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# Edexcel S2 - 27th June 2016 AM watch

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1. Please could someone explain question 4f https://57a324a1a586c5508d2813730734...%20Edexcel.pdf thank you
2. (Original post by veldt127)
Edexcel D2 is on Wednesday :'(
Oh snap I feel sorry for you I thought I was unlucky with S2!
3. (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 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.
What have I done wrong for part e?

Probability of the tear being longer than 1mm for one customer is o.575

Probability of tear being longer than 1mm for 3 or more customers is binomial with n = 5, p = 0.575
Then P(X>=3) = P(x=3) + P(x=4) + P(x=5)?

I got 0.64
4. (Original post by AlphaArgonian)
Oh snap I feel sorry for you I thought I was unlucky with S2!
I'm not too stressed thanks to doing well in AS papers, but I also have Physics on Tuesday which is going to be hell
5. Hey. Can I have some help with this question? (January 2002)

The question says "more than 196 passengers turn up" so its >196. Then, wouldn't it mean that (less than or equal to) 4 do not turn up?

The mark scheme says simply < 4, so I am confused.
6. (Original post by SeanFM)
Afraid not - k is a whole number - actually, let me double check that. I think I've messed it up

Yes - k should be a whole number. Phew.
k = 2, but surely for your last question, it's 1-0.69 not 0.69?

EDIT: damn nah you're right lol nvm
7. (Original post by SeanFM)

Correct. The binomial distribution (a discrete one) takes no values between 6 and 6.9, so those two things are equivalent.
So it's not <=7, as 6.9 is closer to 7 than to 6?
8. (Original post by veldt127)
I'm not too stressed thanks to doing well in AS papers, but I also have Physics on Tuesday which is going to be hell
Fair enough, good luck!
9. (Original post by Aliceeee12)
Hello Everyone,

Hope revision is well

Please could somebody help me in the June 2014 R paper question 5c
https://57a324a1a586c5508d2813730734...%20Edexcel.pdf

How come they poisson to normal approximate rather than binomial to normal approximate because isnt binomial more accurate?
You don't know what the probability of a defect occuring is so you can't use binomial. You're told the rate but the mean is 25 so you need to approximate to normal.
10. (Original post by ktbunny)
thank you so much!
i just realised i need someone to explain the answer to another question, could you help me again?

Attachment 556965
2b)

the answer is 1 Because it is continuous and much like a normal distribution the probability of a single number is 0 therefore the probability that it does equal that number is 0 so then 1-0 = 1.
11. (Original post by apzoe)
Hey. Can I have some help with this question? (January 2002)

The question says "more than 196 passengers turn up" so its >196. Then, wouldn't it mean that (less than or equal to) 4 do not turn up?

The mark scheme says simply < 4, so I am confused.
Because you're binomial for part (a) should be number who do NOT show up, so it is less than 4
12. John picks 10 real numbers randomly from 12 to 17. Find the probability that at least 5 of thesenumbers are greater than 15.5.
13. Do we need to know the proofs of E(X) and Var(X) for binomial or just the proofs of these for uniform distribution?
14. 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

Attachment 556929

MS: Attachment 556931

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15. (Original post by apzoe)
So it's not <=7, as 6.9 is closer to 7 than to 6?
No, you don't round. round or anything like that here. In a discrete distribution, you can only have x =0, 1, 2, 3, 4, 5, 6, 7,... etc.

So if I said X=< 6.9, then it is true that X =< 6. It is not true that X =< 7, because if X were 7 then X would not be =< 6.9. So you round down when the inequalities are that way round.

Converse is when X >= 7.1 - you can infer that X >= 8.
16. (Original post by Mattematics)
You don't know what the probability of a defect occuring is so you can't use binomial. You're told the rate but the mean is 25 so you need to approximate to normal.
17. Hey there, does anyone have a good resources for learning the thousands of definitions that there are in S2?
thanks
18. (Original post by Nikhilm)
Because you're binomial for part (a) should be number who do NOT show up, so it is less than 4
I mean that the mark scheme says less than 4, while I was thinking it should be less than or equal to 4.

So that P(X>196) = P(Y<=4) (not = P(Y<4))
19. (Original post by Mattematics)
What have I done wrong for part e?

Probability of the tear being longer than 1mm for one customer is o.575

Probability of tear being longer than 1mm for 3 or more customers is binomial with n = 5, p = 0.575
Then P(X>=3) = P(x=3) + P(x=4) + P(x=5)?

I got 0.64

But probability of being longer than 1mm I make to be 0.602.
20. (Original post by SSD07)
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??
The perimeter is 20cm, so the sum of the two sides is half the perimeter, which is 10cm. so x+y =10, consider the end points x=1; when x=1 x+y=10, so y=9.

When X=7, y=3, so it adds up to 10. so 1<x<7 and 3<y<9

Now either x or y can be the longer side, depending on their length. You're looking for the longer side to be more than 6cm, so that's when x>6 and y>6. When is y>6 when x<... (remember x+y=10). Then add the probabilities.

--------------------------
Or think about it like this 1<X<7, so 3<Y<9,

X~U[1,7] and Y~U[3,9], then work out the probabilities of both being greater than 6.

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