Q:

Let X be a r.v. has the probability function given by: Find the expectation

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Let X be a r.v. has the probability function given by:

4 3 1- 3- x
1/15 1/5 2/5 1/3 (x)p

Find the expectation of the random variable

Y=2X+1  &  Z=X^2+3X

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Since   
 E(Y)=∑〖yP(x)〗
∴E(Y)=E(2X+1)=2E(X)+1
But
E(X)=∑_x x P(x)=(-3)(1/3)-1(2/5)+1(1/5)+3(1/15)=-1
∴E(Y)=2(-1)+1=-1
To obtain the expectation of Z we evaluate:
E(X^2 )=9(1/3)+(2/5)+(1/5)+9(1/15)=21/5
∴E(Z)=E(X^2 )+3E(X)
=21/5+3(-1)=6/5=1.2

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