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

MATH103 April 2012

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Question 05 (b)
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 $\arctan({\sqrt {3}})=\pi /3$. (b) What is the average delay?

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
The average of a random variable X with probability density function p(t) is given by $\mu =\int _{-\infty }^{\infty }tp(t)\,dt.$

Hint 2
Using the solution to Problem 5a, along with hint 1 for this problem, we have that the mean should be given by: $\int _{0}^{\sqrt {3}}{\frac {3}{\pi }}{\frac {t}{1+t^{2}}}\,dt.$ Substitute $u=1+t^{2}$ to evaluate the integral.

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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. Using the value of k we found in the previous question, and the formula for the mean given in the hint, we know the mean is found using the following integral. $\int _{0}^{\sqrt {3}}{\frac {3}{\pi }}{\frac {t}{1+t^{2}}}\,dt$ We can then solve the integral to find the mean. ${\begin{aligned}\mu &=\int _{0}^{\sqrt {3}}{\frac {3}{\pi }}{\frac {t}{1+t^{2}}}\,dt\\&={\frac {3}{2\pi }}\left.\ln(1+t^{2})\right|_{0}^{\sqrt {3}}\\&={\frac {3}{2\pi }}\ln(4)\\&={\frac {3\ln(2)}{\pi }}\end{aligned}}$ (or $\mu ={\frac {\ln(2)}{\arctan {\sqrt {3}}}}$).

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