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

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

Reply 164

luke010203

1

I think I got about 60, any guesses on what UMS this will be??

Posted from TSR Mobile

Reply 165

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

Reply 166

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)

Reply 167

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.

Reply 168

Suits101

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
Final answer

Reply 169

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..

Reply 170

Suits101

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)

Reply 171

KB_97

10

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?

Reply 172

For Poisson, would e^-lambda=/0 for P(X=x) be allowed as the second B1? The other being the upper limit.

Reply 173

sam_97

OP

6

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.

Reply 174

Suits101

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

(edited 7 years ago)

Reply 175

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

Reply 176

kyconny

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.

Reply 177