Normally distributed rain drops problem

Question

About 50% of raindrops land downtown, downtown is a perfectly circular space around the city centre. Assuming the coordinates of the raindrops are independent and distributed according to the standard normal about the city centre. What is the percentage of rain drops that land within a radius twice that of downtown?

Since around 50% of the raindrops land downtown, clearly the percentage landing within a radius twice that of downtown must be less than or equal to 50, but beyond this I am not sure how to approach/solve this problem, can someone help? Thanks!

Answer 1

I believe you can use standard z-tables (e.g. https://en.wikipedia.org/wiki/Standard_normal_table) to work out the various percentages for the radius in question. The z-table will tell you the probability contained by the standard normal curve from 0 to whatever z-value you choose. Now, if you make the equivalence of rain percentage to probability and z to radius, then you should be able to solve the problem.

For example, if you look at the z-table for the standard normal, you will see that the z-value for 0.25 (25% - remember this is going to be 1/2 the total 50% because it is measured from the center), is about z=0.68. Now, if you double this (double the radius), you will get $z=2(0.68)=1.36$. This z-value corresponds to 0.413 or 41.3%. Then, you'll need to double this to get 82.6% which I believe should correspond to the final answer.

I hope this helps.

Answer 2

Let us look at a normal distribution in one dimension first, then proceed into two dimensions.

According to your statement, 50% of the raindrops land in downtown. Without loss of generality, let's assume that 'downtown' is one kilometer in radius. Let's model that in one dimension: $$Pr(\mu-n\sigma\leq X\leq\mu+n\sigma)=\Phi(n)-\Phi(-n)=0.5$$ In this case, looking at a z-table gives $0.67<n<0.68$. Now, we'll double n to find the standard deviation for the raindrops that land exactly two kilometers from downtown. We get in that case $n\approx1.35$. Now we plug that back in: $$Pr(\mu-2n\sigma\leq X\leq\mu+2n\sigma)\approx\Phi(1.35)-\Phi(-1.35)\approx0.823$$ again using the z-table.

Now, since we are talking about two dimensions rather than one, we square the resultant probability to obtain $(0.823)^2\approx0.677$.

Hope this works! Not sure if I'm correct, I'm no expert. (Not 100% confident about the last step.)

Answer 3

I don't think the answers here are right, perhaps due to fault of vague wording in the question. I think it is not $R$ that is normally distributed, but $X$, $Y\overset{\text{i.i.d.}}{\sim}\mathcal{N}(0,1)$ such that $R\sim\text{Rayleigh}(1)$ with CDF $\mathbb{P}(R\leq r)=1-\text{e}^{-\frac{r^2}{2}}$. Therefore, if we let the radius of the city be $a$, then $\mathbb{P}(r\leq a)=1-\text{e}^{-\frac{a^2}2}=1/2$, so $a=\sqrt{2\ln2}$, and $\mathbb{P}(r\leq 2a)=1-\text{e}^{-2\ln2}=1-\left(\text{e}^{\ln2}\right)^2=1-\left(\frac12\right)^2=\frac34 $which corresponds to $\boxed{75\%}$.