Will I go to UC Berkeley? I forget where I got this data (I think from About.com College), but here are some statistics regarding University of California, Berkeley admissions: the 25th percentile SAT Reasoning Test score was 1870, and the 75th percentile was 2230. Let's say that my performance is a standard distribution with mean 2200 and standard deviation 50. Based on test performance alone, what's the probability that I'll be accepted? Note that I'm a high school student with actually no statistical background besides Wikipedia, so bear with me here.
Here's one way that I thought of. Let's assume the Berkeley admissions is a standard distribution, so the probability of my score, $x$, being near the majority is $P_b(x)=c_b + e^{-a_bx^2}$, where $c$ is the mean and $a$ ensures that $\int^{2400}_0P_b(x)\,dx = 1$. Therefore with the two quartiles and $a$ and $c$ we have two equations with two unknowns, and we will pretty much know $P_b(x)$. Then I will find $P_s(x)$, which is the probability that I will get a certain score $x$ between 0 and 2400. Afterwards, the probability of me being accepted $$\approx \int^{2400}_{0}P_b(x)P_s(x)\,dx.$$
Again, I have no real statistical knowledge (I'll have some next year), so is my reasoning sound?
 A: Not quite.
Your $P_b(x)$ in your notation (putting aside questions on function form) is $P(SAT | Admitted)$
$P(SAT | Admitted)P(MySAT)$ doesn't give you what you want.
What you are looking for is $P(Admitted | SAT)P(MySAT)$
Getting $P(Admitted | SAT)$ is an application of Bayes:
$$ P(B|A) = {P(A|B)P(B) \over P(A)} = {P(A|B)P(B) \over P(A|B)P(B) + P(A|!B)P(!B)} $$
Translating:
$$ P(Admitted | SAT) = {P(SAT | Admitted)P(Admitted) \over P(SAT)}$$
$$ = {P(SAT | Admitted)P(Admitted) \over P(SAT | Admitted)P(Admitted) + P(SAT | !Admitted)P(!Admitted)}$$
You have  


*

*$P(SAT | Admitted)$ - provided in your question as 25/75 percentiles, you need to assume a functional form for this  

*$P(Admitted)$ - google says this is 0.18


You do not have:  


*

*$P(SAT)$  


Note that while you have assumed a $P(My SAT)$ you actually need the distribution of the population of all UCB applicants, not just yourself. Specifically you do not have $P(SAT| !Admitted)$. 
If you are able to obtain that, then you can calculate $P(Admitted | SAT)$ and from there $P(Admitted | SAT)P(MySAT)$
There may be some abuse of notation above.

To answer your follow-up, yes, that is referencing the un-admitted population. You cannot assume the complement. A simplified version of the problem may help. Let's take a look at SAT $\ge$ 2230.
Since the 75 percentile of admitted students is 2230, that means 25% of admitted students have an SAT $\ge$ 2300 and thus
$P(SAT \ge 2300 | Admitted) = 0.25$
You can easily see the incongruity with taking the compliment and saying that 75% of non-admitted students have an SAT score greater than 2300
$P(SAT \ge 2300 | !Admitted) = 0.75$
And yes, by functional form, I mean your assumed normal. So to clarify (and clean up my notation from above).


*

*$f_{SAT}(SAT=x | Admitted)$ is a assumed probability distribution function with 25/75 percentiles 1870/2230.
If you assume this is a normal (ignoring that SATs are capped at 2400), you can set the cumulative distribution equal to 0.25 and 0.75 (or integrate the pdf)1 and solve for the mean and standard deviation to arrive at $\approx N(2050,267)$  

*$P(Admitted) = 0.18$

*You will need to acquire or assume $f_{SAT}(SAT=x)$ or $f_{SAT}(SAT=x | !Admitted)$. For a start, you can consider the overall SAT distribution of all US students (though this is unlikely to be the distribution of UCB applicants)


This results in a function
$$ P(Admitted | SAT=x) = {{f_{SAT}(SAT=x | Admitted)P(Admitted)} \over f_{SAT}(SAT=x) }$$
And then you can calculate
$$ P(Admitted | SAT=My SAT)P(SAT=My SAT) $$
1 Solve the system of equations
$F_{SAT}(SAT = 1870 | Admitted) = \int_{-\infty}^{1870}f_{SAT}(SAT=x | Admitted)dx = 0.25$
$F_{SAT}(SAT = 2230 | Admitted) = \int_{-\infty}^{2230}f_{SAT}(SAT=x | Admitted)dx = 0.75$
