Science:Math Exam Resources/Courses/MATH101/April 2009/Question 05

MATH101 April 2009

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Question 05
Let X be a random variable with probability density function: $f(x)={\begin{cases}kx(1-x^{4})&{\text{ if }}0\leq x\leq 1,\\0&{\text{ if }}x<0{\text{ or }}x>1.\end{cases}}$ (a) Find the value of k. (b) Find the mean $\mu$. (c) Find an algebraic equation satisfied by the median m.

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
For ƒ to be a probability density function it needs to satisfy two conditions: ƒ(x) ≥ 0 for all x, $\int _{-\infty }^{\infty }f(x)\,dx=1.$ Choose k such that both conditions are fulfilled.

Hint 2
The formula for the mean $\mu$ is given by $\displaystyle \mu =\int _{-\infty }^{\infty }xf(x)\,dx.$

Hint 3
The median m is that number which satisfies $\int _{-\infty }^{m}f(x)\,dx={\frac {1}{2}}=\int _{m}^{\infty }f(x)\,dx.$

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.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (a) Using hint 1 we see that k needs to be a positive constant. We can calculate k directly by integrating: ${\begin{aligned}\int _{-\infty }^{\infty }f(x)\,dx&=\int _{-\infty }^{0}0\,dx+\int _{0}^{1}kx(1-x^{4})\,dx+\int _{1}^{\infty }0\,dx\\&=k\int _{0}^{1}(x-x^{5})\,dx\\&=k\left.\left({\frac {x^{2}}{2}}-{\frac {x^{6}}{6}}\right)\right|_{0}^{1}\\&=k\left({\frac {1}{2}}-{\frac {1}{6}}\right)\\&={\frac {k}{3}}.\end{aligned}}$ The only way that this integral has value 1 is when k = 3.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (b) Using the second hint and the fact that k = 3 the calculation is straight forward Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \mu &= \int_{-\infty}^\infty xf(x)\,dx \\ &= \int_0^1 (x)3x(1-x^4)\,dx \\ &= 3 \int_0^1 (x^2-x^6)\,dx \\ &= 3\left.\left(\frac{x^3}3 - \frac{x^7}7\right)\right|_0^1 \\ &= 3\left(\frac13 - \frac17\right) \\ &= \frac47. \end{align} }

Solution 3

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (c) Following the third hint we need to find that value m which satisfies $\int _{-\infty }^{m}f(x)\,dx={\frac {1}{2}}=\int _{m}^{\infty }f(x)\,dx.$ Since $\int _{-\infty }^{0}f(x)\,dx=0=\int _{1}^{\infty }f(x)\,dx$ it is clear that m needs to be a number between 0 and 1. Therefore the algebraic equation that m satisfies is the following; ${\begin{aligned}{\frac {1}{2}}&=\int _{-\infty }^{m}f(x)\,dx\\&=\int _{0}^{m}3x(1-x^{4})\,dx\\&=3\left.\left({\frac {x^{2}}{2}}-{\frac {x^{6}}{6}}\right)\right|_{0}^{m}\\&=3\left({\frac {m^{2}}{2}}-{\frac {m^{6}}{6}}\right).\end{aligned}}$

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