what is the median of the CDF with the form $F(x) = 1 - e^{-(x/3)^2}$, for $x \gt 0$

Question

Consider The distribution of lifetimes, X(in months), of a particular type of component. The Cumulative distribution function (CDF) has the form

$$F(x)=\begin{cases} 1 - e^{-(x/3)^2},& \text{if} ~ x \gt 0\\\\ 0,&\text{otherwise}.\\\\ \end{cases}$$

How to calculate the median?

please make sure I got $F(x)$ correct as written. – Amzoti Mar 05 '13 at 22:27 — Amzoti, Mar 05 '13 at 22:27

Aang · Accepted Answer · 2013-03-05T22:32:08.480

3

HINT: Median is the point $x_0$ where cdf $F(x_0)=0.5$

EDIT: $$1-e^{-\frac{{x_0}^2}{9}}=0.5\iff e^{-\frac{{x_0}^2}{9}}=0.5\iff \frac{{x_0}^2}{9}=-\ln0.5\iff x_0=3(\ln 2)^{1/2}$$

edited Mar 05 '13 at 22:32

answered Mar 05 '13 at 22:25

Aang

score 2 · Answer 2 · edited Mar 05 '13 at 22:29

2

$F(x_0) = 1 - e^{-(x_0/3)^2}=0.5\implies -x_0^2/9=ln(0.5)\implies x_0=3 \sqrt{\ln(2)}$

edited Mar 05 '13 at 22:29

Aang

answered Mar 05 '13 at 22:28

Bravo

2 Answers2