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# AQA A2 Mathematics MS2B Statistics 2B - 21st June 2016

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1. (Original post by MahuduElec)
Actually! sorry to say but it was 16.7 100%. VarX = 12.26 -3.08^2 = 2.7736

VarF=Var(10X)=100Varx(X)= 100 x 2.7736

SD = root(277.36) = 16.7 3sf, this is exactly what I did know that i remember the 16.65
O of course!

Damn that rule of Var(aX) = a^2Var(X)!

(Original post by sam_97)
It 100% asked for standard deviation I think it was 18.2 not 18.7, but yeah 18.2 lied inside the confidence interval
Yes you're right!

Phew that was lucky!
2. (Original post by MahuduElec)
Actually! sorry to say but it was 16.7 100%. VarX = 12.26 -3.08^2 = 2.7736

VarF=Var(10X)=100Varx(X)= 100 x 2.7736

SD = root(277.36) = 16.7 3sf, this is exactly what I did know that i remember the 16.65

I think your mistake was doing VarF=10VarX, it was 10^2VarX
I got the same as you
3. (Original post by Suits101)
Oh I see what you mean...

It was a straight line but you wouldn't treat it as a rectangular distribution.

In my opinion marks are for:

Method 1: splitting into shapes with integration

Area of triangle (1)
Integrating second function with limits (1) and correct integral with limits substituted (1)

Method 2: pure integration

Integrating function 1 with limits (1)
Integrating function 2 with limits (1)
Correct integration and substitution of limits (1)
Hm okay then buddy. I'm hopeful to get at least 1 if not 2 of those method marks then as I did integrate. Cheers! :-)
4. (Original post by Suits101)
Do you know how many marks these were worth?

1(e) - last part of Q1
3(b) - forgot to multiply by 2, book question
3(e)(ii) - I got 5.27p but apparently it asked for variance not SD

Also (sorry!) what did you put about Gerald's claim? I said CI is an element of 18.7 therefore no valid/proper conclusion can be made - also what did you do for last part of 6 as I just calculated a mean and said it doesn't support belief because answer was > 18.7?
1E was 3 marks, you'll lose 1 on 3b (I did this too) and probably 1 on 3eii

I said his claim cannot be supported because of exactly what you said
5. I think I got about 60, any guesses on what UMS this will be??

Posted from TSR Mobile
6. (Original post by luke010203)
I think I got about 60, any guesses on what UMS this will be??

Posted from TSR Mobile
mid to high 70s I'd guess
7. unofficial markscheme

1a. 0.703 (1)
b.0.769 (3)
c.0.133 (3)
d.0.144 (3)
e.0.268 (3)

2. mean =31.7 sd=1.719 (4)

3a. 0.35 (1)
b. 0.158 (3)
c. 3.08 (4)
d. 2.77 (2)
e. mean dosent = variance and something else (1)
f. mean= 31p sd=17p (3)

4a. k=10 (1)
b. 0.7 (1)
c.0.05 (2)
d.0.0289 (3)

5a. chi value =2.3388 reject Ho (11)
b. some wordy stuff (2)

6a, -2.08 accept HA (7)
b. (14.9, 18.55) (7)
c. wordy stuff (2)

7. differentiate and plot (5)
b. mean=3 (4)
8. for the 2 people choosing books, did a lot of people not times the answer by 2? for 3 marks I'd be surprised if you didn't have to times it by anything. My logic was that there are 2 people, so 2 combinations, I used a tree diagram to confirm.
9. (Original post by AngelicAmazing)
1E was 3 marks, you'll lose 1 on 3b (I did this too) and probably 1 on 3eii

I said his claim cannot be supported because of exactly what you said
I think I may lose 2 on 3(e)(ii)?

I'm thinking marks are for:

0.45 identified
0.35 x 0.45 x 2
10. (Original post by Busted838)
unofficial markscheme

1a. 0.703 (1)
b.0.769 (3)
c.0.133 (3)
d.0.144 (3)
e.0.268 (3)

2. mean =31.7 sd=1.719 (4)

3a. 0.35 (1)
b. 0.158 (3)
c. 3.08 (4)
d. 2.77 (2)
e. mean dosent = variance and something else (1)
f. mean= 31p sd=17p (3)

4a. k=10 (1)
b. 0.7 (1)
c.0.05 (2)
d.0.0289 (3)

5a. chi value =2.3388 reject Ho (11)
b. some wordy stuff (2)

6a, -2.08 accept HA (7)
b. (14.9, 18.55) (7)
c. wordy stuff (2)

7. differentiate and plot (5)
b. mean=3 (4)
Lot's of stuff wrong on this..
11. (Original post by Busted838)
unofficial markscheme

1a. 0.703 (1)
b.0.769 (3)
c.0.133 (3)
d.0.144 (3)
e.0.268 (3)

2. mean =31.7 sd=1.719 (4)

3a. 0.35 (1)
b. 0.158 (3)
c. 3.08 (4)
d. 2.77 (2)
e. mean dosent = variance and something else (1)
f. mean= 31p sd=17p (3)

