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    (Original post by BecauseFP)
    The values of a for the question where the variance of the probability density function was given.
    i) a=6
    ii) a=1.5
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    For Q7 I think 41 may be the answer for 7(ii). I've put the solution below I case anyone wants to double-check/verify?

    1 out of every 3 trains are late
    Therefore 1/3 out of every 1 train is late, so p = 1/3 and n=150 (from the question)

    Let X distribute binomially
     \therefore X \rightarrow B\Big(150, \dfrac{1}{3}\Big)

     \Rightarrow \mu = 50, \sigma^2 = \dfrac{100}{3}

     \Rightarrow X~N(50,100/3) approx.

     \Rightarrow \dfrac{X - 0.5 - 50}{\sqrt{100/3}} < -1.645

     X = 41
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    (Original post by BecauseFP)
    The values of a for the question where the variance of the probability density function was given.
    I got 1.5 :/
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    (Original post by lllllllllll)
    for q7 i think 41 may be the answer for 7(ii). I've put the solution below i case anyone wants to double-check/verify?

    1 out of every 3 trains are late
    therefore 1/3 out of every 1 train is late, so p = 1/3 and n=150 (from the question)

    let x distribute binomially
     \therefore x \rightarrow b\big(150, \dfrac{1}{3}\big)

     \therefore \mu = 50, \sigma^2 = \dfrac{100}{3}

     \rightarrow \dfrac{x - 0.5 - 50}{\sqrt{100/3}} < -1.645

     x = 41
    late trains? Did i ****ing miss a question??
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    (Original post by lllllllllll)
    For Q7 I think 41 may be the answer for 7(ii). I've put the solution below I case anyone wants to double-check/verify?

    1 out of every 3 trains are late
    Therefore 1/3 out of every 1 train is late, so p = 1/3 and n=150 (from the question)

    Let X distribute binomially
     \therefore X \rightarrow B\Big(150, \dfrac{1}{3}\Big)

     \therefore \mu = 50, \sigma^2 = \dfrac{100}{3}

     \rightarrow \dfrac{X - 0.5 - 50}{\sqrt{100/3}} < -1.645

     X = 41
    For 7(ii) i got 37, I don't think it was about late trains though?
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    (Original post by jacobe)
    I got 1.5 :/
    the correct answer? well done!
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    (Original post by jacobe)
    For 7(ii) i got 37, I don't think it was about late trains though?
    Idk, I forgot as well so I thought it was "late trains" :P
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    (Original post by lllllllllll)
    For Q7 I think 41 may be the answer for 7(ii). I've put the solution below I case anyone wants to double-check/verify?

    1 out of every 3 trains are late
    Therefore 1/3 out of every 1 train is late, so p = 1/3 and n=150 (from the question)

    Let X distribute binomially
     \therefore X \rightarrow B\Big(150, \dfrac{1}{3}\Big)

     \therefore \mu = 50, \sigma^2 = \dfrac{100}{3}

     X \rightarrow N(50,100/3) approx.

     \rightarrow \dfrac{X - 0.5 - 50}{\sqrt{100/3}} < -1.645

     X = 41
    Hmm, with the general consensus being 37/38, I don't see it being 41. Are you sure those are the correct values for n and p?
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    (Original post by tokopoko)
    the correct answer? well done!


    The only question I think I got wrong was the relationship between X and x. I had no idea...
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    (Original post by rich1334)
    Hmm, with the general consensus being 37/38, I don't see it being 41. Are you sure those are the correct values for n and p?
    I used a Normal approx to the Poisson? Anyone else do this?
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    Anyone know the relationship between X and x in the probability density function?
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    (Original post by jacobe)


    The only question I think I got wrong was the relationship between X and x. I had no idea...
    x represents the values that X can take...right?
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    (Original post by tokopoko)
    look at my working again. its when p=0.3.
    Not sure I'm putting my point across in the right way but it doesnt matter now
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    (Original post by rich1334)
    Hmm, with the general consensus being 37/38, I don't see it being 41. Are you sure those are the correct values for n and p?
    Yes, I used the poison approximation and got 37. I can't see a flaw with either methods so I'm kind of scratching my head as to which is correct. Part of me says that they're both not incorrect but since they're approximations then it explains the difference between the two answers
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    (Original post by Molly_xox)
    x represents the values that X can take...right?
    (Original post by jacobe)
    Anyone know the relationship between X and x in the probability density function?
    That's what i got.
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    (Original post by jacobe)


    The only question I think I got wrong was the relationship between X and x. I had no idea...
    1 marker? i wasnt sure either.

    however i recall it is similar to June 11 4iii)

    http://www.ocr.org.uk/Images/62244-q...atistics-2.pdf

    http://www.ocr.org.uk/Images/62144-m...ics-2-june.pdf

    basically x is the value taken by X
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    (Original post by lllllllllll)
    Yes, I used the poison approximation and got 37. I can't see a flaw with either methods so I'm kind of scratching my head as to which is correct. Part of me says that they're both not incorrect but since they're approximations then it explains the difference between the two answers
    I got 37 too, so I think for the sake of our own sanity it's best that we assume our answer is correct
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    (Original post by rich1334)
    I got 37 too, so I think for the sake of our own sanity it's best that we assume our answer is correct
    Same here! Got n <= 37.665 (smth) so hope n=37 is right!
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    To clarify, for the 11 marker, you estimated the population variance from the data given to get the population variance was 18, then you divided 18 by 120 to estimate the sample variance?
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    Here are the answers I got:
    1. Mean=146 ans sd=2
    2i. Something along the lines that the occurrence od dead rabbits is unpredictable and don't occur at a certain pattern
    2ii. Dead rabbits occur independently. In other words, the appearance of one dead rabbit doesn't influence the death of another one
    2iii 0.222 (3sf)
    3i. 6
    3ii 3/2
    3iii x represent the values that X can take
    4. Missed question 4 out
    5. I think accepted h0 I cant remember
    6i. Accepted h1 (I think its the one about the dentist or something
    6ii. It is necessary to assume normal distribution since the variable t's parent distribution wasn't mentioned. CLT can be used since n is very large
    7i. 0.125
    7ii. 37
    8i 3.15%
    8ii. 0.961 judging by other answers I think I got it wrong since I calculated p=3. Bright side is I used binomial distribution
 
 
 
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