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    Need a bit of explainign how this works
    continuous random variable Y
    mean=100
    variance=256

    Find K such that P(100-k more than /equal to Y less than eual to 100+k)
    = 0.516

    ans :k=11.2

    Mark scheme says
    p(Y less than/equal to 100 + k )= 0.516 + 1/2(1-0.516)
    =0.758
    P(Z less than/eual to k/16) = 0.758
    and so on

    i can't see how they arrived to these 2 conclusions

    It also says some candidates stated
    (100 +k -100) root 256
    can be simplified to k/16 also

    I'm not following this :s
    from June 2001 paper
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    (Original post by MIZZ)
    Need a bit of explainign how this works
    continuous random variable Y
    mean=100
    variance=256

    Find K such that P(100-k more than /equal to Y less than eual to 100+k)
    = 0.516

    ans :k=11.2

    Mark scheme says
    p(Y less than/equal to 100 + k )= 0.516 + 1/2(1-0.516)
    =0.758
    P(Z less than/eual to k/16) = 0.758
    and so on

    i can't see how they arrived to these 2 conclusions

    It also says some candidates stated
    (100 +k -100) root 256
    can be simplified to k/16 also

    I'm not following this :s
    from June 2001 paper
    The question is P(100-k<Y<100+k) = 0.516. As the distribution is normal, you know it must be symmetrical. As you know the area 0.516 is split evenly around the value 100 (which is the mean), there must be 0.258 on each side of 100. So on a standardised curve, phi of the z value would equal 0.5 + 0.258 = 0.758. If you look at the normal distribution table, z = 0.70 corresponds to this exactly. Thus by destandardising it (multiplying it by the standard deviation, which is 16) you should obtain k = 11.2.
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    ahhh
    thanks ever so much boygenius
 
 
 
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