4a. k=10 (1)
b. 0.7 (1)
c.0.05 (2)
d.0.0289 (3)

5a. chi value =2.3388 reject Ho (11)
b. some wordy stuff (2)

6a, -2.08 accept HA (7)
b. (14.9, 18.55) (7)
c. wordy stuff (2)

7. differentiate and plot (5)
b. mean=3 (4)
Hi,

3(e) I don't think they'll accept mean =/= variance as this was given in stem of question (they were similar, similar means it can be used) hence I think asnwers should be that people don't borrow books at a constant average rate and Poisson distribution has no upper limit but model X does (6 books)
12. (Original post by MahuduElec)
for the 2 people choosing books, did a lot of people not times the answer by 2? for 3 marks I'd be surprised if you didn't have to times it by anything. My logic was that there are 2 people, so 2 combinations, I used a tree diagram to confirm.
I didn't times by 2. Urgh. I won't get the answer mark or the method mark for x2. Does that mean I'll get 1 for working out 0.35 and 0.45 and x them together?
13. For Poisson, would e^-lambda=/0 for P(X=x) be allowed as the second B1? The other being the upper limit.
14. (Original post by MahuduElec)
for the 2 people choosing books, did a lot of people not times the answer by 2? for 3 marks I'd be surprised if you didn't have to times it by anything. My logic was that there are 2 people, so 2 combinations, I used a tree diagram to confirm.
Yeah man, you have to times by 2 as there are 2! combinations as you said.
15. Hi I'll try an unofficial markscheme:

1
(a) 0.70291
(b) 0.6691
(c) 0.13271
(d) 0.1441
(e) 0.622

2. Mean = 31.7 s2 = 2.96

3
(a) P(X > 3) = 0.3503
(b) P (X < 3) and P (X > 3) = 0.3153
(c) Show E(X) = 3.08, Var(x) = 2.77363
(d) Poisson distribution cannot have an upper limit, people don't borrow books at a constant average rate
(e)(i) Mean = 30.8p
(e)(ii) Standard deviation = 16.7p

4
(a) k = 10
(b) P = 0.70
(c) E(X) = 0.05
(d) Integrate, ans = 1/300
(e) SD(X) = 0.02895

5
(a) Test statistic = 2.3386, CV = 2.706 so accept null
(b) comment

6
(a) Test statistic = -2.20794, reject null
(b)(i) 16.7 +- 1.80
(b)(ii) Claim not valid because 18.2 lies in CI
(c) -

7
(a) Draw graph - DENSITY function, so have to differentiate
(b) 73/24
16. (Original post by KB_97)
I didn't times by 2. Urgh. I won't get the answer mark or the method mark for x2. Does that mean I'll get 1 for working out 0.35 and 0.45 and x them together?
If the mark scheme is consistent with past papers, then yes you will
17. (Original post by Busted838)
unofficial markscheme

1a. 0.703 (1)
b.0.769 (3)
c.0.133 (3)
d.0.144 (3)
e.0.268 (3)

2. mean =31.7 sd=1.719 (4)

3a. 0.35 (1)
b. 0.158 (3)
c. 3.08 (4)
d. 2.77 (2)
e. mean dosent = variance and something else (1)
f. mean= 31p sd=17p (3)

4a. k=10 (1)
b. 0.7 (1)
c.0.05 (2)
d.0.0289 (3)

5a. chi value =2.3388 reject Ho (11)
b. some wordy stuff (2)

6a, -2.08 accept HA (7)
b. (14.9, 18.55) (7)
c. wordy stuff (2)

7. differentiate and plot (5)
b. mean=3 (4)
Your book answer is wrong it should be 0.45 * 0.35 * 2 as there were two arrangements.
18. (Original post by Suits101)
Hi I'll try and unofficial markscheme:
1(a) 0.70291
(b) 0.6691
(c) 0.13271
(d) 0.1441
(e) 0.622

Mean = 31.7 s2 = ?

3(a) 0.3503
(b) 0.3153
(c) Var(x) = 2.77363
(d) Poisson cannot have an upper limit, people don't borrow books at a constant average rate
3(e)(i) 30.8p
3(e)(ii) 16.7p

4(a) k = 10
(b) P = 0.70
(c) E(X) = 0.05
(d) Integrate, ans = 1/300
(e) SD(X) = 0.02895

(a) Test statistic = 2.3386, CV = 2.706 so accept null5(b) comment6(a) Test statistic = -2.20794, reject null6(b)(i) 16.7 +- 1.806(b)(ii) Claim may be valid because 18.7 lies in CI6(c) -7(a) Draw graph - DENSITY function, so have to differentiate7(b) 73/24
Only disagree with 6Bii, the Ci was 14.9-18.5, the mainland mean was 18.2, which lies within the CI so the Ci does not support Gerald's claim
19. (Original post by MahuduElec)
Only disagree with 6Bii, the Ci was 14.9-18.5, the mainland mean was 18.2, which lies within the CI so the Ci does not support Gerald's claim
Sorry that's what I meant check edit
20. (Original post by Suits101)
Sorry that's what I meant check edit
Nice format, However you sill left it as claim may be vaild

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