Since E(X) = summation_{i= 1}^{5} x(i)*p(i)
E(X^2) = summation_{i=1}^{5} (x(i)^2)*p(i)
and Var(X) = E(X^2) - E(X)^2
--------------------------------------------------------------
MATLAB CODE :-
------------------------
x = [1, 6, 8, 10, 12];
p = [0.2, 0.15, 0.1, 0.2, 0.35];
Ex = 0; %Ex = E(X)
for i =1:5
Ex = Ex + x(i)*p(i); %Ex = E(X)
end
disp("E(x) is given below:-");
Ex %Ex = E(X)
Ex2 = 0; %Ex2 = E(X^2)
for j = 1:5
Ex2 = Ex2 + (x(i)^2)*p(i); %E(x^2) = Ex2
end
disp("Var(x) is given below :-")
varX = Ex2 - (Ex)^2 %Var(X) = E(X^2) - E(X)^2
-------------------------------------------------------------------
OUTPUT :-
----------------
E(x) is given below:-
Ex =
8.1000
Var(x) is given below :-
varX =
186.3900
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