# Edexcel Further Statistics 1 - UNOFFICIAL MARK SCHEME [18 June 2019]Watch

Poll: How many marks do you think you got?
73-75 (15)
5.19%
70-72 (43)
14.88%
67-69 (48)
16.61%
64-66 (45)
15.57%
61-63 (40)
13.84%
58-60 (27)
9.34%
55-57 (19)
6.57%
52-54 (14)
4.84%
49-51 (6)
2.08%
46-48 (3)
1.04%
42-45 (4)
1.38%
39-41 (5)
1.73%
36-38 (3)
1.04%
33-35 (5)
1.73%
30-32 (2)
0.69%
(10)
3.46%
#1

1a) X~B(40, 0.02), P(X >= 3) = 0.0457
1b) Y~NB(3, 0.02), P(Y=40) = 0.00281
1c) E(Y) = 150

2a) X~Po(20/3), P(X>4) = 0.794
2b) P(exactly 3/4 5 min breaks had no calls) = 0.0219
2c) P(1 call in 5m break)*P(1 call in 15m break) = 0.0106

3) X~N(3, 2.6/80), P(X > 3.25) = 0.0828

4a) B(6, 0.55) - Although I suspect B(6, p) got you full marks
4b) v=3, X2 = 8.31 > 7.82, So reject H0 (not 100% on the numbers here, would appreciate confirmation of 8.31)
4c) s=17.67, t=9.62 (or the other way around maybe)
4d) v=4, X2 = 8.749 < 9.488, Accept H0
4e) Wild not cultivated, because poisson

5a) λ=7.5, critical region: X<=2, X>=14
5b) P(Type I) = 0.0418
5c) X~Bin(8, 0.0418), P(X = 2) = 0.0414
5d) P(Type 2 | λ=6.3) = 0.945

6a) 1/ln2
6b) (2ln2 - 1)/(ln2)^2 OR 2/ln2 - 1/(ln2)^2
6c) P(X=3) = 1/24ln2 OR 0.0601

7a) P(B=4) = 8/81, P(B<=5) = 211/243
7b) E(B2) = 15
7c) E(B2) = 15, E(eR) = 19.297, so red is better than blue

I've compared these answers with several people so I'm pretty confident these are all correct
Last edited by freebirdk; 4 weeks ago
9
4 weeks ago
#2
I used the central limit theorem for Q5 - can I still get any marks? This is because it said "mean".
Last edited by Statistician384; 4 weeks ago
1
4 weeks ago
#3
Last question, expectation of red was 1.5, wouldn’t score be e^ 1.5= 4.4... compared to 9 of blue?
2
4 weeks ago
#4
(Original post by hitec25)
Last question, expectation of red was 1.5, wouldn’t score be e^ 1.5= 4.4... compared to 9 of blue?
Score is e^X so expected score is E(e^X). Not the same as e^(E(X))
3
#5
You calculated eE(R) which is not the same as E(eR), for the same reason that e^(x+y) =/= e^x + e^y

Same thing with part b, E(X^2) is 15. I did it by rearranging the Var(X) formula to get E(X^2) = Var(X) + (E(X))^2, and then subbing in mean and variance.
(Original post by hitec25)
Last question, expectation of red was 1.5, wouldn’t score be e^ 1.5= 4.4... compared to 9 of blue?
Last edited by freebirdk; 4 weeks ago
0
4 weeks ago
#6
(Original post by freebirdk)
You calculated eE(R) which is not the same as E(eR), for the same reason that e^(x+y) =/= e^x + e^y
Fair
0
4 weeks ago
#7
(Original post by hitec25)
Last question, expectation of red was 1.5, wouldn’t score be e^ 1.5= 4.4... compared to 9 of blue?
Say 2 marks for finding each expectations 😂??
0
4 weeks ago
#8
Of 3 months
(Original post by Statistician384)
I used the central limit theorem for Q5 - can I still get any marks? This is because it said "mean".
0
4 weeks ago
#9
You reckon I’d get 1, 2 or 3 marks for finding E(X^2)=15 but getting E(e^X) wrongly as 4.44?
0
4 weeks ago
#10
4d was write the hypothesis then4e was perform the test and 4f was wild/cultivated
There's marks per question here if anyone wants: (question 4 is slightly wrong, 4b shouldn't be there and the rest of Q4 should move up one part)
https://www.thestudentroom.co.uk/sho...&postcount=238
0
4 weeks ago
#11
I got 0.055 for P(type 2)
0
4 weeks ago
#12
direction of inequality for critical region is incorrect i believe
0
4 weeks ago
#13
(Original post by freebirdk)

1a) X~B(40, 0.02), P(X >= 3) = 0.0457
1b) Y~NB(3, 0.02), P(Y=40) = 0.00281
1c) E(Y) = 150

2a) X~Po(20/3), P(X>4) = 0.794
2b) P(exactly 3/4 5 min breaks had no calls) = 0.0219
2c) P(1 call in 5m break)*P(1 call in 15m break) = 0.0106

3) X~N(3, 2.6/80), P(X > 3.25) = 0.0828

4a) B(6, 0.55) - Although I suspect B(6, p) got you full marks
4b) v=3, X2 = 8.31 > 7.82, So reject H0 (not 100% on the numbers here, would appreciate confirmation)
4c) s=17.67, t=9.62 (or the other way around maybe)
4d) v=4, X2 = 8.749 < 9.488, Accept H0
4e) Wild not cultivated, because poisson

5a) λ=7.5, critical region: X>=2, X<=14
5b) P(Type I) = 0.0418
5c) X~Bin(8, 0.0418), P(X = 2) = 0.0414
5d) P(Type 2 | λ=6.3) = 0.945

6a) 1/ln2
6b) (2ln2 - 1)/(ln2)^2 OR 2/ln2 - 1/(ln2)^2
6c) P(X=3) = 1/24ln2 OR 0.0601

7a) P(B=4) = 8/81, P(B<=5) = 211/243
7b) E(B2) = 15
7c) E(B2) = 15, E(eR) = 19.297, so red is better than blue

I've compared these answers with several people so I'm pretty confident these are all correct
Signs are wrong way round for critical region, its x<=2 and x>= 14, your part d proves it since doing the opposite region with lamda = 6.3 is what you got.
0
4 weeks ago
#14
can we get the mark distributions up too please
there on the exam discussion thread.
0
#15
(Original post by UrBusted)
Signs are wrong way round for critical region, its x<=2 and x>= 14, your part d proves it since doing the opposite region with lamda = 6.3 is what you got.
Thank you! Fixed
0
4 weeks ago
#16
for 5c, question was probability >=2
0
4 weeks ago
#17

for calculating E(e^R) (which is the same as E(e^X))
Last edited by Levelent; 4 weeks ago
5
#18
I think you found P(rejecting H0 | λ=6.3), not P(accepting H0 | λ=6.3), which is Type II and 1-what you calculated.
(Original post by tbt27)
I got 0.055 for P(type 2)
1
4 weeks ago
#19
(Original post by freebirdk)
I think you found P(rejecting H0 | λ=6.3), not P(accepting H0 | λ=6.3), which is Type II and 1-what you calculated.
I used Lambda = 2.1 and the CTL 😬
0
4 weeks ago
#20
thats power
(Original post by tbt27)
I got 0.055 for P(type 2)
0
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