A random variable X has the following probability mass funct

A random variable X has the following probability mass function:

Problem 3: A random variable X has the following probability mass function x 0 1 2 3 4 p(x) 0.24 0.4 1 0.26 0.08 0.01 a. Find the expected value of the random variable X. h. Find the expected value of 2X-7. c. Find the expected value of X^2. d. Find the variance of X. e. Find the variance 01 2-3X.

Solution

(a) expected value= E(X)=sum of x*f(x)

=0*0.24+1*0.41+2*0.26+3*0.08+4*0.01

=1.21

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(b) E(2X-7)

=2*E(X) -7

=2*1.21-7

=-4.58

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(c) E(X^2) =sum of x^2*f(x)

=0*0.24+1*0.41+2^2*0.26+3^2*0.08+4^2*0.01

=2.33

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(d) Variance = E(X^2) -[E(X)]^2

=2.33-1.21^2

=0.8659

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(e)Var(2-3X)

=9*Var(X)

=9*0.8659

=7.7931