Respuesta :
Answer:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779[/tex]
And the deviation would be:
[tex] Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54[/tex]
Step-by-step explanation:
For this case we have the following distribution given:
X 0 1 2 3 4 5 6
P(X) 0.3 0.25 0.2 0.12 0.07 0.04 0.02
For this case we need to find first the expected value given by:
[tex] E(X) = \sum_{i=1}^n X_i P(X_I)[/tex]
And replacing we got:
[tex] E(X)= 0*0.3 +1*0.25 +2*0.2 +3*0.12 +4*0.07+ 5*0.04 +6*0.02=1.61[/tex]
Now we can find the second moment given by:
[tex] E(X^2) =\sum_{i=1}^n X^2_i P(X_i) [/tex]
And replacing we got:
[tex] E(X^2)= 0^2*0.3 +1^2*0.25 +2^2*0.2 +3^2*0.12 +4^2*0.07+ 5^2*0.04 +6^2*0.02=4.97[/tex]
And the variance would be given by:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779[/tex]
And the deviation would be:
[tex] Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54[/tex]
