Science:Math Exam Resources/Courses/MATH103/April 2012/Question 05 (a)

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MATH103 April 2012

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Question 05 (a)
A professor never finishes his class on time. The time of delay is given by the probability density function ${\begin{aligned}p(t)={\frac {k}{1+t^{2}}},{\text{ where }}0<t<{\sqrt {3}}\end{aligned}}$ You can use the facts that $\displaystyle \arctan(0)=0$ and $\arctan({\sqrt {3}})=\pi /3$. (a) What is the 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 p to be a probability density function it needs to satisfy two conditions: p(t) ≥ 0 for all t, $\int _{-\infty }^{\infty }p(t)\,dt=1.$ Choose k such that both conditions are fulfilled.

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Solution

In order to satisfy the second condition given in the hint, we must have the following equality hold: $1=\int _{0}^{\sqrt {3}}{\frac {k}{1+x^{2}}}\,dx$ However, this is easy enough to solve, especially if we remember that $\int {\frac {1}{1+x^{2}}}\,dx=\arctan {x}$. $1=\int _{0}^{\sqrt {3}}{\frac {k}{1+x^{2}}}\,dx=k(\arctan {\sqrt {3}}-\arctan(0))={\frac {k\pi }{3}},$ hence $k={\frac {3}{\pi }}$. (Or $k={\frac {1}{\arctan {\sqrt {3}}}}$, we used $\arctan({\sqrt {3}})=\pi /3$.)

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