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    ah ill definitely loose a few marks then... hopefully only a few
    2 for first part of 9 - and say 3 for the last bit of 7 with another 2 for error carried forward?
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    7 partt 2 varience = 67 - mewsquared =67-64= 3

    also iv its the probability of P(x>or equal to 2)= 1-P(x< or equal to 1)

    = 1-[e^-3m multiplied by (1+3m)]
    and then u plug the 1.29 and 1.3 to get 0.898 and 0.901 so 0.9 must lie between the two...
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    yeh... i did that - sorry if i didnt explain it very well... by no means those answers are right - just what i put
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    It's a good place to start, though.
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    Also the question about comment on the claim... I think it would be valid to say u do not have enough evidence to reject the claim since u rejected hnaught but that doesnt make the claim correct either... there was a similar qu. in june 06 or jan/june 07 i think
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    What was the reason that the last part of the final question was an approximation?
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    (Original post by Gallant92)
    I accepted H0 for 4 time wasn't underestimated

    I rejected H0 for 5, mean number of customers had increased.

    q 7 I put central limit theorem
    err can't remember all the figures but none that i'm sure are different except 5.

    7 ii var was 3 because you have to minus mean^2 (64)

    ?
    for question 4 time was underestimated... n i think u have to justify central limit theorem for qu7 by saying n was more than 30 and it was not normal distribution...
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    oh for the comment i wrote there wasnt really enough evidence - 1 mark and such an ambiguous question
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    Hopefully Mr. M may pop along and sort out any confusion? 0: ) I may pm him, It doesn't half seem a bit cheeky though.
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    eeek how'd everyone do q 8 iii??
    i did 1/2[ P(type 2 | p = 0.4) + P(type 2 | p = 0.7)] and got something like 0.633.... i did get 0.3669 initially but then crossed it out in the last minute
    nothing makes any sense ahhh.....
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    6) last part i got 19.1 as my standard deviation...what did every1 else get?
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    (Original post by korobeiniki)
    Hopefully Mr. M may pop along and sort out any confusion? 0: ) I may pm him, It doesn't half seem a bit cheeky though.
    I'm pretty sure Mr M hates stats, so don't get your hopes up.
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    (Original post by PhyMath)
    I'm pretty sure Mr M hates stats, so don't get your hopes up.
    Oh fair enough, have to wait 'til monday and get a teacher to go through then.

    (Original post by musicluva)
    eeek how'd everyone do q 8 iii??
    i did 1/2[ P(type 2 | p = 0.4) + P(type 2 | p = 0.7)] and got something like 0.633.... i did get 0.3669 initially but then crossed it out in the last minute
    nothing makes any sense ahhh.....
    I got 0.183 for this as P (Type II | p=0.7) = 0, i think, as 0.7 is what is being tested, and P(Type II | p=0.4) was found to be 0.3669 in 8ii
    so 0.5(0 + 0.3669) = 0.183 (3sf)
    anyone care to step in on that, I had a feeling it may be wrong when I first wrote it.

    (Original post by examhunter)
    6) last part i got 19.1 as my standard deviation...what did every1 else get?
    Yeah I got that. Was a bit worried as my probability value for the standardised "z" value came to 0.6999 instead of 0.7
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    (Original post by PhyMath)
    I'm pretty sure Mr M hates stats, so don't get your hopes up.
    :p:

    All true.
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    Hi

    I have taught this to my further maths lot this term so i did a solution sheet for them. feel free to have a look.

    Mr Ward
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    (Original post by ward309)
    Hi

    I have taught this to my further maths lot this term so i did a solution sheet for them. feel free to have a look.

    Mr Ward
    You've made a small error on the unbiased estimator for the variance in the first question, you seem to have used the wrong value for sigma x^2, using 1154.88 rather than 1154.58.

    Thanks a lot though, really helpful, the rest of the answers tally up with everything I got, so if my written answers are substantial I might be on for a decent score. :woo:
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    (Original post by QED)
    You've made a small error on the unbiased estimator for the variance in the first question, you seem to have used the wrong value for sigma x^2, using 1154.88 rather than 1154.58.

    Thanks a lot though, really helpful, the rest of the answers tally up with everything I got, so if my written answers are substantial I might be on for a decent score. :woo:
    Agreed, apart from the little mistake with the variance I got almost identical answers. I made a couple of ridiculous mistakes like not doing 1 minus the probability though so I threw away a couple of marks.
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    I haven't looked at these yet but I made the ridiculous mistake in the exam of writing P(X>7) = P(X>=6). Despite my meticulous checking that somehow escaped me. B******. Only 2 marks I guess.
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    (Original post by ward309)
    Hi

    I have taught this to my further maths lot this term so i did a solution sheet for them. feel free to have a look.

    Mr Ward
    Very useful - thanks
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    the formula for variance of a crv is integral of x^2f(x)dx - mu^2 which is 67-64=3
 
 
 
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