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    can someone tell me how to do Q6d pls........step by step cos ive done it again and again and got wrong answer. thx
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    q 6d of what??
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    sorry i get u that jan o2 s2
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    (Original post by zxcvbnm)
    sorry i get u that jan o2 s2
    yeah and Q5d+e, 7d on June 03.........pls. stupid mrhughes and ajimal websites don't have solutions.

    i really need help.
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    6d jan 02

    Ho : p=0.4 H1 : p>0.4

    n=10 p=0.4 X-B(10,0.4)

    P(X>=8) = 1- P(X<=7)
    = 1- 0.9877 (FROM TABLES)
    = 0.0123
    which is less than 5% so u reject Ho
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    5d
    it is a rectangular distribution form 180 to 200 ml and height 0.05

    so it look like this 0.05 ---------------------------
    180 x 200

    suppose x is in the middle as shown in diagram above
    then (200-x)(0.05)=2(x-180)(0.05)
    solve this to get x= 186.67 (approx)

    5e

    this means that 1/3 of cups have less than 0r equal to 186.67 ml
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    (Original post by zxcvbnm)
    5d
    it is a rectangular distribution form 180 to 200 ml and height 0.05

    so it look like this 0.05 ---------------------------
    180 x 200

    suppose x is in the middle as shown in diagram above
    then (200-x)(0.05)=2(x-180)(0.05)
    solve this to get x= 186.67 (approx)

    5e

    this means that 1/3 of cups have less than 0r equal to 186.67 ml

    hmm thx but don't understand it, but its no big deal, i doubt it'll come up and if it does its only 5 marks.....
    i've forgotten stuff bout skewness..........if mean<median<mode....is that negative skew or positive?
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    7d

    from 7c u know that F(x)=x^3 + 3X^2 + 3X +1
    so
    P(-0.3<X<0.3)= F(0.3)-F(-0.3)
    = 1 - 0.343 = 0.657

    Note that F(0.3) is 1 coz the formuls for F(x) is valid from 0 to -1 and for all values above 0 the F(x)=1

    i hope this will help u ....by the way what is ajmals website can u plz give me the adress
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    the diagram is crappy that y i might have problems but try to make ur own on a paper ... u will understand it
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    (Original post by zxcvbnm)
    7d

    from 7c u know that F(x)=x^3 + 3X^2 + 3X +1
    so
    P(-0.3<X<0.3)= F(0.3)-F(-0.3)
    = 1 - 0.343 = 0.657

    Note that F(0.3) is 1 coz the formuls for F(x) is valid from 0 to -1 and for all values above 0 the F(x)=1

    i hope this will help u ....by the way what is ajmals website can u plz give me the adress

    thx so much, i get the idea of it...but for 7c i got F(x)= x^3 + 3x^2 + 3x.....where did u get +1 from?
    www.ajimal.com
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    i think u have applied wrong limits.. the limits are x and -1 NOT x and 0
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    (Original post by zxcvbnm)
    i think u have applied wrong limits.. the limits are x and -1 NOT x and 0

    oh dammit, how am i supposed to know?! i just assumed limits were 0 and -1.......why r u supposed to use x? and why is it x and -1?
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    (Original post by wonkey)
    oh dammit, how am i supposed to know?! i just assumed limits were 0 and -1.......why r u supposed to use x? and why is it x and -1?
    ok, i understand y its X and -1 now.........
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    remember one rule that the limits for F(x) for every case are
    - infinity to +infinity

    but u ignore the areas that we are not interseted in.

    basically the easy rule is always apply the limits as x and (the lowest number)
 
 
 
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