If a random variable X has the probability distribution given by P(X=0)=3 C^3 , P(X=2)=5 C-10 C^2 and P(X=4)=4 C-1 , then the variance of that distribution is

Step-by-step solution

Given, aligned & P(X=0)=3 C^3 & P(X=2)=5 C-10 C^2 & and P(X=4)=4 C-1 & We know that, & P(X)=1 & 3 C^3+ (5 C-10 C^2 )+(4 C-1)=1 & 3 C^3-10 C^2+9 C-2=0 & aligned array ll & (C-1) (3 C^2-7 C+2 )=0 & (C-1)(3 C-1)(C-2)=0 & C=1, 1 3^ , 2 & C= 1 3 array Now, Hence, variance = X_P^2- ( X_P )^2 aligned & = (0^2 1 9 +4 5 9 +16 1 3 )- ( 10 9 + 4 3 )^2 & = ( 20 9 + 16 3 )- ( 66 27 )^2= 60+144 27 - 484 81 & = 204 27 - 484 81 = 612-484 81 = 128 81 aligned