Respuesta :
Answer:
[tex] E(X) =\sum_{i=1}^n X_i P(X_i)[/tex]
Replacing the values that we have:
[tex] 1 = 0.98*a + 0.02(a-42) = 0.98a +0.02a -0.84[/tex]
And solving for a we got:
[tex] 1.84 = a[/tex]
So then the premium value for the insurance on this case should be 1840 dollars.
Explanation:
For this case we can define the random variable X as the gain ( in thousand of dollars) of insurance company
We assume that the premium clase charge and amount of a to the company and we know from the info given that:
[tex] p(X=a) = 1-0.02 = 0.98[/tex]
[tex] p(X = a-42) = 0.02[/tex]
[tex] E(X) = 1[/tex] represent the expected gain in thousand of dollars
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.
And using the definition for a discrete random variable we know that :
[tex] E(X) =\sum_{i=1}^n X_i P(X_i)[/tex]
Replacing the values that we have:
[tex] 1 = 0.98*a + 0.02(a-42) = 0.98a +0.02a -0.84[/tex]
And solving for a we got:
[tex] 1.84 = a[/tex]
So then the premium value for the insurance on this case should be 1840 dollars.
