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Finding the median from a c.d.f.

F(x)= 0, x<0
1/12x^2 0<x<3
2x - 0.25x^2 -3 3<x<4
1, x>4

I know to find the median i use the fact that F(m)=0.5
But there are two parts to the F(x), so which do I use.

Reply 1

nuodai

Is the median not

m

such that

Unparseable latex formula:

\displaystyle \int^m_{-\infty} \mbox{f}(x)\ \mbox{d}x = 0.5

?

In that case, you have either

Unparseable latex formula:

\displaystyle \int^m_0 \dfrac{x^2}{12}\ \mbox{d}x = 0.5

.

If this gives a value of m > 3, then it'd be

Unparseable latex formula:

\displaystyle\int^3_0\dfrac{x^2}{12}\ \mbox{d}x + \int^m_3 (2x - \frac{x^2}{4} -3)\ \mbox{d}x = 0.5

In turn if this gives a value of m > 4, then it must be

Unparseable latex formula:

\displaystyle\int^3_0\dfrac{x^2}{12}\ \mbox{d}x + \int^4_3 (2x - \frac{x^2}{4} -3)\ \mbox{d}x + \int^m_4 1\ \mbox{d}x = 0.5

Reply 2

zafaru123

Thanks a lot for that..very helpful.

Reply 3

MushyPeas

F(x)= 0, x<0
1/12x^2 0<x<3
2x - 0.25x^2 -3 3<x<4
1, x>4

Median is where F(m)=0.5
When you have multiple-part questions you put the highest number of the boundary into the first bit
eg. using 1/12x^2 0<x<3 put x=3 into 1/12x^2 and see if it equal or bigger than 0.5, if it is then the median lies in this section, if not then use the next section and equal it to 0.5