The Student Room Group

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

Scroll to see replies

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] 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)
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


Posted from TSR Mobile


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
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


:facepalm:
drew F(x) feelsbadman
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


Posted from TSR Mobile
Reply 145
Original post by Suits101
1(b) 0.669 worked out using 1 - [P(X] 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.
Reply 146
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!
Could somebody please explain how you were suppose to go about 7b for me please.
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!:-)
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


Posted from TSR Mobile


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
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
Could somebody please explain how you were suppose to go about 7b for me please.
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! :smile:
(edited 7 years ago)
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:/


Posted from TSR Mobile
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
Original post by luke010203
Can you remember what sigma was because I got the wrong values for my confidence interval:/


Posted from TSR Mobile


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? :smile:
Original post by luke010203
Can you remember what sigma was because I got the wrong values for my confidence interval:/

Posted from TSR Mobile


(x-xbar)^2 was around 22.6
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! :smile:


But for 7a was the graph not rectangular between 1 and 4 at f(x)=0.25?
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? :smile:


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
(edited 7 years ago)
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:frown: hopefully some method marks tho


Posted from TSR Mobile
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)
Final answer (1)

Method 2: pure integration

Integrating function 1 with limits (1)
Integrating function 2 with limits (1)
Correct integration and substitution of limits (1)
Final answer (1)
Original post by luke010203
I have no idea what I did wrong to get the values I did:frown: hopefully some method marks tho


Posted from TSR Mobile


hopefully :frown:
Reply 159
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 :smile:

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

Quick Reply