Solving Quadratics to Find the Interquartile Range

[chwala]

Homework Statement

A continous random variable, $X$ has the following probability density function;

$f_{x} = \begin{cases} \dfrac{2}{25} (5-x), 0≤t≤5 \\ \\0 , Otherwise \end{cases}$

Relevant Equations

understanding of probability distribution.

I do not have solution for this; looking forward to your insight. $$F_{X}=\int_0^m \dfrac{2}{25} (5-x)dx$$ ... ending up with $$\dfrac{2}{25} (5m-\dfrac{m^2}{2})=\dfrac{1}{4}$$ and $$\dfrac{2}{25} (5m-\dfrac{m^2}{2})=\dfrac{3}{4}$$ we shall end up with two quadratic equations. Solving them gives us; $$4m^2-40m+25=0$$ $$m=0.669$$ and $$4m^2-40m+75=0$$ $$m=2.5$$ Therefore our interquartile range is given by; $$IQR=2.5-0.669=1.83$$ to $3$ decimal places.

 

[pasmith]

I think I would start with \frac{2}{25}\left(5m - \frac{m^2}2\right) = q \in [0,1] and rearrange it into (m-5)^2 = 25(1- q). It is obvious that we want the negative root, so <br /> m = 5(1 - \sqrt{1 - q}). Then the interquartile range is <br /> 5\left(1 - \sqrt{\tfrac14}\right) - 5\left(1 - \sqrt{\tfrac34}\right) = \frac52 \left(\sqrt{3} - 1\right) \approx 1.830.