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

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1. (Original post by kyle18)
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.
1(b) 0.669 worked out using 1 - [P(X=0) + P(X=1)] with lambda = 2.3

6(b)(i) agree

3(d) agree with all reasons but independent is not accepted as it's given in question (saying it will give you no mark but you won't lose a mark for saying it as it's still right)
2. (Original post by luke010203)
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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6(a) SD was given so z values for hypothesis test
6(b)(i) SD was unknown so had to use t values for confidence interval
3. (Original post by sam_97)
I think you might be thinking about the median there, the question was about the mean.

(Original post by MahuduElec)
That is the method for the median. since Fx is a cumulative function, Fx=0.5 gives the median value when solving for X
5. (Original post by Suits101)
6(a) SD was given so z values for hypothesis test
6(b)(i) SD was unknown so had to use t values for confidence interval
Can you remember the island data from the table??
For island B I think it was 116.9 and n was 7, so the mean was 16.7 but I can't remember what the standard deviation was to calculate the confidence interval

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6. (Original post by Suits101)
1(b) 0.669 worked out using 1 - [P(X=0) + P(X=1)] with lambda = 2.3

6(b)(i) agree

3(d) agree with all reasons but independent is not accepted as it's given in question (saying it will give you no mark but you won't lose a mark for saying it as it's still right)
Yeah 1b) is 0.669 just redone it, oh well I shouldn't have lost all 3 marks.
3d) okay thank you.
7. (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?
1)e) 3 marks

3)b) 3 marks

3)e)ii) 3 marks in total for part i) and ii). Definitely asked for SD but I got 16.6p

My explanations about Gerald's claim were pretty much the same as yours, I just made a comment about the upper confidence limit in the first one as well. I'm glad you have mentioned the last part actually as I was unsure if I had done the right thing!
I used the rectangular distribution between 1 and 4 to calculate the E(X)=2.5 in that region, then integrated xf(x) between 4 and 6 and added 4 to it to calculate the E(X)=31/6 in that region. Then I averaged these two numbers to get somebody like 3.83. I didn't think I had got it right in the exam, but I'm curious about method marks for this question.
I hope everybody else found the exam okay btw!:-)
9. (Original post by luke010203)
Can you remember the island data from the table??
For island B I think it was 116.9 and n was 7, so the mean was 16.7 but I can't remember what the standard deviation was to calculate the confidence interval

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Yes that's right.

You calculated s which is used as sigma (standard deviation) that was given in question as sigma(x-x bar) squared
10. (Original post by sam_97)
1)e) 3 marks

3)b) 3 marks

3)e)ii) 3 marks in total for part i) and ii). Definitely asked for SD but I got 16.6p

My explanations about Gerald's claim were pretty much the same as yours, I just made a comment about the upper confidence limit in the first one as well. I'm glad you have mentioned the last part actually as I was unsure if I had done the right thing!
Thanks!

So then I was right about my answer because everyone is saying it asked for variance? Makes me feel much better!

I forgot about the first CI - did 18.7 lie in it because if so then I think I just got away with that haha!

(Original post by benjammy)
I used the rectangular distribution between 1 and 4 to calculate the E(X)=2.5 in that region, then integrated xf(x) between 4 and 6 and added 4 to it to calculate the E(X)=31/6 in that region. Then I averaged these two numbers to get somebody like 3.83. I didn't think I had got it right in the exam, but I'm curious about method marks for this question.
I hope everybody else found the exam okay btw!:-)
It was not a rectangular distribution, it was a cumulative distribution function - that's why!
11. (Original post by Suits101)
Yes that's right.

You calculated s which is used as sigma (standard deviation) that was given in question as sigma(x-x bar) squared
Can you remember what sigma was because I got the wrong values for my confidence interval:/

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12. Oh yh, i do remember doing sd for 3eii. Sorry about that to anyone who I told it was variance, i just remember the 27.77 but now remember doing the square root of that
13. (Original post by luke010203)
Can you remember what sigma was because I got the wrong values for my confidence interval:/

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Sorry I can't! I want to say something like 22.7 though.

(Original post by MahuduElec)
Oh yh, i do remember doing sd for 3eii. Sorry about that to anyone who I told it was variance, i just remember the 27.77 but now remember doing the square root of that
Few! So it was 5.27 or something?
14. (Original post by luke010203)
Can you remember what sigma was because I got the wrong values for my confidence interval:/

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(x-xbar)^2 was around 22.6
15. (Original post by Suits101)
Thanks!

So then I was right about my answer because everyone is saying it asked for variance? Makes me feel much better!

I forgot about the first CI - did 18.7 lie in it because if so then I think I just got away with that haha!

It was not a rectangular distribution, it was a cumulative distribution function - that's why!
But for 7a was the graph not rectangular between 1 and 4 at f(x)=0.25?
16. (Original post by Suits101)
Sorry I can't! I want to say something like 22.7 though.

Few! So it was 5.27 or something?
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
17. (Original post by MahuduElec)
(x-xbar)^2 was around 22.6
I have no idea what I did wrong to get the values I did hopefully some method marks tho

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18. (Original post by benjammy)
But for 7a was the graph not rectangular between 1 and 4 at f(x)=0.25?
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)
19. (Original post by luke010203)
I have no idea what I did wrong to get the values I did hopefully some method marks tho

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hopefully
20. (Original post by Suits101)
Thanks!

So then I was right about my answer because everyone is saying it asked for variance? Makes me feel much better!

I forgot about the first CI - did 18.7 lie in it because if so then I think I just got away with that haha!
It 100% asked for standard deviation

I think it was 18.2 not 18.7, but yeah 18.2 lied inside the confidence interval

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Updated: June 22, 2016
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