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# Edexcel S2 - 27th June 2016 AM

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1. Hi, can anyone tell me how to work out k?
2. (Original post by Student403)
P much
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
3. (Original post by mupsman2312)
Am I the only one who got that k = 1/40?
Oh, wait - I see what I did wrong: after finding that a = -12, I then completely messed-up substituting 3 into F(x)... Oh, dear! I do now agree that k = 1/9.
4. (Original post by Student403)
a is -ve
How did you work out k?
5. (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
You've made one mistake for definite: for 3.)b.). The distribution is zero between 9 and 10, so you go 0.25(9-7) for 0.5 as the answer.

Other than that, I got everything the same as you, except for 7 a and b because I can't remember what I got for those and didn't have a chance to copy all my
6. (Original post by Armpits)
How did you work out k?
If I recall, what I did was set F(2) equal to zero, and work from there. I might've also used F(3)=1
7. (Original post by Armpits)
How did you work out k?
F(2) = 0
F(3) = 1

simultaneous from there

(Original post by Mathemagicien)
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
I did get that binomial and normal

Bolded part I don't recognise

i had something like (55.5 - 0.2n)/0.4sqrt(n) = 1.75
8. Guys, wasn't the E(X) question 53 pi? Because it's the E(X) of a normal distribution so b-a)/2? Where b=81 pi and a=25 pi?
9. i got k=1/18 but show b=8 and i got a =-12 help
10. judging by the poll distribution, i can guess the grade boundaries will be low.
68 - A*
62 - A
54 - B
48 - C
42 - D
36 - E
11. (Original post by Oxyfrost)
If I recall, what I did was set F(2) equal to zero, and work from there. I might've also used F(3)=1
Since the pdf was quadratic, I assumed median was 2.5 due to symmetricity, set the cdf = 0.5, and put x=2.5 in the eq. Then I attempted to solve simultaneously with integrated pdf = 1.

Do you think I'll get method marks?
Guys, wasn't the E(X) question 53 pi? Because it's the E(X) of a normal distribution so b-a)/2? Where b=81 pi and a=25 pi?
no

you're looking for pi*E(X^2) since E(A) = E(piR^2) = pi*E(R^2)

Var(R) = E(R^2) - E(R)^2

You can work out Var(R) and E(R) simply from the stuff in the formula book and work from there
13. What are people thinking for boundaries? Lower than last year? About the same?
14. (Original post by Student403)
i had something like (55.5 - 0.2n)/0.4sqrt(n) = 1.75
Wait, that looks like what I got

I think I did something retarded like multiplying the LHS by root n instead of dividing it...

Well thanks for helping, I feel like such an idiot
15. How many marks do I lose

I wrote 18k= 2 so k= 9 (loool I'm such an idiot)

And for the 8 marker I did the normal distribution up to the quadratic where y=n^0.5 . So out of the 8 how many marks will I get

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16. (Original post by Mathemagicien)
Wait, that looks like what I got

I think I did something retarded like multiplying the LHS by root n instead of dividing it...

Your method was fine then. Should just be a couple of A1s lost
Guys, wasn't the E(X) question 53 pi? Because it's the E(X) of a normal distribution so b-a)/2? Where b=81 pi and a=25 pi?
I did this but we're probably wrong
18. How did you find k for Q4? I proved b = 8.
19. (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.
That's right! Edexcel can be very sneaky, can't they? I'm just so glad that I managed to see it... I've made that same mistake many times before, and so I was really looking out for it this time!
20. (Original post by smartalan73)
Anyone else check the front the paper after reading the first question to make sure it was S2 and not S1?
I did lol

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