Let X be a ramdom variable with the following probability distribution
x: 1, 2, 3, 4
P(X=x): 0.4, 0.3, 0.2, 0.1
a. Find E(X) and Var(X)
b. is E(X) a possible value of X
a. E(X)=1∗0.4+2∗0.3+3∗0.2+4∗0.1=2.E(X)=1*0.4+2*0.3+3*0.2+4*0.1=2.E(X)=1∗0.4+2∗0.3+3∗0.2+4∗0.1=2.
Var(X)=12∗0.4+22∗0.3+32∗0.3+42∗0.1−22=1.9.Var(X)=1^2*0.4+2^2*0.3+3^2*0.3+4^2*0.1-2^2=1.9.Var(X)=12∗0.4+22∗0.3+32∗0.3+42∗0.1−22=1.9.
b. E(X) is a possible value of X.
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