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

MATH103 April 2012

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[hide] 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?

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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.

[show] Hint 1
The average of a random variable X with probability density function p(t) is given by $\mu =\int _{-\infty }^{\infty }tp(t)\,dt.$

[show] 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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[show] Solution
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 $\mu ={\frac {\ln(2)}{\arctan {\sqrt {3}}}}$ ).