Answer to Question #173274 in Statistics and Probability for Denisse Bisuña

Question

Mean and Variance of the Probability Distribution

Find the mean and the variance of the following probability distribution.

X P(X)

1 0.237

2 0.317

3 0.178

4 0.157

5 0.070

Answer 1

Mean.

From the question, probabilities do not add up to one, but I will work on it the way they are.

$E(X)=\sum_{x=1}^5X(P(X))$

$=1(0.237)+2(0.317)+3(0.178)+4(0.157)+5(0.07)$

=2.383

Variance

$Var(X)=E(X^2)-[E(X)]^2$

$E(X^2)=\sum_{x=1}^5X^2(P(X))$

$=1(0.237)+4(0.317)+9(0.178)+16(0.157)+25(0.07)$

$=7.369$

$Var(X)=7.369-2.383^2=1.69$