You don't workout E(x^2) by squaring E(X)
you either var(x)+E(X)^2 or integrate
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Edexcel S2  27th June 2016 AM
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 Follow
 881
 27062016 12:30

 Follow
 882
 27062016 12:31
(Original post by igotohaggerston)
Here are the real answers I am very confident they are all right
1.
a) Mean = 1.41
Variance = 1.4419
b) mean approximately equals the variance therefore Poisson is suitable
c)
i) 0.2510
ii)0.7769
d) 0.1378
e)0.1516
2.
a) 60
b)
i) 0.0133
ii)0.0159
c) 0.2369>0.01 therefore do not reject H0 his claim is not supported
3.
a) f(R)={0.25 , 5<R<9
{0 , otherwise
b) P(7<R<10)= (107)x0.25= 0.75
c)E(A)=E(piR^2)=PiE(R^2) = 151pi/3
4.
a) show by differentiating to find the pdf then differentiating again and setting it equal to zero to find maximum k cannot be zero the for solving what is in the bracket gives you that b=8.
b) I think there are a lot of ways to do this question, I solved for a first but k=1/9
5.
n=225 ignore the negative value because the standard deviation is always positive and plus you can't get a fraction of a student.
6.
a) (4,4)(6,6)(8,8)
(4,6)(4,8)(6,4)
(6,8)(8,4)(8,6)
b) m 4 5 6 7 8
P(M=m) 1/4 3/10 29/100 3/25 1/25
c)Y~B(n,1/25)
P(Y>=1)>0.9
1P(Y=0)>0.9
P(y=0)<0.1
0.96^n<0.1
taking logs( don't forget to switch the inequality)
you get n>56....
therefore the smallest value for n is 57
7.
a) E(x)=1.2
b)Var(x)=0.24
c) P(X>1.5)=0.325
d)£41.88
e) He should not as 40 <41.88Post rating:3 
 Follow
 883
 27062016 12:32
(Original post by Student403)
1) 1.41, 1.4419, 0.2510, 0.7769, 0.1378, 0.1516
2) 60, 0.0133, 0.0159, 0.2396
3) 0.5, 151pi/3
4) B = 8, A = 12 (didn't need this), k = 1/9
5) 225 (i got 342.25 too but put 225 as my final answer. Not sure of the correct reason)
6) 9 samples, 0.25(4), 0.3(5), 0.29(6), 0.12(7), 0.04(8) n = 57
7) mean 1.2, var 0.24, prob 0.325, 41.88GBP, no he should not remove staples 
 Follow
 884
 27062016 12:32
(Original post by Armpits)
Wasn't N 342.5 or something like that?
But that gives you the opposite sign z value. (i.e. if the z value needed was 1.75, 342.5 gives you 1.75 or vice versa).
I'm wondering how many marks would you lose for writing 342.5, 
 Follow
 885
 27062016 12:33
(Original post by Armpits)
Why 225?
1) whole number whereas 342.25 is not
2) 225 works completely plugging in to the normal distribution expression. Whereas you have to take the negative square root of 342.25 for it to work. (i.e. if you take 342.25 normally, you will get a z value which obviously can't have a cumulative probability of about 0.9, by observation of the curve) 
 Follow
 886
 27062016 12:34
(Original post by igotohaggerston)
Here are the real answers I am very confident they are all right
1.
a) Mean = 1.41
Variance = 1.4419
b) mean approximately equals the variance therefore Poisson is suitable
c)
i) 0.2510
ii)0.7769
d) 0.1378
e)0.1516
2.
a) 60
b)
i) 0.0133
ii)0.0159
c) 0.2369>0.01 therefore do not reject H0 his claim is not supported
3.
a) f(R)={0.25 , 5<R<9
{0 , otherwise
b) P(7<R<10)= (107)x0.25= 0.75
c)E(A)=E(piR^2)=PiE(R^2) = 151pi/3
4.
a) show by differentiating to find the pdf then differentiating again and setting it equal to zero to find maximum k cannot be zero the for solving what is in the bracket gives you that b=8.
b) I think there are a lot of ways to do this question, I solved for a first but k=1/9
5.
n=225 ignore the negative value because the standard deviation is always positive and plus you can't get a fraction of a student.
6.
a) (4,4)(6,6)(8,8)
(4,6)(4,8)(6,4)
(6,8)(8,4)(8,6)
b) m 4 5 6 7 8
P(M=m) 1/4 3/10 29/100 3/25 1/25
c)Y~B(n,1/25)
P(Y>=1)>0.9
1P(Y=0)>0.9
P(y=0)<0.1
0.96^n<0.1
taking logs( don't forget to switch the inequality)
you get n>56....
therefore the smallest value for n is 57
7.
a) E(x)=1.2
b)Var(x)=0.24
c) P(X>1.5)=0.325
d)£41.88
e) He should not as 40 <41.88
I probably haven't explained that very well thoughPost rating:2 
 Follow
 887
 27062016 12:34
Last edited by redraccoon; 27062016 at 12:35.Post rating:1 
 Follow
 888
 27062016 12:35
How did you work out k?

