The Edexcel S2 (24/05/12 - AM) Revision Thread

Guys, quick question before I call it a day: When there's a 2-staged CDF, how do you work out the median? (+rep available!)

Yes.

You can do a neat trick and flip these to make the question easier if it comes up too.

Let's say that X~B(10, 0.7) and you're asked for P(X>=7)

You could, from that, say X~B(10,0.3) where the P(X<=3)

This is because with Binomial there is either success or failure.

I'm pretty sure that's right - please let me know if it isn't.

Guys, quick question before I call it a day: When there's a 2-staged CDF, how do you work out the median? (+rep available!)

EDIT: It's late - I misread your question - I'll leave my flawed explanation of how to work out the second part of a pdf to CDF anyways haha!

Do the lower part first by integrating as such between x and the lower bound.

Let's say 0<=x<=1 has value 1/2

so you get 1/2x between x and 0 for that part of F(x)

For the second part, say, 1<x<=2 you need to integrate but ADD F(highest bound of previous part) in this case you add F(1) which is 1/2(1) = 1/2

So say the second part of pdf is x^2 you get x^3/3 between x and 1

This totals x^3/3 - 1/3 + 1/2 = x^3/3 + 1/6 for the second part of the CDF

I'm not sure I am making myself clear here haha.

Edit #2:

To answer your question directly, as stated, find where 0.5 lies by putting in the max value for each bound. If you get 0.4 using the max bound on the CDF then it cannot lie in that bound as it is <0.5 etc

Then the cdf part you have identified as where the median lies within = 0.5

Okay so if lets say the first interval is F(x) = 0.5x, 0<X<0.8

Then obvs 0.5*0.8=0.4 so X<=0.8 is not the median.

Then for the next interval, if F(x) = 0.1x, 0.8<X<1.0 (I KNOW IT DOESN'T WORK!)

Then do we put 0.1m=0.5 so m=5, or do we put 0.1m= 0.5-0.4 = 0.1 (0.4 from previous question).

I think you just simply put it equal to 0.5 but I just wanna confirm.

Tbh, I think im over-complicating things and just confusing myself too much

sweet brah, you gonna be trolling fp2/fp3 threads?

this is easy but i gotta spend the next week actually learning M2

i got m2 as well...then m4 the next day

fp2 fp3 thread is dead as the moderator is always cleaning up my junk

*shhhhh* No rockstars on this thread please! Studious and quiet people only

gonna start m2 papers friday night and beast mode it this weekend probs

i did the same for m1 and got 90something when i was completely new to it and i don't do physics so confident i can pull out a decent score

gonna start m2 papers friday night and beast mode it this weekend probs

i did the same for m1 and got 90something when i was completely new to it and i don't do physics so confident i can pull out a decent score

does that mean you don't have to do all the maths exercises in the textbook?

Do you mean continuous uniform distribution ?

That's just $\frac{a+b}{2}$. You can derive it "by symmetry"

I think that means that because the distribution is symmetrical E(X) must be in the middle of a and b.

how do you know whether to go from binomial to poisson or normal ? when approximating..

Binomial to Poisson: n is large and p is small such that np is approx. 10 (i.e you can look it up on the Poisson tables)

Also use this approx. when the MEAN=np is approximately close to the VARIANCE=npq

Binomial to Normal: n is large, p is close to 0.5
Poisson to Normal: n is large such that n cant be searched on the Poisson tables

does that mean you don't have to do all the maths exercises in the textbook?

thought of this when i was in bed - i've never seen a question on it before but knowing my luck i'll ask just in case, do we need to know how to work out the variance if the p.d.f is in two parts.

to do the mean you do the mean of each part then add them together so would you do the same for variance?