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# Edexcel S2 - 27th June 2016 AM Watch

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1. (Original post by Waffles15)
thats what i got
E(r) as 7.
then E(A) as 49 pi
You don't workout E(x^2) by squaring E(X)
you either var(x)+E(X)^2 or integrate
2. (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)= (10-7)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
1-P(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
For 3b it should be (9-7)/(9-5) = 1/2 since R goes up to 9 and not 10. I think.
3. (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
Why 225?
4. (Original post by Armpits)
Wasn't N 342.5 or something like that?
That was the other solution to the quadratic equation and which is what I wrote down.
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,
5. (Original post by Armpits)
Why 225?
well I said 2 things

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)
6. (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)= (10-7)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
1-P(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
3b. It's 1/2 isn't it ? As the probability of R is 1/4 between 5-9. So when you find p(7<R<10) it's actually just the same as p(7<R<9) which is 1/2.
I probably haven't explained that very well though
7. (Original post by Waffles15)
thats what i got
E(r) as 7.
then E(A) as 49 pi
I think what you had to do (or what I did) was write a as pi r^2 so you were looking for e(a) which is e(pi r^2) which is pi e(r^2) and then find e(r^2) by doing var(x) + e(x)^2 with e(x) being 7 and var being 4/3 I think which gives you 151/3 and then × by pi for 151/3 pi.
8. How did you work out k?
9. what was the function for Q4 anyone?
10. (Original post by Mathemagicien)
P much
11. (Original post by Armpits)
Wasn't N 342.5 or something like that?
That will be -1.75 so 225 is the correct one
12. (Original post by KFC_Fleshlight)
55.5-0.2n / (0.16n)^0.5 = 1.75
55.5-0.2n = 0.4*1.75n^0.5
3080.25-0.04n^2 = 0.49n
308025 - 4n^2 = 49n
4n^2+49n - 308025 = 0
.....
....
.... n = 82. something or 66. something
how did people get 225?
You made a mistake when squaring:

(55.5 - 0.2n)^2 =(3080.25 - 22.2n + 0.04n^2)
13. (Original post by AMarques)
For 3b it should be (9-7)/(9-5) = 1/2 since R goes up to 9 and not 10. I think.
14. anyone else got k=1/81 ?? a = 12? or is a= -12?
15. (Original post by crepole)
anyone else got k=1/81 ?? a = 12? or is a= -12?
a is -ve
16. 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?
17. 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.
18. (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)= (10-7)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
1-P(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
Agree with all except 3)b), which as others have pointed out, should be 0.5 I believe, since when x>9, f(x) = 0
19. (Original post by jonnypdot)
what was the awnser for var(x) do you find e(x^2)-(e(x))^2
Yes - it was 1.68 - (1.2)^2
= 1.68 - 1.44 = 0.24.
20. Anyone else check the front the paper after reading the first question to make sure it was S2 and not S1?

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