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    Hello Just a question on Binomial Distribution:

    THe Random variable X has the distribution B(n,0.4)

    Give that n = 10, find the smallest value of x such that P(X>x) <0.05

    (b) Find the smallest value of n such that P(X=0) isless than 0.001


    Thanks
    From John
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    (Original post by Neo1)
    Hello Just a question on Binomial Distribution:

    THe Random variable X has the distribution B(n,0.4)

    Give that n = 10, find the smallest value of x such that P(X>x) <0.05

    (b) Find the smallest value of n such that P(X=0) isless than 0.001


    Thanks
    From John
    P(X>x) = 1-P(X =< x)
    1-P(X =< x) < 0.05
    P(X =< x) > 0.95
    use table for n=10 from here onwards (with p=0.4)

    b) P(X=0) < 0.001
    nC0 * 0.6^n < 0.001
    0.6^n < 0.001
    n = log(0.001)/ log(0.6) (rounded up)
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    I didnt get part (b) very well do you mind explaining please and I havent done logs yet so what other method could I use


    Thanks a lot
    From John
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    b) P(X=0) < 0.001

    binomial formula is [for X~B(n,p) ]:
    P(X=r) = n!/(r!(n-r)! * p^r * (1-p)^(n-r)

    n!/0!n! * 0.6^n < 0.001

    0.6^n < 0.001
    n = log(0.001)/ log(0.6) (rounded up)

    there's no "proper" way to do it apart from logs. having said that you could type ANS = 0.6 into your calculator. then type ANS*0.6 and see how many time you have to press "equals" before the display shows a number less than 0.001.
 
 
 
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