Edexcel S2 - 27th June 2016 AM

thats what i got
E(r) as 7.
then E(A) as 49 pi

So your business here is just to drive up our grade boundaries?

Wasn't N 342.5 or something like that?

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?

For 3b it should be (9-7)/(9-5) = 1/2 since R goes up to 9 and not 10. I think.

anyone else got k=1/81 ?? a = 12? or is a= -12?

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

what was the awnser for var(x) do you find e(x^2)-(e(x))^2

Am I the only one who got that k = 1/40?

a is -ve

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

How did you work out k?

How did you work out k?

I'm an idiot (and I have a headache), I completely forgot how to do q5 - I got Z=1.75, but I thought the variable X (number of students owning laptops) was binomial (n, 1/5), which was approximated by N(n/5, 4n/25)...

then I said P(X>55) = P(2(root n)Z/5 + n/5 > 55.5) = P(Z>5[55.5-n/5]/2(root n) ) = P(Z=1.75)

Could you be kind and tell me where have I gone [horribly, stupidly] wrong (or what I should have done)? Else I'm going to spend the whole summer trying to figure it out