# Edexcel A Level Further Mathematics Paper 3B (9FM0 3B) - 14th June 2024 [Exam Chat]

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Original post by rebecca9801
Jesus christ how did you remember all these numbers????

I had 30 minutes remaining so I chucked the score distributions into my calculator (fx CG-50). I used matrices to store multiple lines of data in the same line so all of the numbers would stay in the calculator's memory.
Original post by kyan.cheung
Hi all, I've recorded the score distributions and my answers to each questions. Please tag me if you notice any issues!

1.

(6 marks total)

a. (3 marks) Var(X)=4.9
b. (3 marks) Var(X^2)=97.69
2. (7 marks total)
ai. P(X=3) = 0.2090
aii. P(X<2)=0.3084 (2 marks total)
b. (4 marks) The Critical region is A<=1 or A>=12
c. (1 mark) P(Type I)=0.03744
3. (6 marks total)
a. (1 mark) Hypotheses - H0 colour and year group not associated, H1 colour and year group are associated
bi. Black Y10-12 because they are respectively the least frequent favourite colour and year group
ii. 5.383 (2 marks total)
c. (3 marks) (5-1)(3-1)=8 degrees of freedom with critical value 20.090, so reject H0 and conclude there is association between year group and favourite colour
4. (12 marks total)
a. (1 mark) X~Geo(1/6)
b. (2 marks) Show P(X<=3)=91/216
c. (5 marks) E[x]=6, Var(X)=30 -> Var(Xmean)30/64, P(5.6<Xmean<7.2)=0.6806
d. (4 marks) P(X>=16)=test value=0.064905...>0.05 so dont reject H0
5. (10 marks total)
a. (1 mark) H0 p=0.03 and H1 p>0.03
b. (2 marks) Size of test a 0.1821
c. (3 marks) Size of test b 0.2083
di. Power of a 0.7232
ii. We expect 94.877 compontents tested with test b (3 marks total)
e. Use test A as it has lower size and higher power (1 mark)
6. (16 marks total)
a. (6 marks) Variance 6
b. (4 marks) P(X<=2)=203/255
c. (4 marks) Y is a geometric distribution
d. (2 marks) P(X1+X2>5)=729/4096
7. (18 marks total)
a. (2 marks) Sam has prob 0.5886 of winning >=4 times
b. (4 marks) Don't remember what it asked but the answer was 0.04364 apparently
c. (6 marks) Show sam expects to win 1/25(n^2-16n). Sketch - basically set W~B(n, 1/5), then expected winnings is E[w^2+w-n] by sum rule. Then use E[w^2]=Var(W)+E^2[w]=4n/25+n^2/25 and simplify.
d. (4 marks) Sam has probability 0.3518 of not losing money
e. (4 marks) The other person has expected profit r^2-4r
RE predictions: with many people forgetting contingency tables and struggling on question 7, I think grade boundaries will be ~52 for A*.
Wasn’t Var(X^2) 160.9?
Original post by sssdssss
Wasn’t Var(X^2) 160.9?

No, that's E[X^4].
Can u use poison approximation for Q5
Original post by kyan.cheung
No, that's E[X^4].
E(X^4) was 176.9 ,

[E(X)^2]]^2 was 16

176.9-16 = 160.9
no?
Original post by sssdssss
E(X^4) was 176.9 ,
[E(X)^2]]^2 was 16
176.9-16 = 160.9
no?

No, to find Var(X2) you find E[X4]-(E[X2])2 so the expression should be 176.9-8.92.
Original post by kyan.cheung
No, to find Var(X2) you find E[X4]-(E[X2])2 so the expression should be 176.9-8.92.
Why wouldn’t it be

E(X^4) -[E(X)^2]^2
Original post by Abc1321
Can u use poison approximation for Q5

thats what i did and im hoping you can or im finished, do you mean like x--po(0.03n)
https://discord.gg/WdYyXZPD
Come discuss the paper.
Original post by sssdssss
Why wouldn’t it be
E(X^4) -[E(X)^2]^2

it is but your second term is wrong its not 16
Original post by sssdssss
Why wouldn’t it be
E(X^4) -[E(X)^2]^2

Set Y=X2, then we want to find Var(Y). We know that's E[Y2]-E2[Y], and substituting Y=X2 back in we get Var(Y)=E[(X2)2]-E2[X2].
(edited 2 months ago)
Did anyone do a binomial expansion to get P(X<2) on the p.g.f question?
is there unofficial ms
Original post by craig222
thats what i did and im hoping you can or im finished, do you mean like x--po(0.03n)

Yes
Original post by anonymoussbbq
if the show that for part 7c was a typo would they remove the whole q😂

Just to be clear, 7c was not a typo.
If you're interested, I've attached a type-up of my solution to 7c.
(edited 2 months ago)
Dyu think 235-240 is a good estimate for grade boundaries FS1,FM1
Original post by kyan.cheung
Just to be clear, 7c was not a typo.
If you're interested, I've attached a type-up of my solution to 7c.

That’s mental fair play mate
Original post by kyan.cheung
I had 30 minutes remaining so I chucked the score distributions into my calculator (fx CG-50). I used matrices to store multiple lines of data in the same line so all of the numbers would stay in the calculator's memory.

ohhhh makes sense. Idk how you had 30m left though damn. Gonna ask you something then since you're smart:

For the last question proving that E(X) = 1/25(N^2 ....):
I wrote down X~B(n, 1/5)
I made a table of 3 rows:
1) X: 1, 2, 3, 4 ... n
2) Money paid to Sam: 2, 4, 6, 8 ... 2n
3) P(X=x): n!/x!(n-x)! p^x (1-p)^n-x

Then I did the sum of each pair multiplied together: 2 x n!/(n-1)! 0.2^1 0.8^n-1 + 4 x n!/2!(n-2)! 0.2^2 0.8^n-2 ... + ... 2n x n!/n!(n-n)! 0.2^n

Then I used Sn = a(1/r^n) / 1-r

It didn't simplify well and I didn't get the final answer (it's wrong).

But do you think I might've gotten some marks for that or is it 100% wrong?
Original post by Ray1kay
Did anyone do a binomial expansion to get P(X<2) on the p.g.f question?

I did maclaurin but I think both would've worked
Original post by Ray1kay
Did anyone do a binomial expansion to get P(X<2) on the p.g.f question?

Yeah I did, both that and mclaurins came out with the same answer