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^

Algebra ->  Probability-and-statistics -> 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^      Log On


   



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) About Me  (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=mu
= E(X)=sum+%28x%2Ap%28x%29%29=-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
V%28Y%29=sigma%5E2=E%28Y%5E2%29-E%5E2%28Y%29=E%28y%5E2%29-mu%5E2
=V%283%2Ax%5E3%29=9%2AV%28X%5E3%29=9%2A%28E%28%28x%5E3%29%5E2%29-E%5E2%28x%5E3%29%29
=9%2A%28E%28x%5E6%29-E%5E2%28x%5E3%29%29
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E%28X%5E6%29=sum%28x%5E6%2Ap%28x%29%29=(-4)^6*(0.2)+(-1)^6*(0.2)+2^6*(0.2)+3^6*(0.2)+4^6*(0.2)=1797.2
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mu=E%28X%5E3%29=sum%28x%5E3%2Ap%28x%29%29=(-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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mu%5E2=E%5E2%28X%5E3%29=%286.8%29%5E2=46.24
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V%28Y%29=sigma%5E2=9%2A%28E%28x%5E6%29-E%5E2%28x%5E3%29%29=9%2A%281797.2-46.24%29=15758.64