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# Another important S2 question!! watch

1. When ur given a cdf with several parts (multi-cdf), which part do u use to find the median?

leaving for my exam in an hour!
2. F(M)= 1/2
3. dont think u understand my question. I know u let the cdf equal 0.5, but which part of the cdf do u let equal 0.5? or do u let the whole thing equal 0.5?
4. no, u cant do that
u have to look at the ranges for x n see which one the median is more likely to be in
yup..http://thestudentroom.co.uk/t121618-4.html
go rite at the bottom of the page
5. right, so i should draw a graph of the cdf and just guess? sorry, dont really understand wht u went in tht other thread
6. I thought you just added them together to = 1/2?
7. u dont have to draw a graph..i just meant if u have a graph it can b observed from there
usually tho, its just common sense by looking at the ranges for each bit of the cdf
8. is it possible to work it out for each part and then somehow eliminate the ones which are wrong?
9. (Original post by bestdeceptions)
I thought you just added them together to = 1/2?
thats when u need the mean...u calculate the mean for each bit n add them
10. For median, calculate maximum and minimums of each cdf...from there one can see in which function any percentile lies.
11. (Original post by Faiz)
is it possible to work it out for each part and then somehow eliminate the ones which are wrong?
im not a 100% sure, but i tried on one of the questions..n the answer came out out of the range of that bit of cdf
so yeah if u do that, u can eliminate the cdf's for which its coming out of range

but to b honest i dont really think they will throw something at us thats too ambiguous
dont worry, u'll be able to spot where it shud lie!
12. i think tht makes sense. So say if for the first part, the max is 0.2 and the min is 0, the median wont lie in tht part?

if the max of the second part is 1 and the min 0.2, will the median lie in tht part?
13. Absotively (I think)
14. yeah..its usually in the largest of the ranges

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