Answer to Question #310396 in Statistics and Probability for Glenn

Question

Calculate the mean and the variance of the discrete random variable X which can take only values 1, 2, and 3, given that P(1) = 10/33, P(2) = 1/3, and P(3) = 12/33.

P(X) X-P(X) X²P(X)

Solve for the mean and the variance of the discrete random variable X which can take only the values 2, 4, 5 and 9, given that P(2) = 9/20, P(4) = 1/20, P(5) = 1/5 and P(9) = 3/10.

X P(X) X∙P(X) X2∙P(X)

Answer 1

a)

Mean of X.

$\mu$_x=$\sum$ X.P(X)

=1(10/33)+2(1/3)+3(12/33)

=2.06

Variance of X.

Var(X) =$\sum$ (X-$\mu$_x)².P(X)

=(1-2.06)²(10/33)+(2-2.06)²(1/3)+(3-2.06)²(12/33)

=0.663

b)

Mean of X.

$\mu$_x=∑ X.P(X)

=2(9/20)+4(1/20)+5(1/5)+9(3/10)

=4.8

Variance of X.

Var(X) =∑ (X-$\mu$)².P(X)

=(2-4.8)²(9/20)+(4-4.8)²(1/20)+(5-4.8)²(3/10)

=8.86