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

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1. (Original post by Suits101)
This is what I remember and it MAY NOT BE RIGHT

1(a) 0,7029
1(b) 0.669
1(c) 0.1327
1(d) 0.144
1(e) 0.462 (should be 0.622?)

2. Mean = ? s2 = ?

3(a) 0.350
3(b) 0.1575
3(c) Var(x) = 2.7736
3(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) 5.27p

4
(a) k = 10
(b) P = 0.70
(c) E(X) = 0.05
(d) Integrate, ans = 1/300
(e) SD(X) = 0.0289

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

7(a) Draw graph - DENSITY function, so have to differentiate
7(b) 61/6 but I got it wrong
For question 6(a) the hypothesis test, I got:
Test statistic = -2.20794
Critical Value = -2.0537
Reject Ho

6(b) the confidence interval I got:
12.61759
20.78241

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2. (Original post by 2014_GCSE)
0.703
0.669
0.133
0.144
0.622

mean - 31.7 o^2 - can't remember

0.35
0.158
2.77
30.8
16.7
10
0.7
0.05
1/300
root3 / 60
14.9
18.5

ANSWER DUMP. didn't do the last 4 marker.
Agree with all - I got what you got for 3(e)(ii) but apparently it asked for variance not SD
3. (Original post by 2014_GCSE)
0.703
0.669
0.133
0.144
0.622

mean - 31.7 o^2 - can't remember

0.35
0.158
2.77
30.8
16.7
10
0.7
0.05
1/300
root3 / 60
14.9
18.5

ANSWER DUMP. didn't do the last 4 marker.
i recognise all of these
last one was 3.04 something, sd^2 was 2.96

what did people get for all the wordy answers?
4. (Original post by df97)
Hey guys! I had time at the end of the exam so I wrote down most (not all) of the answers I put on my calculator. Obviously some will likely be wrong but here's what I put so we can compare:
1) 0.703, 0.870, 0.133, 0.144, 0.622
2) mean=31.7 sd^2=2024.4
3) 0.35, 0.315, 2.77, 31p, £2.77
4) 0.7, sd(x)=0.0289
5) Z=2.34
6) (can't remember what I put for crit value of island A lizards), Mu(B)=15.1cm
7) For graph I had straight horizontal line from (1, 1/4) to (4, 1/4) and then straight line going down to (6,0) and 0 otherwise.
E(X)=3.04
Let me know if you think any are wrong (i think 1b might be). Thanks!
for Q3, since it's an average, you can have .8 of a penny no? If it asks for a value then then I agree round to 2sf.

eg when you take the average amount of books, you get 3.08
5. (Original post by df97)
For 3(b) I did 2(0.35 x 0.45) as there are two possible ways of people taking out the books (either first guy got out more than 3 and second got out less than 3 or vice versa). Not sure if I'm right but either way we have very similar answers which is good news!
For 3(e)(ii) I did 2.77 x 10^2 as var(aX)=a^2var(X), likewise I'm unsure on this one too.

Thanks

(Original post by luke010203)
For question 6(a) the hypothesis test, I got:
Test statistic = -2.20794
Critical Value = -2.0537
Reject Ho

6(b) the confidence interval I got:
12.61759
20.78241

Posted from TSR Mobile
Agree with 6(a)

6(b) I got 16.7 +- 1.80?
6. (Original post by MahuduElec)
About the same yes, the last Q in both had sneaky tricks
Yeah I think the first part of the last question may have thrown people. The probability questions seemed quite long winded as well.
7. (Original post by MahuduElec)
for Q3, since it's an average, you can have .8 of a penny no? If it asks for a value then then I agree round to 2sf.

eg when you take the average amount of books, you get 3.08
Exactly what I did.
8. (Original post by Suits101)
6(b) I got 16.7 +- 1.80?
I got this too
9. (Original post by sam_97)
I got this too
10. (Original post by MahuduElec)
for Q3, since it's an average, you can have .8 of a penny no? If it asks for a value then then I agree round to 2sf.

eg when you take the average amount of books, you get 3.08
I agree. I put 30.8 and then rounded it to 31p, hopefully I'll still get the mark for putting both.
11. (Original post by tanyapotter)
The confidence interval in question 6, for Island B.
12. For the last question, why didn't you just put

1/4(x-1) = 0.5

and re-arrange for x=3?

Therefore E(X) = 3

I know this is wrong because it's not 4 marks worth but I don't know why it's wrong?
13. (Original post by df97)
I agree. I put 30.8 and then rounded it to 31p, hopefully I'll still get the mark for putting both.
You will get the mark
14. (Original post by 2014_GCSE)
For the last question, why didn't you just put

1/4(x-1) = 0.5

and re-arrange for x=3?

Therefore E(X) = 3

I know this is wrong because it's not 4 marks worth but I don't know why it's wrong?
0.5 is the Median value of F(x), not the mean
15. (Original post by Suits101)

Thanks

Agree with 6(a)

6(b) I got 16.7 +- 1.80?
Pretty sure your 6B is correct
16. (Original post by 2014_GCSE)
For the last question, why didn't you just put

1/4(x-1) = 0.5

and re-arrange for x=3?

Therefore E(X) = 3

I know this is wrong because it's not 4 marks worth but I don't know why it's wrong?
I think you might be thinking about the median there, the question was about the mean.
17. (Original post by sam_97)
You will get the mark
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?
18. (Original post by 2014_GCSE)
For the last question, why didn't you just put

1/4(x-1) = 0.5

and re-arrange for x=3?

Therefore E(X) = 3

I know this is wrong because it's not 4 marks worth but I don't know why it's wrong?
That is the method for the median. since Fx is a cumulative function, Fx=0.5 gives the median value when solving for X
19. q1) a) 0.703 b) 0.967 (dunno if right) c) 0.133 d) 0.144 e) 0.622
q2) I used Z so could be wrong as thinking needed to use t. mean= 31.7 var= 4.3264
3a) 0.35 b) 0.1575 c) mean proof of 3.08, 2.7736 d) independent, possion has no top limit, random doesn't occur at a constant rate etc. e)i) 30.8 e)ii) 4.08 (i think this is what i got)
4a) k=10 b) 0.7 c) 0.35 d) 1/300 e) (root3)/60
5a) 2.3386 chi calc, 2.706 chi crit, hence accept H0. 5b) some logical explanation linking Observed and expected and then the suggested engine
6a) accept H0 i think 6bi) 14.9 to 18.5 6bii) explanation 6c) mean is above 18.2 hence doesn't support his claim.
7a) f(x) = 1/4 from x=1,2,3,4. then negative gradient from (4,1/4) to (6,0)
7b) 73/24

I believe most of these are right. 1b, 2, and 6bi i need conformation of these results then we'll have an unofficial mark scheme.
grade boundaries i think will be
A* 68-69 A 63-64 B 57-58 C 51-52
Full UMS would be at 74/75 as easier than last year so would expect about 2 above. Hope this has helped.
20. For question 6 what were the values for the means of island A/B/C and their sample sizes? and what was the standard deviation you had to use in the confidence interval? I think it said it was unknown so had to use the t value

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Updated: June 22, 2016
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