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Edexcel S2 - 27th June 2016 AM

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Original post by SeanFM
a. b = 23/6

b. k = 2

c E(X) = 1.39. E(X^2) = 2.70. Var(X) = 0.77. Deviation = 0.88 (answers within that range, depending on rounding etc)

d. IQR question removed - takes too much time.

e. 0.69.






Well done :h: the last bit should be 2 s.f but doesn't matter really. Just something to be wary of in the exam if they specify how many s.f you need.

hey man do you mind explaining the first part for me
Original post by taysc
Screen Shot 2016-06-26 at 16.50.07.png

Sean explained the first part to before. I thought I understood it but still managed to get the last part of the question wrong.

I did 1 - (225)/(t+15)^2 = 0.1

Can someone explain to me why its wrong. If you can make the explanation really simple, even child like. It would be really helpful


This is the probability it survives for less than t days
Original post by SeanFM
Not necessarily - I think it's just a method mark (hence the M) for having the 'right method'. The cao means 'correct answer only' so that's where the two methods will be differentiated - the one where it has been applied to the right one will get you that A1 mark.


Ah alright I see, Ill just make sure to check the upper limits of each function and just work from there. Just wanted to clarify. Thank you! :smile:
This question
6aii

I don't understand why we are looking at the upper critical value and not the lower critical value as it says minimum

ImageUploadedByStudent Room1466955202.704696.jpg

MS: ImageUploadedByStudent Room1466955220.947715.jpg




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June 15 (IAL) Q2b?
Reply 405
Original post by NotNotBatman
Consider both sides of the rectangle:
Look at the method used for question 11:

http://www.madasmaths.com/archive/maths_booklets/statistics/rectangular_distribution.pdf


i still really don't get it
Original post by Ben4
Do we need to know any proofs? I heard we need to know how to prove the variance of a uniform distribution... is this correct?


Yeah you can be asked to work out the variance through integration (for uniform distribution) , which is basically the proof
I have a question about the S2 exam tomorrow, so I was doing past papers and some questions ask you to work E(X^2) for a CDF yet, solutions show the answer being gathered from the VAR(X) formula in the S2 formula booklet. Do you use the Var formula in the booklet for E(X^2) or am I nuts??
Original post by ktbunny
I hope it is not too much trouble, but please could someone explain how they think through some questions to help my understanding!

s2jan15ialq1.png
1b)

Attachment not found

2c)

and

Attachment not found


7d)

Thank you very much in advance!




Hope this helps mate!

Appologies for hand writing.

https://gyazo.com/af6ccadea92e33a2fa1cc2830439e94d
https://gyazo.com/082b793a25a8162518a2bf5585ed8183
Reply 409
Original post by NotNotBatman
Consider both sides of the rectangle:
Look at the method used for question 11:

http://www.madasmaths.com/archive/maths_booklets/statistics/rectangular_distribution.pdf


is it because you don't know which side is the shorter side??? so it could be either which is why you add the probabilities for both??
Original post by кяя
Is this the last A level exam in general? From the people I know S2 is last. 27th June is damn late... Can't believe how quick college flew by. Seems like yesterday I was going to open days.


Along with FP3 it is one of the last A level maths exams, however I believe there is a physics exam the day after.
Original post by XxNattyboyxX
hey man do you mind explaining the first part for me



Original post by SeanFM
A distribution X is used to model the length, in mm, of the first tear of a uPhone charger, and the PDF is specified below. However, Wapple have been careless and have lost some key details of the PDF.

kx(1x)kx(1-x) if 0x<0.5 0 \leq x < 0.5

23403x20\frac{23}{40} - \frac{3x}{20} if 0.5xb 0.5 \leq x \leq b

0 otherwise.

a) Given that b is the minimum value greater than 0 such that f(b) = 0, find the value of b.




So the b is the upper limit of the second part of the pdf. (f(x)).

You are told that, after 0 (since f(0) = 0) b is the first value such that f(b) = 0.

