Respuesta :
Answer:
[tex] E(X)=\sum_{i=1}^n X_i P(X_i) = 1*0.25 +2 *0.5 +3*0.25= 2[/tex]
We can calculate the moment of second order and we got:
[tex] E(X^2)=\sum_{i=1}^n X^2_i P(X_i) = 1^2*0.25 +2^2 *0.5 +3^2*0.25= 4.5[/tex]
And we can calculate the variance like this:
[tex]Var(X) = E(X^2) -[E(X)]^2 = 4.5 -(2)^2 =0.5[/tex]
And the deviation would be:
[tex] Sd(X) = \sqrt{0.5}=0.707[/tex]
Step-by-step explanation:
For this case we have the following distribution:
X 1, ,2, 3
P(X) 0.25, 0.5, 0.25,
We can calculate the expected value with this formula:
[tex] E(X)=\sum_{i=1}^n X_i P(X_i) = 1*0.25 +2 *0.5 +3*0.25= 2[/tex]
We can calculate the moment of second order and we got:
[tex] E(X^2)=\sum_{i=1}^n X^2_i P(X_i) = 1^2*0.25 +2^2 *0.5 +3^2*0.25= 4.5[/tex]
And we can calculate the variance like this:
[tex]Var(X) = E(X^2) -[E(X)]^2 = 4.5 -(2)^2 =0.5[/tex]
And the deviation would be:
[tex] Sd(X) = \sqrt{0.5}=0.707[/tex]
