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    Hi,

    I've got a question about probability density functions.

    We are given that the area if a circle has prob. density function:

    f(a) = 2/Pi - 2a/(Pi)^2 for 0<a<Pi
    0 otherwise

    We also know that E(a) = Pi/3 and Var(a) = Pi^2 / 18

    The question is: let S be the sum of the areas of 36 of these circles, randomly chosen. Find P(S>20)

    My workings:

    E(S) = 36E(a) = 36 x Pi/3 = 12Pi

    Var(S) = 36Var(a) = 2Pi^2 (this is 36Var(a) not 36^2 Var(a) since we are taking 36 different circles, not multiplying the area of one circle by 36)


    Then since sample is large (greater than 30), by central limit theorem we can assume that S is approx normally distributed so has distribution N(12Pi, 2Pi^2)

    P(S>20) = 1 -  P(Z&lt;\dfrac{20-12\Pi}{\sqrt{2\Pi^2}})

    = 1 - P(Z< -3.984...)

    = 1 - (1 - P(Z< 3.984...))

    but we can't look up 3.984 in our tables, do we just assume P(Z<3.984) = 1 ?


    Thanks for the help xx
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    To 3 dec.pl. it comes out as 1.000 since the z value exceeds the critical value for p=0.9995 which should be in tables or attached to them. So you don't have much choice other than to assume it's one.

    I do find it rather an odd question to be asked though; producing such an extreme value. Does make me wonder if there is an error in the book.
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    (Original post by ghostwalker)
    To 3 dec.pl. it comes out as 1.000 since the z value exceeds the critical value for p=0.9995 which should be in tables or attached to them. So you don't have much choice other than to assume it's one.

    I do find it rather an odd question to be asked though; producing such an extreme value. Does make me wonder if there is an error in the book.
    Thanks, I understand now Taking it to be 1.000 does give the correct answer, so perhaps it's just a weird question.
 
 
 
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