Will I go to UC Berkeley?

Question

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?

Answer 1

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
$F_{SAT}(SAT = 1870 | Admitted) = \int_{-\infty}^{1870}f_{SAT}(SAT=x | Admitted)dx = 0.25$
$F_{SAT}(SAT = 2230 | Admitted) = \int_{-\infty}^{2230}f_{SAT}(SAT=x | Admitted)dx = 0.75$