# A-Level S2 Maths Edexcel 25th June 2018 Unofficial Mark Scheme

Watch
Announcements

Q1.

a) i) 0.8843

a) ii) 0.5662

b) 0.33

c) 0.0436418928

Q2.

a) 1: Fixed number of trials, 2: Probability of outcome is constant

b) 0.09375

c) 0.3437

d) 0.1694

Q3.

a) 2.5

b) 17/40

c) F(t) = {

0 for x<1

1/4t^2 - 1/2t + 1/4 for 1<x<2

7/16t^2 - 1/16t^3 - 1/2t for 2<x<4

1 for x>4

d) t = 1.894

e) 0.9375

f) 4/15

Q4.

a) alpha = -2, beta = 10

b) 107/240

c) 12/107

Q5.

a) As 0.1239 < 0.05, there is insufficient evidence at the 5% significance level to reject the null hypothesis in favour of the claim

b) X</=10 and X>/=25

c) 8 lies in the critical region, therefore there is sufficient evidence

d) 0.0367

e) n = 31

Q6.

a) show that (differentiate the c.d.f. to give the p.d.f., then differentiate the p.d.f., equate to zero and solve for a, by factoring k out)

b) 623/1280

a) i) 0.8843

a) ii) 0.5662

b) 0.33

c) 0.0436418928

Q2.

a) 1: Fixed number of trials, 2: Probability of outcome is constant

b) 0.09375

c) 0.3437

d) 0.1694

Q3.

a) 2.5

b) 17/40

c) F(t) = {

0 for x<1

1/4t^2 - 1/2t + 1/4 for 1<x<2

7/16t^2 - 1/16t^3 - 1/2t for 2<x<4

1 for x>4

d) t = 1.894

e) 0.9375

f) 4/15

Q4.

a) alpha = -2, beta = 10

b) 107/240

c) 12/107

Q5.

a) As 0.1239 < 0.05, there is insufficient evidence at the 5% significance level to reject the null hypothesis in favour of the claim

b) X</=10 and X>/=25

c) 8 lies in the critical region, therefore there is sufficient evidence

d) 0.0367

e) n = 31

Q6.

a) show that (differentiate the c.d.f. to give the p.d.f., then differentiate the p.d.f., equate to zero and solve for a, by factoring k out)

b) 623/1280

9

reply

Report

#6

(Original post by

you parts 4b and 4c are wronf I believe, they are 107/240 and 12/107

**GausIsTheBoss**)you parts 4b and 4c are wronf I believe, they are 107/240 and 12/107

0

reply

Report

#7

(Original post by

I got n as 40?

**ghfghgfg**)I got n as 40?

np was 200*0.2 which was fourty, which would've been the mean (and then you would approximate it to a normal distribution of mean 40 and standard deviation of root 32)

10

reply

Report

#8

I got the critical region correct, but for some reason said 8 is not in the critical region. How much marks is lost from that?

2

reply

Report

#9

(Original post by

I got the critical region correct, but for some reason said 8 is not in the critical region. How much marks is lost from that?

**Mypenjustranout**)I got the critical region correct, but for some reason said 8 is not in the critical region. How much marks is lost from that?

0

reply

Report

#10

for the approximation, you can only do

binomial=>normal

poisson=>normal

i did binomial approx and got 50 and most people got that

binomial=>normal

poisson=>normal

i did binomial approx and got 50 and most people got that

4

reply

Report

#11

(Original post by

Q1.

a) i) 0.8843

a) ii) 0.5662

b) 0.33

c) 0.0436418928

Q2.

a) 1: Fixed number of trials, 2: Probability of outcome is constant

b) 0.09375

c) 0.3437

d) 0.1694

Q3.

a) 2.5

b) 17/40

c) F(t) = {

0 for x<1

1/4t^2 - 1/2t + 1/4 for 1<x<2

7/16t^2 - 1/16t^3 - 1/2t for 2<x<4

1 for x>4

d) t = 1.894

e) 0.9375

f) 4/15

Q4.

a) alpha = -2, beta = 10

b) 29/48

c) 0.05

Q5.

a) As 0.1239 < 0.05, there is insufficient evidence at the 5% significance level to reject the null hypothesis in favour of the claim

b) X</=10 and X>/=25

c) 8 lies in the critical region, therefore there is sufficient evidence

d) 0.0367

e) n = 31

Q6.

a) show that (differentiate the c.d.f. to give the p.d.f., then differentiate the p.d.f., equate to zero and solve for a, by factoring k out)

b) 623/1280

**Sammy Arab**)Q1.

a) i) 0.8843

a) ii) 0.5662

b) 0.33

c) 0.0436418928

Q2.

a) 1: Fixed number of trials, 2: Probability of outcome is constant

b) 0.09375

c) 0.3437

d) 0.1694

Q3.

a) 2.5

b) 17/40

c) F(t) = {

0 for x<1

1/4t^2 - 1/2t + 1/4 for 1<x<2

7/16t^2 - 1/16t^3 - 1/2t for 2<x<4

1 for x>4

d) t = 1.894

e) 0.9375

f) 4/15

Q4.

a) alpha = -2, beta = 10

b) 29/48

c) 0.05

Q5.

a) As 0.1239 < 0.05, there is insufficient evidence at the 5% significance level to reject the null hypothesis in favour of the claim

b) X</=10 and X>/=25

c) 8 lies in the critical region, therefore there is sufficient evidence

d) 0.0367

e) n = 31

Q6.

a) show that (differentiate the c.d.f. to give the p.d.f., then differentiate the p.d.f., equate to zero and solve for a, by factoring k out)

b) 623/1280

heres my answers bro

0

reply

Report

#12

(Original post by

for the approximation, you can only do

binomial=>normal

poisson=>normal

i did binomial approx and got 50 and most people got that

**arjun0212**)for the approximation, you can only do

binomial=>normal

poisson=>normal

i did binomial approx and got 50 and most people got that

0

reply

Report

#14

(Original post by

YES same here !!

**nisha.sri**)YES same here !!

0

reply

Report

#15

(Original post by

Did you do the correction? As in -0.5

**daisie**)Did you do the correction? As in -0.5

Edit: oh I just tried my calculation again and got 31.

2

reply

Report

#16

(Original post by

Why can't you approximate directly with normal? surely because np and npq > 5, its suitable to approximate with normal?

**wendychan**)Why can't you approximate directly with normal? surely because np and npq > 5, its suitable to approximate with normal?

np=40

sd=root32

0

reply

Report

#17

(Original post by

I got like 39.99 which basically rounds to 40. I did the approximation aswell as a continuity correction.

**Mypenjustranout**)I got like 39.99 which basically rounds to 40. I did the approximation aswell as a continuity correction.

3

reply

**GausIsTheBoss**)

you parts 4b and 4c are wronf I believe, they are 107/240 and 12/107

Given alpha is -2, that suggests the train could possibly leave 2 minutes early, at 08:28, and this should be considered in the calculation.

(I'm assuming you calculated the probability that the train left between 0 and 5 minutes after 08:30)

2

reply

Report

#19

0

reply

Report

#20

(Original post by

I got 30.998 or something like that so I rounded to 31. I did the continuity correction as n-0.5-mean

**daisie**)I got 30.998 or something like that so I rounded to 31. I did the continuity correction as n-0.5-mean

How much marks would be lost for this mistake. I did everything else correct, but just wrote the wrong number at the end.

I must have just thaught in my head 30.99 rounds to 40 :/

0

reply

X

### Quick Reply

Back

to top

to top