Math Help: P(40 ≤ X ≤ 50, 20 ≤ Y ≤ 25) & P(4(X-45)^2+100(Y-20)^2 ≤ 2)

[acgold]

Suppose that X and Y are independent random variables, where X is normally distributed with mean 45 and standard deviation 0.5 and Y is normally distributed with mean 20 and standard deviation 0.1.

(a) Find \ P(40 \leq X \leq 50, \ 20 \leq Y \leq 25). Ans. ~0.5
(b) Find \ P(4(X-45)^2+100(Y-20)^2 \leq 2). Ans. ~0.632

Part (a) is easy. I used Maple to find the double integral of the joint density function from 40 \leq X \leq 50, \ 20 \leq Y \leq 25. and I get 0.5

Part (b) is my problem. How do I find the limits of integration? I tried solving for X and Y but I couldn't get that to work. Am I missing something easy here? Please please please help me! This assignment is due tomorrow and I've been stuck on this problem for days.

 

[Tide]

Your region is a small ellipse centered on (45, 20) with a semimajor radius of 1/sqrt(2) and semiminor radius of 1/sqrt(50).

 

[ehild]

Suppose that X and Y are independent random variables, where X is normally distributed with mean 45 and standard deviation 0.5 and Y is normally distributed with mean 20 and standard deviation 0.1.


(b) Find \ P(4(X-45)^2+100(Y-20)^2 \leq 2). Ans. ~0.632

Notice that
4(X-45)^2 = (\frac{X-45}{0.5})^2= u^2 \mbox{ and } 100(Y-20)^2= (\frac{Y-20}{0.1})^2=v^2.
u and v are the standard variables for X and Y, respectively. They are normally distributed with zero mean and standard deviation 1.

The condition u^2+v^2\leq 2 means that the point(u,v) is inside a circle of radius \sqrt{2}

P(u^2+v^2 \leq 2) = \frac{1}{2\pi}\int\int\exp(-\frac{u^2+v^2}{2})du dv
Use polar coordinates and then your problem reduces to calculate

P(r\leq\sqrt{2})=\int_0^{2\pi}\int_0^{\sqrt(2)}{\exp(-r^2/2)rdrd\phi


ehild

 

[acgold]

Thanks for the help guys. Unfortunately, I couldn't get this particular problem finished in time to turn it in but I appreciate the help anyway. At least I understand it somewhat now. Thanks again!