac {x}{2}}\right)dx\\&=\int _{0}^{2}\left(x^{2}{\frac {4}{3}}x+{\frac {4}{9}}\right)\left(1{\frac {x}{2}}\right)dx\\&=\int _{0}^{2}\left({\frac {x^{3}}{2}}+{\frac {5}{3}}x^{2}{\frac {14}{9}}x+{\frac {4}{9}}\right)dx\\&=\left[{\frac {x^{4}}{8}}+{\frac {5}{9}}x^{3}{\frac {7}{9}}x^{2}+{\frac {4}{9}}x\right]_{0}^{2}\\&=\left({\frac {2^{4}}{8}}+{\frac {5}{9}}(2)^{3}{\frac {7}{9}}(2)^{2}+{\frac {4}{9}}(2)\right)\left({\frac {0^{4}}{8}}+{\frac {5}{9}}(0)^{3}{\frac {7}{9}}(0)^{2}+{\frac {4}{9}}(0)\right)\\&={\frac {16}{8}}+{\frac {40}{9}}{\frac {28}{9}}+{\frac {8}{9}}0=2+{\frac {20}{9}}\\&=\mathbf {\frac {2}{9}} \end{aligned}}$
Alternatively, using the other formula for the variance, we get the same value:
 ${\begin{aligned}V&=\int _{0}^{2}x^{2}\left(1{\frac {x}{2}}\right)dx\left({\frac {2}{3}}\right)^{2}\\&=\int _{0}^{2}\left(x^{2}{\frac {x^{3}}{2}}\right)dx{\frac {4}{9}}\\&=\left[{\frac {x^{3}}{3}}{\frac {x^{4}}{8}}\right]_{0}^{2}{\frac {4}{9}}\\&=\left({\frac {2^{3}}{3}}{\frac {2^{4}}{8}}\right)\left({\frac {0^{3}}{3}}{\frac {0^{4}}{8}}\right){\frac {4}{9}}={\frac {8}{3}}2{\frac {4}{9}}\\&=\mathbf {\frac {2}{9}} \end{aligned}}$

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MER QGH flag, MER QGQ flag, MER QGS flag, MER QGT flag, MER Tag Expected value and median, MER Tag Variance and standard deviation, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag

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