MATH105 April 2010
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Question 01 (n)

Find the variance of the random variable X whose density function is
 $f(x)={\frac {1}{2{\sqrt {x}}}}\quad 1\leq x\leq 4$.
(You may need the fact that the expected value of the random variable X above is 7/3.)

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

The definition of the variance is
 $\mathrm {Var} (X)=\int _{a}^{b}(x{\overline {X}})^{2}f(x)\,dx$

Hint 2

The calculation becomes a little easier when you use the identity
 $\mathrm {Var} (X)=\mathbb {E} (X^{2})(\mathbb {E} (X))^{2}.$

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Solution 1

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From the definition of the variance we can directly compute its value.
 ${\begin{aligned}\mathrm {Var} (X)&=\int _{a}^{b}(x{\overline {X}})^{2}f(x)\,dx=\int _{1}^{4}\left(x{\frac {7}{3}}\right)^{2}{\frac {1}{2{\sqrt {x}}}}\,dx\\&=\int _{1}^{4}\left({\frac {x^{2}}{2{\sqrt {x}}}}{\frac {7}{3}}{\frac {x}{\sqrt {x}}}+{\frac {49}{9}}{\frac {1}{2{\sqrt {x}}}}\right)\,dx\\&=\int _{1}^{4}\left({\frac {1}{2}}x^{3/2}{\frac {7}{3}}x^{1/2}+{\frac {49}{18}}x^{1/2}\right)\,dx\\&=\left.\left({\frac {1}{2}}{\frac {2}{5}}x^{5/2}{\frac {7}{3}}{\frac {2}{3}}x^{3/2}+{\frac {49}{18}}2x^{1/2}\right)\right_{1}^{4}\\&={\frac {1}{5}}\,32{\frac {14}{9}}\,8+{\frac {49}{9}}\,2\left({\frac {1}{5}}{\frac {14}{9}}+{\frac {49}{9}}\right)\\&={\frac {34}{45}}.\end{aligned}}$

Solution 2

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Alternatively, we can also use the identity
 $\mathrm {Var} (X)=\mathbb {E} (X^{2})(\mathbb {E} (X))^{2}.$
We're already given $\mathbb {E} (X)=7/3$ so we just need $\mathbb {E} (X^{2})=\int _{1}^{4}x^{2}f(x)dx$.
We compute this integral, finding $\mathbb {E} (X^{2})=\int _{1}^{4}x^{2}{\frac {1}{2{\sqrt {x}}}}dx=\int _{1}^{4}{\frac {1}{2}}x^{3/2}dx={\frac {1}{5}}x^{5/2}_{1}^{4}={\frac {1}{5}}(321)=31/5.$
Thus $\mathrm {Var} (X)={\frac {31}{5}}({\frac {7}{3}})^{2}=34/45$.

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MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Variance and standard deviation, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag

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