Respuesta :
Answer:
[tex] E(X) = 1*0.52+ 2*0.25 +3*0.23= 1.71[/tex]
Now we can calculate the second moment with the following formula:
[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*0.52+ 2^2*0.25 +3^2*0.23= 3.59[/tex]
And the variance is given by:
[tex] Var(X) = E(X^2) -E(X)[/tex]
And replacing we got:
[tex] Var(X) = 3.59 -[1.71]^2 = 0.6659[/tex]
And the standard deviation is just the square root of the variance:
[tex]Sd(X) = \sqrt{0.6659}= 0.816[/tex]
Step-by-step explanation:
Previous concepts
For this case we define the random variable X =" how many children the couple will have" and we know the following distribution:
X 1 2 3
P(X) 0.52 0.250 0.230
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 can find the expected value with the following formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]
And replacing we got:
[tex] E(X) = 1*0.52+ 2*0.25 +3*0.23= 1.71[/tex]
Now we can calculate the second moment with the following formula:
[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*0.52+ 2^2*0.25 +3^2*0.23= 3.59[/tex]
And the variance is given by:
[tex] Var(X) = E(X^2) -E(X)[/tex]
And replacing we got:
[tex] Var(X) = 3.59 -[1.71]^2 = 0.6659[/tex]
And the standard deviation is just the square root of the variance:
[tex]Sd(X) = \sqrt{0.6659}= 0.816[/tex]