Respuesta :
Answer:
[tex] E(X) = 1*0.3 +2*0.5 +3*0.1 +4*0.1 = 2[/tex]
In order to find the variance first we need to 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)=1^2*0.3 +2^2*0.5 +3^2*0.1 +4^2*0.1 = 4.8[/tex]
And then the variance would be given by:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 4.8 -(2)^2 = 0.8[/tex]
Step-by-step explanation:
Previous concepts
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).
Solution to the problem
For this case we have the following probability distribution:
X 1 2 3 4
P(X) 0.3 0.5 0.1 0.1
And we can calculate the expected value with this formula:
[tex] E(X) =\sum_{i=1}^n X_i P(X_i)[/tex]
And replacing we got:
[tex] E(X) = 1*0.3 +2*0.5 +3*0.1 +4*0.1 = 2[/tex]
In order to find the variance first we need to 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)=1^2*0.3 +2^2*0.5 +3^2*0.1 +4^2*0.1 = 4.8[/tex]
And then the variance would be given by:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 4.8 -(2)^2 = 0.8[/tex]