You automatically know that b is greater than 0.5, otherwise the pdf would not make sense, so you have to look at the second part and realise that the PDF in that part is a decreasing function, so there will come a point where f(x) = 0, and that x will be b. After f(x) is 0, you go past b, it doesn't become negative as you'd expect - it actually becomes 0, because after b, any x value falls into 'otherwise' so the graph flatlines. So you're looking for b, the y intercept.

So that happens when
23403b20=0b=236\frac{23}{40} - \frac{3b}{20} = 0 \Rightarrow b = \frac{23}{6}

Kind of hard to see to be honest. But don't worry, I don't think you'd get asked something like that, particularly in S2 - it's more like C3 stuff.
Reply 412
This year I did C3, C4, M1, S1, M1 and S2 tomorrow - is it fine to get my exams officer to place S2 in A2 Maths and keep AS FM all AS modules?
Original post by Yasmin3101
I have a question about the S2 exam tomorrow, so I was doing past papers and some questions ask you to work E(X^2) for a CDF yet, solutions show the answer being gathered from the VAR(X) formula in the S2 formula booklet. Do you use the Var formula in the booklet for E(X^2) or am I nuts??


Just like Var(X)=E(X2)(E(X))2 Var(X) = E(X^2) - (E(X))^2 , you can rearrange that to give E(X2)=Var(X)+(E(X))2 E(X^2) = Var(X) + (E(X))^2 . which is useful especially if you know a formula for E(X) and Var(X) already. (eg for uniform distribution).
Original post by Yasmin3101
I have a question about the S2 exam tomorrow, so I was doing past papers and some questions ask you to work E(X^2) for a CDF yet, solutions show the answer being gathered from the VAR(X) formula in the S2 formula booklet. Do you use the Var formula in the booklet for E(X^2) or am I nuts??



E(X^2) Will be multiply all values in the PDF By X^2 before integration. Because the Var(X) Formula is multiply it by X^2 - x̄^2 think about how the equation works. Var(x) = E(X^2)-E(X)^2 Therefore to find E(X^2) You must only multiple all values by X^2 Then integrate.

If asked form a CDF you must initally differentiate, then multiply by x^2 then go forward to integrate between b and a.
Original post by SeanFM
So the b is the upper limit of the second part of the pdf. (f(x)).

You are told that, after 0 (since f(0) = 0) b is the first value such that f(b) = 0.

You automatically know that b is greater than 0.5, otherwise the pdf would not make sense, so you have to look at the second part and realise that the PDF in that part is a decreasing function, so there will come a point where f(x) = 0, and that x will be b. After f(x) is 0, you go past b, it doesn't become negative as you'd expect - it actually becomes 0, because after b, any x value falls into 'otherwise' so the graph flatlines. So you're looking for b, the y intercept.

So that happens when
23403b20=0b=236\frac{23}{40} - \frac{3b}{20} = 0 \Rightarrow b = \frac{23}{6}

Kind of hard to see to be honest. But don't worry, I don't think you'd get asked something like that, particularly in S2 - it's more like C3 stuff.

ah thanks man, makes sense now :smile:
]This question
6aii

I don't understand why we are looking at the upper critical value and not the lower critical value as it says minimum

ImageUploadedByStudent Room1466955202.704696.jpg

MS: ImageUploadedByStudent Room1466955220.947715.jpg






Posted from TSR Mobile
Original post by AlphaArgonian
Along with FP3 it is one of the last A level maths exams, however I believe there is a physics exam the day after.


Edexcel D2 is on Wednesday :'(
Hello Everyone,

Hope revision is well

Please could somebody help me in the June 2014 R paper question 5c
https://57a324a1a586c5508d2813730734691051ac35fd.googledrive.com/host/0B1ZiqBksUHNYZ3M4QzJ0N19IeHc/June%202014%20(R)%20QP%20-%20S2%20Edexcel.pdf

How come they poisson to normal approximate rather than binomial to normal approximate because isnt binomial more accurate?
Reply 419


thank you so much!
i just realised i need someone to explain the answer to another question, could you help me again? :h:

s2jan15ialq2.png
2b)

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