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# Edexcel S2 - 27th June 2016 AM

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1. (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.
2. (Original post by SeanFM)

But probability of being longer than 1mm I make to be 0.602.
Hmm.

P(X > 1) = 1 - P(X =< 1)

= 1 - [ P(X = 1) + P(X = 0) ]
= 1 - [ 0.425 + 0 ]

?
3. (Original post by NotNotBatman)
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.
makes sense now! thank you so much
4. Does anyone know why p(x>1) is 1-F(1) because isnt it that p(x<1) = F(1)
5. (Original post by economicss)
Please could someone explain question 4f https://57a324a1a586c5508d2813730734...%20Edexcel.pdf thank you
6. (Original post by Aliceeee12)
Does anyone know why p(x>1) is 1-F(1) because isnt it that p(x<1) = F(1)
p(x=1)=0 since continuous so p(x>1) = 1 - p(x<1) = 1 -F(1)
7. (Original post by Mattematics)
Hmm.

P(X > 1) = 1 - P(X =< 1)

= 1 - [ P(X = 1) + P(X = 0) ]
= 1 - [ 0.425 + 0 ]

?
I make the CDF to be (23x - 3x^2 )/40 - (49/480)
8. (Original post by apzoe)
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))
Because 'more' does not mean 'more and equal to', it just means 'more'.
9. (Original post by MaxWalker1)
p(x=1)=0 since continuous so p(x>1) = 1 - p(x<1) = 1 -F(1)
Thankyou! So if it is p(x>2) is that 1-F(2) or 1-F(1)
10. (Original post by Aliceeee12)
Does anyone know why p(x>1) is 1-F(1) because isnt it that p(x<1) = F(1)

11. (Original post by NotNotBatman)

Isnt the formula p(x<1) = F(1) instead of less than equal?
12. (Original post by Aliceeee12)
Thankyou! So if it is p(x>2) is that 1-F(2) or 1-F(1)
p(x>2) = 1 - F(2) for '''continuous''' data
13. (Original post by NoahMal)
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.
thank you so much! i am so bad at this general knowledge about how distributions work, but i now understand more
14. (Original post by MaxWalker1)
p(x>2) = 1 - F(2) for '''continuous''' data
Much appreciated tyy
15. (Original post by iMacJack)
This is the probability it survives for less than t days
I'm really sorry I still don't understand
16. (Original post by apzoe)
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))
If 197 DO turn up, then 3 haven't turned up (198, 199, 200). Therefore you're looking for P(Y =< 3) which is P(Y < 4)
17. (Original post by Ayman!)
Conditional probability applied to continuous distributions. Quite common as of late in stats papers - came up in my S3
do you know any S2 papers where this comes up? / Have any questions on it, just so i can practice some questions
18. started this on friday evening, didnt know what binomial was and now just got 65 on gold paper 4. two all nighters in a row sure work miracles lool
19. (Original post by Aliceeee12)
Isnt the formula p(x<1) = F(1) instead of less than equal?
No, , it wouldn't really matter in continuous distribution, but just stick to less than or equal.
20. (Original post by SSD07)
do you know any S2 papers where this comes up? / Have any questions on it, just so i can practice some questions
On S2? I've only done S2 papers since Jan 13 and it came up twice - IAL June 15 and Jan 13

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