SOLUTION: A random variable X can have values -4,-1,2,3 and 4, each with probability 0.2 .
find a) the density function
b) the mean and
c) the variance of the random Y = 3x^
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Question 334471: A random variable X can have values -4,-1,2,3 and 4, each with probability 0.2 .
find a) the density function
b) the mean and
c) the variance of the random Y = 3x^3
Answer by jrfrunner(365) (Show Source): You can put this solution on YOUR website!
a) density function:
X=-4 p(X)=0.2
X=-1 p(X)=0.2
X=2 p(X)=0.2
X=3 p(X)=0.2
X=4 p(X)=0.2
all other X, P(X)=0
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b) mean=
= E(X)==-4*(.2)-1*(0.2)+2*(0.2)+3*(0.2)+4*(0.2)=0.8
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c) the variance of the random Y = 3x^3
=
=
--
=(-4)^6*(0.2)+(-1)^6*(0.2)+2^6*(0.2)+3^6*(0.2)+4^6*(0.2)=1797.2
--
=(-4)^3*(0.2)+(-1)^3*(0.2)+2^3*(0.2)+3^3*(0.2)+4^3*(0.2)=6.8
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