Hey there! Sign in to join this conversationNew here? Join for free

Edexcel S2 - 27th June 2016 AM Watch

  • View Poll Results: How did you find the Edexcel S2 exam?
    Very hard
    145
    14.71%
    Hard
    331
    33.57%
    Normal
    293
    29.72%
    Easy
    152
    15.42%
    Very easy
    65
    6.59%

    Offline

    2
    ReputationRep:
    These questions are so ******** Why are the grade boundaries so high for this
    Offline

    2
    ReputationRep:
    (Original post by fpmaniac)
    How do I do question 1b and c

    https://57a324a1a586c5508d2813730734...%20Edexcel.pdf
    48 hours = 2 days

    Therefore X~Po(1.6) and since the next car is caught within the next 48 hours then we know that there is at least 1 car that speeds within the next 48 hours

    therefore P(X=>1) = 1 - P(X=0)
    Offline

    2
    ReputationRep:
    gnight everybody I'm going to bed

    good luck
    Offline

    2
    ReputationRep:
    (Original post by apzoe)
    It's just that there have been several exercises already in past papers with questions like "given that.. find that..".

    I know you have to apply the formula P(B|A)=P(A intersection B)/P(A). It's just about using it correctly, right?
    In the question above, just work out prob that between 5 and 8 mins she isn't served.

    You don't always have to use the conditional probabilities formula (in fact you never have to, but in many cases it wouldn't make it easier).
    • Very Important Poster
    • Thread Starter
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by apzoe)
    It's just that there have been several exercises already in past papers with questions like "given that.. find that..".

    I know you have to apply the formula P(B|A)=P(A intersection B)/P(A). It's just about using it correctly, right?
    Yes, that is correct. Sometimes you have to 'interpret' the intersect though. Like in this case:

     P(t> 15 | t>10) = \frac{P(t>15 \cap t>10)}{P(t>10)} = \frac{P(t>15)}{P(t>10)} as  (t>15 \cap t>10) = t>15 as t> 15 implies that t>10, whereas t>10 doesn't imply that t>15
    Online

    5
    ReputationRep:
    Is cumulative distribution function the same as cumulative probability density function?
    Offline

    1
    ReputationRep:
    https://www.youtube.com/watch?v=BJPcDV0rx88

    June 2013, Question 3b.

    Can anyone explain to me why P(X>n)? I thought it would be X<n..

    Thanks
    Offline

    2
    ReputationRep:
    (Original post by fpmaniac)
    How do I do question 1b and c

    https://57a324a1a586c5508d2813730734...%20Edexcel.pdf
    For Part c:

    We're given that 4 were caught

    1 on the first day and 2 on the second so calculate these using X~Po(0.8) as each occurrence happened within a 1 day period.

    Since we're given that 4 happened we can use P(AnB)/P(B)
    which is simply the probabilities of X=1 on the first day multiplied by the probability X=3 on the second day all over the prob of 4 accidents within the 2 day period, where X~(1.6)

    Hope this makes sense
    Offline

    3
    ReputationRep:
    (Original post by M3WIZARD)
    Any thoughts on tomorrow's paper? - Hopefully it's going to be a 6 Question paper ending on a nice easy mean or median sampling distribution
    boundaries would be 74/75 for an A then
    Offline

    2
    ReputationRep:
    (Original post by SeanFM)
    Yes, that is correct. Sometimes you have to 'interpret' the intersect though. Like in this case:

     P(t&gt; 15 | t&gt;10) = \frac{P(t&gt;15 \cap t&gt;10)}{P(t&gt;10)} = \frac{P(t&gt;15)}{P(t&gt;10)} as  (t&gt;15 \cap t&gt;10) = t&gt;15 as t> 15 implies that t>10, whereas t>10 doesn't imply that t>15
    Yes I see now. Thank you for the example btw. I solved it and got the same result as you did.
    Sadly there are only 2 examples of such questions in regular past papers I could find to practice on, but now it makes more sense.

    Thank you ^^
    • Very Important Poster
    • Thread Starter
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by Moral010)
    Is cumulative distribution function the same as cumulative probability density function?
    It sounds like it might be. I've never heard of it though, and a quick google doesn't seem to turn up many relevant results for 'cumulative probability density function'. But if you've been taught S2 in a class then it's probably the same thing.
    Offline

    2
    ReputationRep:
    (Original post by Moral010)
    Is cumulative distribution function the same as cumulative probability density function?
    Never heard of "cumulative probability density function".

