14.3 - Finding Percentiles

14.3 - Finding Percentiles

At some point in your life, you have most likely been told that you fall in the something-something percentile with regards to some measure. For example, if you are tall, you might have been told that you are in the 95th percentile in height, meaning that you are taller than 95% of the population. When you took the SAT Exams, you might have been told that you are in the 80th percentile in math ability, meaning that you scored better than 80% of the population on the math portion of the SAT Exams. We'll now formally define what a percentile is within the framework of probability theory.

Definition. If \(X\) is a continuous random variable, then the \((100p)^{th}\) percentile is a number \(\pi_p\) such that the area under \(f(x)\) and to the left of \(\pi_p\) is \(p\).

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That is, \(p\) is the integral of \(f(x)\) from \(-\infty\) to \(\pi_p\):

\(p=\int_{-\infty}^{\pi_p} f(x)dx=F(\pi_p)\)

Some percentiles are given special names:

  • The 25th percentile, \(\pi_{0.25}\), is called the first quartile (denoted \(q_1\)).
  • The 50th percentile, \(\pi_{0.50}\), is called the median (denoted \(m\)) or the second quartile (denoted \(q_2\)).
  • The 75th percentile, \(\pi_{0.75}\), is called the third quartile (denoted \(q_3\)).

Example 14-5

A prospective college student is told that if her total score on the SAT Exam is in the 99th percentile, then she can most likely attend the college of her choice. It is well-known that the distribution of SAT Exam scores is bell-shaped, and the average total score is typically around 1500. Here is a picture depicting the situation:

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The student would like to know what her total score, \(\pi_{0.99}\), needs to be in order to ensure that she falls in the 99th percentile. Data from the 2009 SAT Exam Scores suggests that the student should obtain at least a 2200 on her exam. That is, \(\pi_{0.99}=2200\).

Example 14-6

Let \(X\) be a continuous random variable with the following probability density function:

\(f(x)=\dfrac{1}{2}\)

for \(0<x<2\). What is the first quartile, median, and third quartile of \(X\)?

Solution

Because the p.d.f. is uniform, meaning it remains constant over the support, we can readily find the percentiles in one of two ways. We can use the p.d.f. directly to find the first quartile, median, and third quartile:

Alternatively, we can use the cumulative distribution function:

Example 14-7

Let \(X\) be a continuous random variable with the following probability density function:

\(f(x)=\dfrac{1}{2}(x+1)\)

for \(-1<x<1\). What is the 64th percentile of \(X\)?

Solution

To find the 64th percentile, we first need to find the cumulative distribution function \(F(x)\). It is:

\(F(x)=\dfrac{1}{2}\int_{-1}^x(t+1)dt=\dfrac{1}{2} \left[\dfrac{(t+1)^2}{2}\right]^{t=x}_{t=-1}=\dfrac{1}{4}(x+1)^2\)

for \(-1<x<1\). Now, to find the 64th percentile, we just need to set 0.64 equal to \(F(\pi_{0.64})\) and solve for \(\pi_{0.64}\). That is, we need to solve for \(\pi_{0.64}\) in the following equation:

\(0.64=F(\pi_{0.64})=\dfrac{1}{4}(\pi_{0.64}+1)^2\)

Multiplying both sides by 4, we get:

\(2.56=(\pi_{0.64}+1)^2\)

Taking the square root of both sides, we get:

\(\pi_{0.64}+1=\pm \sqrt{2.56}=\pm 1.6\)

And, subtracting both sides by 1, we get:

\(\pi_{0.64}=-2.6 \text{ or } 0.60\)

Because the support is \(-1<x<1\), the 64th percentile and must be 0.6, not −2.6.


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