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    (Original post by Clarity Incognito)
    Essentially, you need to integrate equation one between Xo and its bottom limit. This gives you the cumulative distribution from Xo to the bottom limit.

    For this second equation, you need to integrate equation 2 between Xo and its bottom limit. This gives you the cumulative distribution from Xo to the bottom limit BUT because there is also the extra area of equation 1, you integrate equation 1 between the entirety of its limits. This therefore adds on. In this case, the area from equation 1 is 2/3. Add this to (x/3) - 1 and you get (x/3) - 1/3
    aw **** Hope it will be just -1
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    (Original post by Moa)
    aw **** Hope it will be just -1
    Don't worry about it too much, seems like you hammered the rest. You've even put the paper up yourself with the question, it's not -1, I even went and did that part again just to make sure I didn't end up lying to you!
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    (Original post by Clarity Incognito)
    Don't worry about it too much, seems like you hammered the rest. You've even put the paper up yourself with the question, it's not -1, I even went and did that part again just to make sure I didn't end up lying to you!
    lets hope for the best
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    How many marks do you guys think I'll receive for summing up the sampling distributions instead of listing them out individually i.e. 1/64 + 3(3/4)^3*(1/4) + ..... = 1

    Also, what does a -6 look like to you? Thanks!
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    my mistakes were quite a few.
    first the probability on the robot question was wrong by my reason of breakdowns being independent was correct. so im guessing there would be a -1 there.
    for the critical region question, i took the value of x for which P(X<=x) was greater than .9950, rather than the value which was the closest. how many marks will i lose for this(how much will i get out of 5?). my critical values were 2 and 17
    and the most stupid mistake was the mean on no. 4 for the 2nd equation i integrated 1/9 only instead of 3/9. the integrations were correct. so was the method, except that i copid the wrong equation on the next page when doing the sum. so i got a 1/23/36 instead of a 2/5/12. that was a 3 mark question. how much am i gonna lose?
    PLEASE HELP!
    (and what do you guys think the grade boundary will be?)
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    i thought that 2(a) was 1/3 as p(x<0)=F(0)???

    for that robot thing, isnt the last asnwer also 0.05363....as they are asking about twice?? whats the reason for the anwer anyway?
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    (Original post by Moa)
    Here are my answers:
    answers

    question 1.
    a) A~B(20,0.05)
    b) 0.3585
    c) 0.0026
    d) mean=1 Var(x)=0.95

    question 2.
    a) 1/12
    b) f(x)=1/6 for -2<=x<=4
    f(x)=0 otherwise

    c) Uniform rectangular
    d) Var(x)=3
    e) 0

    question 3.
    a) 0.7788
    b) 0.3297
    c) 0.0536
    d) 0.03297 (independence)

    question 4.
    a) k=1/9
    b) F(x)=0 for x<=0
    F(x)=1/9( x^3/3 -x^2 +2x) for 0<x<=3
    F(x)=x/3 -1 for 3<x<=4
    F(x)=1 for x>=4

    c) mean=2.4167
    d) I only used the equation for 0-3 range, change of sign indicates root in between

    question 5.
    a) 0.3328
    b) 0.0485

    question 6.
    a) x<=2 x>=16
    c) 0.85%

    question 7.
    a) mean=1.75 Var(x)=0.1875
    b) list of all samples
    c) increasing mean, prob: 1/64, 9/64, 27/64, 27/64
    Thanks for the answers

    Anyone:

    Question 7
    I listed all possible samples
    it came to 8 things

    111 = 0.25^3
    112 = 0.25^2 x 0.75
    122 = 0.25 x 0.75^2
    222 = 0.75^3
    221 = 0.25 x 0.75^2
    211 = 0.25^2 x 0.75
    121 = 0.25^2 x 0.75
    212 = 0.25 x 0.75^2

    so why in the last part are we left with 4 answers only?
    this might be answering my own question but is it because the ones in red and ones in blue repeat? so we're actually left with 4 different combinations?

    in this case how many marks will i lose for just listng all 8 samples and their probabilities?