    At A2 we only studied the probability density function and the cumulative distribution function. Have you been taught something more or are you combining them and confusing yourself?
    Offline

    20
    ReputationRep:
    You know when you have lots of different functions on a cumulative distribution function and they ask you to find median or quartile and you can only pick one function to be equal to the median or quartiles.
    How do you know which one to pick!?
    Thank you!


    Posted from TSR Mobile
    • Very Important Poster
    • Thread Starter
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by Maths help pls)
    https://www.youtube.com/watch?v=BJPcDV0rx88

    June 2013, Question 3b.

    Can anyone explain to me why P(X>n)? I thought it would be X<n..

    Thanks
    You could go down the (x=<n) route, but you would have to use 0.95 (1-0.05).

    Basically, the shop owner is trying to make it unlikely that he will run out of what he's selling (i.e P(X=<n) =<0.95) 9r P(X>n) (selling more than n copies) has probability less than 0.05
    • Very Important Poster
    • Thread Starter
    Offline

    21
    ReputationRep:
    Very Important Poster
    (Original post by Bloom77)
    You know when you have lots of different functions on a cumulative distribution function and they ask you to find median or quartile and you can only pick one function to be equal to the median or quartiles.
    How do you know which one to pick!?
    Thank you!


    Posted from TSR Mobile
    Trial and error. Only one of the 'sections' will have the right value.

    I hope this post explains it.

    Say the CDF is defined with 3 functions -
     f(x) a \leq x &lt; b

g(x) b \leq x &lt; c

h(x) c \leq x &lt; d

    Where f, g and h are different functions and a, b and c are used for different ranges.

    Say you're trying to find where F(x) = 0.5, you know it's going to be either from f, g, or h. Firstly, test f(b) (the upper limit of the domain of f(x)). If it is less than 0.5, then you know that it's not going to be in that function. If f(b) is greater than 0.5, then x is going to be somewhere between a and b.

    Say it isn't between a and b, the next one is to test g(c) and apply the same logic.

    Then you find the right function. Similar logic can be applied to any value of F(x).
    Online

    5
    ReputationRep:
    (Original post by apzoe)
    Never heard of "cumulative probability density function".

    At A2 we only studied the probability density function and the cumulative distribution function. Have you been taught something more or are you combining them and confusing yourself?
    Probably confusing myself, they were just named differently in this revision website... Thanks
    Offline

    20
    ReputationRep:
    (Original post by SeanFM)
    Trial and error. Only one of the 'sections' will have the right value.

    I hope this post explains it.

    Say the CDF is defined with 3 functions -
     f(x) a \leq x &lt; b

g(x) b \leq x &lt; c

h(x) c \leq x &lt; d

    Where f, g and h are different functions and a, b and c are used for different ranges.

    Say you're trying to find where F(x) = 0.5, you know it's going to be either from f, g, or h. Firstly, test f(b) (the upper limit of the domain of f(x)). If it is less than 0.5, then you know that it's not going to be in that function. If f(b) is greater than 0.5, then x is going to be somewhere between a and b.

    Say it isn't between a and b, the next one is to test g(c) and apply the same logic.

    Then you find the right function. Similar logic can be applied to any value of F(x).
    Thank you! That makes sense


    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    Name:  Screenshot_2016-06-26-23-41-30.png
Views: 98
Size:  76.3 KBguys, can you help me with q7b June 2013 R?
    Offline

    14
    ReputationRep:
    (Original post by Supermanxxxxxx)
    Time to do an all nightier then


    Posted from TSR Mobile
    You need to be able to think in the morning, so go to bed!
    Offline

    15
    ReputationRep:
    (Original post by kosvengali)
    Name:  Screenshot_2016-06-26-23-41-30.png
Views: 98
Size:  76.3 KBguys, can you help me with q7b June 2013 R?
    We know X represents the correct answer, so by logic, the amount of wrong answers is 20-X

    S represents the final score, which will obviously be the amount he gets - amount he loses

    So, each correct is 4 points, so it's 4X - 1(20-X)

    Which = 5X-20

    Because 1 point's lost for a wrong answer
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brussels sprouts
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.