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I have a random variable $X$ which has the following CDF: $$F(y) = \left\{\begin{array}{ll} 0 & : y \lt 0\\ \frac{y}{30} & : 0 \le y \lt 20\\ \frac{2}{3} + \frac{y-20}{60} & : 20 \le y \lt 40\\ 1 & : y \ge 40 \end{array} \right.$$

To find the median of $X$, I know I need to plug $q_2$ into the CDF and set it equal to 0.5: $$ F(q_2) = 0.5 $$ and solve for $q_2$. But which interval of the CDF should I use? My intuition tells me it would be the third interval since 20 falls in the middle of 0 and 40 and this is the interval that would be used for $F(20)$, but I have a feeling this is wrong. Which one should I use for $q_2$, as well as $q_1$ and $q_3$?

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  • $\begingroup$ A CDF is increasing, so you can check that for instance the CDF is at least 2/3 on the third interval just by plugging in $y=20$. $\endgroup$ – Ian Sep 27 '18 at 17:41
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I found the answer so I'll go ahead and post it here:

The easy way is just to make a guess and plug in the quartile using any of the intervals. If the result falls within the interval you chose, it is correct. If not, pick a different one.

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If your random variable $X$ has the cdf

$$F(y) = \left\{\begin{array}{ll} 0 & : y \lt 0\\ \frac{y}{30} & : 0 \le y \lt 20\\ \frac{2}{3} + \frac{y-20}{60} & : 20 \le y \lt 40\\ 1 & : y \ge 40 \end{array} \right.$$

you're supposed to find the inverse so you get $F(y) = \frac{y}{30} $ for $y$. then the inverse is $30y$. If plug in $F^{-1}(.5) = \frac{1}{2}30 = 15$

I'll just note you switch from $X$ to $y$. You can also see it has $\frac{2}{3} + \frac{y-20}{60} $. If you plug in $y=20$ you get $\frac{2}{3}$

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