    Thanks
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    why is the significant level 0.85
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    (Original post by MIKE ESSIEN IS QUITE SICK)
    why is the significant level 0.85
    I don't think I put 0.85 but I got the critical regions right, I must have made some stupid mistake adding them (or added the wrong numbers) :rolleyes:
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    (Original post by milliondollarcorpse)
    I don't think I put 0.85 but I got the critical regions right, I must have made some stupid mistake adding them (or added the wrong numbers) :rolleyes:
    same
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    (Original post by kosy91)
    i thought that 2(a) was 1/3 as p(x<0)=F(0)???

    for that robot thing, isnt the last asnwer also 0.05363....as they are asking about twice?? whats the reason for the anwer anyway?
    Yep i got that too. isnt the answer 1/3???
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    Good luck everyone, fingers crossed we got good marks
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    (Original post by crazygirl29)
    Yep i got that too. isnt the answer 1/3???
    yes it is a 1/3

    moa's expalnation ia wrong. this is a continuous distribution
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    could someone explain to me why the critical region question had one tail as X>=16? from my tables i had p(X>=15) = 0.0064 and p(x>=16) = 0.0021, therefore as the question asked the probability to be as close to 0.005 in each tail it was surely 15 that was the critical region. ????
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    (Original post by Milan.)
    so we're actually left with 4 different combinations?
    the question asks for the sampling distribution of the mean value of the samples.
    so you need to list all the samples with their mean and the probability of getting that sample.

    111 mean=3/3 prob=1/64
    112 mean=4/3 prob=3/64
    122 mean=5/3 prob=9/64
    222 mean=6/3 prob=27/64
    221 mean=5/3 prob=9/64
    211 mean=4/3 prob=3/64
    121 mean=4/3 prob=3/64
    212 mean=5/5 prob=9/64

    there are 4 different values the mean can take.
    next step is to find the probabilities of getting each mean.
    getting a mean of 1 is 1/64
    getting a mean of 4/3 is 3/64+3/64+3/64 = 9/64
    getting a mean of 5/3 is 9/64+9/64+9/64 = 27/64
    getting a mean of 2 is 27/64

    so the sampling distribution is

    mean.............1.........4/3..........5/3............ 2
    P(mean):....1/64.....9/64......27/64......27/64

    (ignore the dots... its the only way i could space it out lol)
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    ill add my 2 cents to the questions people seem to be also disagreeing on!

    (2a) P(X<0) = 1/3

    (3d) same as part (3b) due to independence

    (4b) F(x)=
    0 for x<0
    1/27(x^3-3x^2+6x) for 0<x<=3
    1/3(x-1) for 3<x<=4
    1 for x>4

    (6b+6c) critical regions X<=2 X>=16 sig level 0.85%
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    (Original post by ben_1991)
    the question asks for the sampling distribution of the mean value of the samples.
    so you need to list all the samples with their mean and the probability of getting that sample.

    111 mean=3/3 prob=1/64
    112 mean=4/3 prob=3/64
    122 mean=5/3 prob=9/64
    222 mean=6/3 prob=27/64
    221 mean=5/3 prob=9/64
    211 mean=4/3 prob=3/64
    121 mean=4/3 prob=3/64
    212 mean=5/5 prob=9/64

    there are 4 different values the mean can take.
    next step is to find the probabilities of getting each mean.
    getting a mean of 1 is 1/64
    getting a mean of 4/3 is 3/64+3/64+3/64 = 9/64
    getting a mean of 5/3 is 9/64+9/64+9/64 = 27/64
    getting a mean of 2 is 27/64

    so the sampling distribution is

    mean.............1.........4/3..........5/3............ 2
    P(mean):....1/64.....9/64......27/64......27/64

    (ignore the dots... its the only way i could space it out lol)
    great... i flopped that then!!
    thank allah it wasn't tooo many marks

    thanks a lot for clearing it up!
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    (Original post by Moa)
    Basic rule for probabilities is:

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

    so

    P(X>=0) = 1-P(X<0)
    P(X>=0)-1 = -P(X<0)
    -(P(X>=0)-1)=P(X<0)

    P(X<0)= -(P(X>=0)-1)

    P(X>=0)=1-P(X<=-1)
    P(X>=0)=1- integrated [(x+2)/6] with limits -1, -2
    P(X>=0)=1-1/12
    P(X>=0)=11/12

    P(X<0)= -(P(X>=0)-1)
    P(X<0)= -(11/12-1)
    P(X<0)= -(11/12-1)
    P(X<0)= 1/12
    for that one, isn't it just F(0)? cos F equations are cumulative anyway.
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    (Original post by BrightGirl)
    for that one, isn't it just F(0)? cos F equations are cumulative anyway.
    Yes P(X<0) = F(0) = (0+2)/6 = 1/3
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    (Original post by ben_1991)
    Yes P(X<0) = F(0) = (0+2)/6 = 1/3
    Yeah :top:

    it was like 1 mark anyway, there couldn't be all that working!
 
 
 
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