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June 22nd 2005 - The Official S2 Discussion Thread! watch

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    (Original post by hajira)
    but thats just the probability of x=10!
    when u r doing hypothesis test u look at either x»a or x«a
    when when u will add all the probabilities from 0-9 to that it wud go higher than 0.01!
    Oh yeah. Should've told him that :o
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    (Original post by mockel)
    Work out n in a similar way to part one: n = 75

    X~B(75,0.04)

    σ² = np(1-p) = (75)(0.04)(0.96)
    σ = 1.70
    I thought of that but i didnt know which value to carry over from q1 because the q wasnt very specific. So i carried over the 'n' value worked out in q1a

    σ² = np(1-p) = (125)(0.024)(0.976)
    σ = √2.928 = 1.71

    For the last q i got 9

    X~B(20,0.75)
    P(X<=x) < 0.01

    X~B(20. 0.25)
    P(X=>20-x) < 0.01
    P(X<=[20-x]-1) > 0.99

    from tables, P(X<=10) > 0.99
    therefore,
    (20-x)-1 = 10
    x = 9
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    (Original post by priyababy)
    were they not referring to using binomial since the outcome was either success or failure...ie defective or non defective

    i know my method was right..as in it should work..although it took longer than 2marks' time worth. im just confused whether they wanted it done the poisson way
    Ur sooo right "baby"
    lol
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    (Original post by hajira)
    well u had to calculate the new n which was n=3/0.04=75
    then using Var(x)= np(1-p)
    it came out as 75*0.04*0.96=2.88
    so s.d (x)=√2.88=1.697

    as for the median, it was F(x)=0.5
    u got 1/4(x²- x4/4) = 0.5
    which on simplification came down to
    x4 - 8x² + 8 = 0
    then usuing the quadratic formula:
    x²= [8 ± √(64 - 32)]/2
    that gave x²=4±2√2
    but x²=4+2√2 was out of range, leaving with x²=4-2√2
    and a grand final answer of x = 1.08

    I didn't know you could use that formula for x's that are to the power of four (although it makes sense now).
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    How many marks was Q1 b?
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    (Original post by Savioahang)
    How many marks was Q1 b?
    4 marks
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    (Original post by Kiril)
    I didn't know you could use that formula for x's that are to the power of four (although it makes sense now).
    Yeah, well you could think of it like this if you want...

    Y = x², and sub in. You get a quadratic in Y.
    Solve in the same way, and at the end, sub x² for Y. Same thing.
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    (Original post by Kiril)
    I didn't know you could use that formula for x's that are to the power of four (although it makes sense now).
    aww well its that they had some harder pure in it which is quite unsual
    but a way of doing that is to say y=x²
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    I got 9 for last aprt of question 7. But I used my graphic calulator to work the prob of 0.75 coz I didnt know how to change it. Do u think I will lose loads of marks for that ????
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    (Original post by mockel)
    You see, that's the logic I used, but afterwards, my mate said he put 10 because...

    (20C10)(0.75)10(0.25)10 < 0.01 (only just!)

    So, I'm just confused.
    Your mate's wrong. I put 9. It was definitely right.

    Look at n = 20, p = 0.25 on the binomial table.

    P(X<=10) = 0.9961

    1 - P(X<=10) = P(X>=11) = 0.0039

    That's the first one that's less than 0.01

    Therefore 11 or more patients would have to not recover, therefore 9 or less patients would have to recover.
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    (Original post by hajira)
    aww well its that they had some harder pure in it which is quite unsual
    but a way of doing that is to say y=x²
    Even so your simplification doesn't seem right, correct me if I'm wrong?
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    Gah, I didn't see that one question was specific about a number of decimal places.

    There goes a mark, completely unnecessarily...

    =(
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    (Original post by Kiril)
    Even so your simplification doesn't seem right, correct me if I'm wrong?
    what do u mean?
    it is rite..that working iv done in that post is the same as saying x² rather than y if u use a substituion.
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    Are you all trying to say that you didn't use a Normal approximation for the lat question?????????????????///
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    (Original post by Kiril)
    Are you all trying to say that you didn't use a Normal approximation for the lat question?????????????????///
    nope, if that was required, the question would have specified. It's more accurate to use the binomial one given
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    (Original post by mockel)
    Work out n in a similar way to part one: n = 75

    X~B(75,0.04)

    σ² = np(1-p) = (75)(0.04)(0.96)
    σ = 1.70

    Ur right
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    (Original post by Featherflare)
    nope, if that was required, the question would have specified. It's more accurate to use the binomial one given
    What if I have used the normal approximation, would I have lost marks?
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    For the last question,
    did you put for:
    (a) X~B(20, 0.75)
    (b) P(x = 6) did you use the binomial fomula?

    (c)... firstly what did people put as their H0 and H1
    did people reject h0 or not reject h0?
    and what did people get as the probabilty?
    i think i did the working out wrong, but i stated both H1 and H0 and said you dont reject H0. if you DONT reject H0, and this is correct but your markings wrong, given everything else in the question, would i get any marks?
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    You never had to work out n for part 1b.

    E(X) = np = 3 and p = 0.04.

    Var(X) = np(1-p) = 3 * 0.96 = 2.88

    SD = √2.88 = 1.70
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    (Original post by Kiril)
    What if I have used the normal approximation, would I have lost marks?
    you should only use normal where p is close to 0.5, so no
 
 
 
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