Respuesta :
Answer:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5[/tex]
We can find the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496 [/tex]
And we can calculate the variance with this formula:
[tex] Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246[/tex]
And the deviation is:
[tex] Sd(X) = \sqrt{1.246}= 1.116[/tex]
Step-by-step explanation:
For this case we have the following probability distribution given:
X 0 1 2 3 4 5
P(X) 0.031 0.156 0.313 0.313 0.156 0.031
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
We can verify that:
[tex] \sum_{i=1}^n P(X_i) = 1[/tex]
And [tex] P(X_i) \geq 0, \forall x_i[/tex]
So then we have a probability distribution
We can calculate the expected value with the following formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5[/tex]
We can find the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496 [/tex]
And we can calculate the variance with this formula:
[tex] Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246[/tex]
And the deviation is:
[tex] Sd(X) = \sqrt{1.246}= 1.116[/tex]