 Follow
 889
 27062016 12:35
what was the function for Q4 anyone?

 Follow
 890
 27062016 12:36
(Original post by Mathemagicien)
So your business here is just to drive up our grade boundaries? 
 Follow
 891
 27062016 12:36
(Original post by Armpits)
Wasn't N 342.5 or something like that? 
 Follow
 892
 27062016 12:36
(Original post by KFC_Fleshlight)
55.50.2n / (0.16n)^0.5 = 1.75
55.50.2n = 0.4*1.75n^0.5
3080.250.04n^2 = 0.49n
308025  4n^2 = 49n
4n^2+49n  308025 = 0
.....
....
.... n = 82. something or 66. something
how did people get 225?
(55.5  0.2n)^2 =(3080.25  22.2n + 0.04n^2) 
 Follow
 893
 27062016 12:36
(Original post by AMarques)
For 3b it should be (97)/(95) = 1/2 since R goes up to 9 and not 10. I think. 
 Follow
 894
 27062016 12:37
anyone else got k=1/81 ?? a = 12? or is a= 12?

 Follow
 895
 27062016 12:37
(Original post by crepole)
anyone else got k=1/81 ?? a = 12? or is a= 12? 
 Follow
 896
 27062016 12:38
Can't believe S2 is the exam that let me down, prob got a b in raw terms, UMS might **** me. What a joke.
How do you work out k? 
 Follow
 897
 27062016 12:39
My answers: (May have missed some)
1.41
1.44
0.2510
0.7769
0.1378
0.1516
60
0.0133
0.0159
0.2396
0.5
158
a=12 > k= 1/9
225
57
41.88
Had to write these in a hurry, due to only having a couple of minutes left. Perspective,I usually get A* on S2 papers, and I've never gotten below an A. 
 Follow
 898
 27062016 12:40
(Original post by igotohaggerston)
Here are the real answers I am very confident they are all right
1.
a) Mean = 1.41
Variance = 1.4419
b) mean approximately equals the variance therefore Poisson is suitable
c)
i) 0.2510
ii)0.7769
d) 0.1378
e)0.1516
2.
a) 60
b)
i) 0.0133
ii)0.0159
c) 0.2369>0.01 therefore do not reject H0 his claim is not supported
3.
a) f(R)={0.25 , 5<R<9
{0 , otherwise
b) P(7<R<10)= (107)x0.25= 0.75
c)E(A)=E(piR^2)=PiE(R^2) = 151pi/3
4.
a) show by differentiating to find the pdf then differentiating again and setting it equal to zero to find maximum k cannot be zero the for solving what is in the bracket gives you that b=8.
b) I think there are a lot of ways to do this question, I solved for a first but k=1/9
5.
n=225 ignore the negative value because the standard deviation is always positive and plus you can't get a fraction of a student.
6.
a) (4,4)(6,6)(8,8)
(4,6)(4,8)(6,4)
(6,8)(8,4)(8,6)
b) m 4 5 6 7 8
P(M=m) 1/4 3/10 29/100 3/25 1/25
c)Y~B(n,1/25)
P(Y>=1)>0.9
1P(Y=0)>0.9
P(y=0)<0.1
0.96^n<0.1
taking logs( don't forget to switch the inequality)
you get n>56....
therefore the smallest value for n is 57
7.
a) E(x)=1.2
b)Var(x)=0.24
c) P(X>1.5)=0.325
d)£41.88
e) He should not as 40 <41.88 
 Follow
 899
 27062016 12:40
(Original post by jonnypdot)
what was the awnser for var(x) do you find e(x^2)(e(x))^2
= 1.68  1.44 = 0.24. 
 Follow
 900
 27062016 12:40
Anyone else check the front the paper after reading the first question to make sure it was S2 and not S1?
Post rating:3
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Updated: August 20, 2016
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