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  • STA256 Lecture 07

Percentile

STA256

• Example:
  • Let $X$ be a Continuous Random Variable with Probability Density Function
  • $f(x)=\begin{array}{cc}2x-2& 1\le x\le 2\\ 0& \text{otherwise}\end{array}$
  • Find the $\underset{0.90\text{ quantile}}{\underbrace{{90}^{\text{th}}\text{ percentile}}}$ of $x$
  • The ${90}^{\text{th}}$ percentile is a value such that $90\mathrm{\%}$ of all observations are below it.
  • Suppose a class wrote a difficult test where the ${95}^{\text{th}}$ percentile was $60$ marks.
    • $95\mathrm{\%}$ of students got less than $60$ on the test.
  • Label the ${90}^{\text{th}}$ percentile as $x_{90}$

    	left=0; right=3;
    	top=3; bottom=0;
    	---
    	y=2x-2|1<=x<=2
    	x=1+ \frac{3}{\sqrt{10}}
    

  • The line is the ${90}^{\text{th}}$ percentile
    • $x_{90}$
    • $P(X\le x_{90})=0.9$
    • ${\int }_1^{x_{90}}2x-2dx=0.9$
    • $(x^2-2x){}_1^{x_{90}}=0.9$
    • $x_{90}^2+2x_{90}+0.1=0$
    • solve
    • $x_{90}=1\pm \frac{3}{\sqrt{10}}$
  • We can show that the Cumulative Distribution Function is:
    • $F(x)=\begin{array}{cc}0& x<1\\ (x-1{)}^2& 1\le x\le 2\\ 1& x>2\end{array}$
    • How to use the Cumulative Distribution Function to find the ${90}^{\text{th}}$ percentile?
    • Plug in $F(0.9)$ because $F(x_{90})=P(X\le x_{90})=0.9$
• #tk see example slide 87

STA258

• Ex:
  • ${80}^{\text{th}}$ percentile is $60$ marks, then $80\mathrm{\%}$ of the class got less than $60$ on the test.
  • The Median is the ${50}^{\text{th}}$ percentile.
• $\to$ Quartiles