# Percentiles of a continuous pdf

• Oct 20th 2009, 10:54 AM
affelix
Percentiles of a continuous pdf
An insurer.s annual weather-related loss, X, is a random variable with density function
1. f(x) ={2.5(200)^2:5/x^3.5 for x > 200
0 otherwise
Calculate the difference between the 25th and 75th percentiles of X.
• Oct 20th 2009, 03:29 PM
mr fantastic
Quote:

Originally Posted by affelix
An insurer.s annual weather-related loss, X, is a random variable with density function
1. f(x) ={2.5(200)^2:5/x^3.5 for x > 200 Mr F say: This function requires clarification. What does the full colon : mean. Do you mean 2.5(200)^(2.5/x^3.5)?
0 otherwise
Calculate the difference between the 25th and 75th percentiles of X.

Let the 25th percentile be \$\displaystyle a\$ and the 75th percentile be \$\displaystyle b\$. Calculate \$\displaystyle b - a\$ where

\$\displaystyle \Pr(X < a) = 0.25\$ and \$\displaystyle \Pr(X < b) = 0.75\$

(it's assumed that you know how to calculate probabilities using a pdf).

If you need more help, please show all you've done and state where you get stuck.