Science:Math Exam Resources/Courses/MATH101/April 2005/Question 07 (a)

MATH101 April 2005

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Question 07 (a)
The length of time in minutes it takes students to solve a certain mathematics problem (on probability) is a continuous random variable whose probability density function is $f(x)={\begin{cases}k\cos(x/10)\sin ^{2}(x/10),&{\text{if }}0\leq x\leq 5\pi ,\\0,&{\text{otherwise.}}\end{cases}}$ (a) Find the value of the positive constant $k$.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
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.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First, note that cos(x/10) ≥ 0 and sin(x/10) ≥ 0 for any 0 ≤ x ≤ 5π. Hence we need k ≥ 0 as well. To satisfy the second condition we require $1=\int _{-\infty }^{\infty }f(x)\,dx=\int _{0}^{5\pi }k\cos \left({\frac {x}{10}}\right)\sin ^{2}\left({\frac {x}{10}}\right)\,dx.$ The integral can easily be solved using the substitution u = sin(x/10), du = 1/10 cos(x/10), so that u(0) = 0 and u(5π) = sin(π/2) = 1. Hence we need to choose k such that ${\begin{aligned}1&=\int _{0}^{5\pi }k\cos \left({\frac {x}{10}}\right)\sin ^{2}\left({\frac {x}{10}}\right)\,dx\\&=\int _{0}^{1}10ku^{2}\,du\\&=\left.{\frac {10ku^{3}}{3}}\right|_{0}^{1}\\&={\frac {10}{3}}k\end{aligned}}$ Hence we choose k = 3/10.

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