Question 331508
Suppose that Y has density function
f(y) = ky(1-y), 0<=y<=1,
f(y) = 0, elsewhere. 
a, find the value of k that makes f(y) a probability density function
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K* integal (0 to 1)(y(1-y) dy ={{{ K*( Y^2/2-y^3/3)}}} (evaluated for y from 0 to 1)
= K*(1/2-1/3)=1
=k/6=1
K=6
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b, find P(.4<=Y<=1)
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{{{6*(Y^2/2-Y^3/3)}}} evalutated for y=0.4 to 1
{{{6*(1/2-1/3-(0.4^2/2-0.4^3/3))}}}
=6(1/6-16/200+64/3000)=0.648

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c, find P(.4<=Y<1)

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same as b, continuous functions do not have probability (area) at point
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d, find P(Y<=.4|Y<=.8)

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P(Y<=.4|Y<=.8)=P(Y<=0.4 and Y<=0.8)/P(Y<0.8) = P(Y<=0.4)/P(Y<0.8)
P(y<0.4) = {{{ 6*(0.4^2/2-0.4^3/3 -0) = 0.352}}}
P(y<0.8) = {{{6*(0.8^2/2-0.8^3/3 -0) = 0.896}}}
P(Y<=.4|Y<=.8)=0.352/0.896=0.393
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e, find P(Y<.4|Y<.8).
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same as d