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

MATH101 April 2005

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Question 07 (b)
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 $f(x)={\begin{cases}k\cos(x/10)\sin ^{2}(x/10),&{\text{if }}0\leq x\leq 5\pi ,\\0,&{\text{otherwise.}}\end{cases}}$ (b) What is the probability that a student will take less than $5\pi /2$ minutes to complete the problem?

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 a continuous random variable X with probability density function f(x) the probability that an event a ≤ X ≤ b happens is ${\text{Prob}}(a\leq X\leq b)=\int _{a}^{b}f(x)\,dx.$

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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. A student takes at most 5π/2 minutes to complete the question if his completion time is anywhere between 0 and 5π/2. Hence the probability we are looking for is ${\begin{aligned}{\text{Prob}}(a\leq X\leq b)&=\int _{a}^{b}f(x)\,dx\\&=\int _{0}^{5\pi /2}{\frac {3}{10}}\cos \left({\frac {x}{10}}\right)\sin ^{2}\left({\frac {x}{10}}\right)\,dx\\&=\int _{0}^{1/{\sqrt {2}}}3u^{2}\,du\\&=\left.u^{3}\right|_{0}^{1/{\sqrt {2}}}\\&={\frac {1}{2{\sqrt {2}}}}.\end{aligned}}$ We used the substitution u = sin(x/10), du = 1/10 cos(x/10), so that u(0) = 0 and u(5π/2) = sin(π/4) = $\scriptstyle 1/{\sqrt {2}}$ from part (a).

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