Question

A random variable X has the following distribution
Find E(x) , E(x²) and Var (x)

Answer 1

Final Answer : 1) E(X) = 1/2 
                          2) $E(X^{2}) = \frac{11}{6}$ 
                          3) Var(X)= 19/12

Steps:
1) The Expected value of  discrete random variable  X is usually written as  

m=E(X)= $\sum_{all \:\: x}^{} P(X=x)* \: x$ 

According to given values,

m= E(X) = $\frac{1}{3} * (-1) + \frac{1}{6} * (0) + \frac{1}{6} * (1) + \frac{1}{3} * (2) = \frac{1}{2}$

Hence,E(X) = 1/2 


2) We know that 

 [tex]E(X^{2} ) = \sum_{all \:\: x}^{} P(X=x)* \: x^{2} \\ \\=\ \textgreater \ \frac{1}{3} *(-1)^{2} + \frac{1}{6} *(1)^{2} + \frac{1}{6} *(0)^{2} + \frac{1}{3} *(2)^{2} \\ \\ =\ \textgreater \ \frac{11}{6} [/tex]

3) Variance =Tells us about the spreadness of the possible values of the random variable X.

$Var(X) = E(X^{2})- m^{2} \\ \\ =\ \textgreater \ \frac{11}{6} - ( \frac{1}{2} )^{2} = \frac{19}{12 }$