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

This poll is closed

How many marks do you think you got?

73-75 5%

70-72 15%

67-69 17%

64-66 16%

61-63 14%

58-60 9%

55-57 7%

52-54 5%

49-51 2%

46-48 1%

42-45 1%

39-41 2%

36-38 1%

33-35 2%

30-32 1%

<303%

Total votes: 292

freebirdk

The answers I got were:

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(e^R) = 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

(edited 4 years ago)

Scroll to see replies

Reply 1

Statistician384

I used the central limit theorem for Q5 - can I still get any marks? This is because it said "mean".

(edited 4 years ago)

Reply 2

hitec25

Last question, expectation of red was 1.5, wouldn’t score be e^ 1.5= 4.4... compared to 9 of blue?

Reply 3

Lucallah

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))

Reply 4

freebirdk

You calculated e^E(R) which is not the same as E(e^R), 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?

(edited 4 years ago)

Reply 5

hitec25

Original post by freebirdk

You calculated e^E(R) which is not the same as E(e^R), for the same reason that e^(x+y) =/= e^x + e^y

Fair

Reply 6

hitec25

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 😂??

Reply 7

Statistician384

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".

Reply 8

LJ2001

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?

Reply 9

Lemur14

Volunteer Section Leader

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/showpost.php?p=84034170&postcount=238

Reply 10

tbt27

I got 0.055 for P(type 2)

Reply 11

Sanjith Hegde123

direction of inequality for critical region is incorrect i believe

Reply 12

UrBusted

Original post by freebirdk

The answers I got were:

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(e^R) = 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.

Reply 13

Sanjith Hegde123

can we get the mark distributions up too please :smile:

there on the exam discussion thread.

Reply 14

freebirdk

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.