Statistics 16/06/14 MS2B Unit 2B AQA

How did everyone find it? I desperately want to know how to do question 4, with the a, b and k. Especially how you got E(X)^2 = 1/3 (a^2 + b^2 + ab) !? I didn't speak to anyone at my school who managed to do it! Any other answers are useful too!

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Reply 1

kb382

f(x)=1/b-a so E(x) = integral between b and a of x^2/b-a . that is x^3/3(b-a) between b and a. so you get b^3/3(b-a) - a^3/3(b-a). this leads to 1/3(b^3/b-a - a^3/b-a) , then 1/3(b^3-a^3/b-a). thats all i could think of , then i just wrote 1/3(b^2+a^2 + ab) and hoped i would get the mark, i didnt know how to get to the answer from there

Reply 2

mathsRus

I know... I thought at first it was a mistake in the question but I didnt get the end asked answer although I got very close to it. I thought this paper was OK (not too easy not too hard). I got b as 2 and -10 for some reason. Q4 was the hardest I had ever done because it was algebraically challenging.

I think A* - 68
A - 61
B - 54
and so on for every grade being in 7 mark interval.

I think I got probably 67 - 68. How was your exam?

Other people in my school ran out of time.

Reply 3

mathsRus

Original post by kb382

I got answer for E(X) as (b+a)/2

then i integrated for E(X^2) where it was Integral ( x^2/k) limits b and a

after that it was all a waffle

Reply 4

Sam12345....

What you do is integrate 1/kx to get 1/k x2/2 with limits b and a..
Sub them in, and times it by 1/b-a
To get 1/b-a (b^3/3-a^3/)
Then you will get b^3/3(b-a) and a^3/3(b-a)
And you should be able to get the answer with rearranging
by factorising the b^3 -a^3

(edited 9 years ago)

Reply 5

kb382

I hope theyre the grade boundaries. i found timing difficult aswell, i missed out a couple of 3 markers because of timing which was frustrating. timing has let me down on this exam, couldnt finish every question. i did all the long ones first which was no problem, but it was the shorter 3/4 markers which threw me. I also found difficult the question proving the E(X) fell between 1.31 and 1.41 i think it was. I only found my lower limit as 1.4 and upper limit of 1.9 so im hoping method marks gets me through

Reply 6

LizzieB.

I did change in sign for showing the median was between 1.31 and 1=41. It was the only thing I could think of and it seemed to work but I didn't think you'd ever use that in stats so I'm not sure if I'll get any marks

Reply 7

eshaq

How did you guys do the last two Poisson distribution question?

1. used twice, find probability that at least one had no pebbles?

I did it probably wrong but here is what i did: Mean = 2 x 3.2 = 6.4
P(X=0) = e^-(6.4) x 1 = 1.66 x 10^(-3), then i times the answer by two. 2x1.66x10^(-3)

2. Independent, what is the probability of 9 OR 10, so i used : 5+3.2 = 8.2 as mean. Then i used the formula to calculate P(X=9) and P(X=10). At last i summed my two answers to get 0.233 or something.

Reply 8

potassosium

Because it's rectangular just work out the area of a rectangle with sides 1/k and b-a (difference between those two values), but we know the area is 1 so then we know that k is b-a (multiply through by k and you're left with k = b-a). E(X) is worked out by multiplying f(x) by x then integrating between b and a, E(X^2) is worked out by multiplying f(x) by x^2 then integrating. The bit after that was fiddly as you had to divide b^3 - a^3 by b-a and stuff but it does work out in the end.

Reply 9

Sparta_Kane

Original post by eshaq

I did the same as you for 2 and got pretty much the same answer.

For 1 I believe what you had to do is to find P(X=0) which was 0.0408 or something. Then you use the binomial distribution to find the probability of 1 out of 2 of the squares having no pebbles (i.e. 2x0.0408x(1-0.0408) ) then add on the probability of both squares having no pebbles in (0.0408^2) which gives you the answer.

Reply 10

mathsRus

Original post by potassosium

Lol i got to the very last stage and then just wrote the answer so hopefully should get full marks on that question. Btw for mean what numerical value did u get. I got two values of which I didn't know which to use

Reply 11

potassosium

Right this may sound really stupid but can anyone tell me how many questions there were on that exam and what the last question was about? I'm worrying that I missed a question there didn't seem to be that much on it :s

Reply 12

potassosium

Original post by mathsRus

I think I got 1 and it seemed wrong for some reason, but that's probably because I think I've seen others say it was 7 :s

Reply 13

mathsRus

Original post by potassosium

There were 7 questions in total the last being random cumulative variable.... The f(x) one

Reply 14

potassosium

Original post by mathsRus

There were 7 questions in total the last being random cumulative variable.... The f(x) one

Omg thank you, I've done them then, I just thought 7 questions was a bit lower than usual so I panicked ahah

Reply 15

eshaq

Original post by Sparta_Kane

ARGH! I thought about using binomial.

Are you sure you didn't have to times the mean by 2? It said it was used twice...?

Reply 16

mathsRus

Original post by potassosium

Omg thank you, I've done them then, I just thought 7 questions was a bit lower than usual so I panicked ahah

What mark do u think u got?

Reply 17

potassosium

Original post by Sparta_Kane

(P(X=0)*P(X=0)) + 2(P(X>0)*P(X=0)) is what I did and I ended up with 0.0799

(edited 9 years ago)

Reply 18

potassosium

Original post by mathsRus

What mark do u think u got?

If I'm honest I think that paper was pretty good and I've probably only dropped a few so hopefully I'll be on for an A in it, I've been getting As in past papers aswell

Reply 19

Sparta_Kane

Original post by eshaq

ARGH! I thought about using binomial.

Are you sure you didn't have to times the mean by 2? It said it was used twice...?

No, using it twice doesnt mean to multiply it by 2. This is one of those questions AQA like you to spot where you use Poisson to work out the proabability of an event happening then use Binomial to work out the probability of it happening multiple times.