# Urgent Math Help - Normally distributed probability density function

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. Related Introductory Physics Homework Help News on Phys.org
Tide
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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
Homework Helper
acgold said:
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

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!