X    P(X)
--------- 
0    .1
1    .2 
2     ?
4    .3
6    .3

For this to be a probability distribution, the sum
of the probabilities must = 1.  So since the probability
of the other four values of X have sum
.1+.2+.3+.3 = .9, then P(2) = 1.0-.9 = .1 and the
distribution is:

X    P(X)     E(X)=X·P(X)
-------------------------- 
0    .1            0    
1    .2           .2
2    .1           .2
4    .3          1.2
6    .3          1.8
--------------------------   
Sum = 1  ∑E(X) = 3.4

Answer:  ∑E(X) = 3.4 


Edwin