Science:Math Exam Resources/Courses/MATH101/April 2006/Question 06

MATH101 April 2006

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[hide] Question 06
Full-Solution Problems. Justify your answers and show all your work. Simplification of answers is not required. X is a random variable with probability density function $f(x)=2xe^{-x^{2}}$ where $x\geq 0$. Find the median of X.

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.

[show] Hint 1
The median m is defined as that number, which satisfies $\int _{-\infty }^{m}f(x)\,dx={\frac {1}{2}}=\int _{m}^{\infty }f(x)\,dx$

[show] Hint 2
A simple substitution may be helpful to evaluate the integral.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

[show] Solution

Probability was not covered this year, hence this is not material for the April 2012 exam. Since f(x) = 0 for x≤ 0 it follows that m > 0. Now we can directly compute m: ${\begin{aligned}{\frac {1}{2}}&=\int _{-\infty }^{m}f(x)\,dx\\&=\int _{-\infty }^{0}0\,dx+\int _{0}^{m}2xe^{-x^{2}}\,dx\end{aligned}}$ We use the substitution u = x2, du = 2x dx. Then u(0) = 0 and u(m) = m2. Hence ${\begin{aligned}{\frac {1}{2}}&=\int _{-\infty }^{m}f(x)\,dx\\&=\int _{-\infty }^{0}0\,dx+\int _{0}^{m}2xe^{-x^{2}}\,dx\\&=\int _{0}^{m^{2}}e^{-u}\,du\\&=\left.-e^{-u}\right|_{0}^{m^{2}}\\&=-e^{-m^{2}}+e^{0}\\&=1-e^{-m^{2}}.\end{aligned}}$ In other words, m satisfies ${\begin{aligned}{\frac {1}{2}}&=1-e^{-m^{2}}\\e^{-m^{2}}&={\frac {1}{2}}\\-m^{2}&=\ln(2^{-1})\\m^{2}&=\ln 2\\m&=\pm {\sqrt {\ln 2}}\end{aligned}}$ Note that we can rule out the solution $m=-{\sqrt {\ln 2}}$ since we already noted that m is positive. Hence, the solution is $m={\sqrt {\ln 2}}$.

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