A random variable takes only the values 3, 4, 5, 6,..., and the probabilities that it assumes these
values are given by
P(X = k) = 48/4^k for k = 3, 4, 5,....
Find the:
a. mean;
b. variance.
Solution
a) mean = summation [ k*P(k)]
= summation[ k*48/4^k] , where k =2 ,3,4, 5 ...
= 48[ 2/4^2 + 3/4^3 + 4/4^4 +5/4^5 .+ ...]
= (48/16)*14/3
= 14
b) variance = summation [k^2*p(k)] - (mean)^2
= 48[ 2^2/4^2 + 3^2/4^3 + 4^2/4^4 +5^2/4^5 .+ ...] - 14^2
= (48/16)[ 2^2 +3^2/4 +4^2/4^2 + 5^2/4^3 + .....] -14^2
=