SOLUTION: 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 b, find P(.4<=Y<

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
b, find P(.4<=Y<=1)
c, find P(.4<=Y<1)
d, find P(Y<=.4|Y<=.8)
e, find P(Y<.4|Y<.8).
Answer by jrfrunner(365) (Show Source): You can put this solution on YOUR website!
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
----
K* integal (0 to 1)(y(1-y) dy = (evaluated for y from 0 to 1)
= K*(1/2-1/3)=1
=k/6=1
K=6
========================
b, find P(.4<=Y<=1)
----
evalutated for y=0.4 to 1

=6(1/6-16/200+64/3000)=0.648
===================
c, find P(.4<=Y<1)
------
same as b, continuous functions do not have probability (area) at point
===================
d, find P(Y<=.4|Y<=.8)
----
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) =
P(y<0.8) =
P(Y<=.4|Y<=.8)=0.352/0.896=0.393
===================
e, find P(Y<.4|Y<.8).
---
same as d

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