Answer:
the approximate probability that the insurance company will have claims exceeding the premiums collected is [tex]\mathbf{P(X>1100n) = 0.158655}[/tex]
Step-by-step explanation:
The probability of the density function of the total claim amount for the health insurance policy is given as :
[tex]f_x(x) = \dfrac{1}{1000}e^{\frac{-x}{1000}}, \ x> 0[/tex]
Thus, the expected total claim amount [tex]\mu[/tex] = 1000
The variance of the total claim amount [tex]\sigma ^2 = 1000^2[/tex]
However; the premium for the policy is set at the expected total claim amount plus 100. i.e (1000+100) = 1100
To determine the approximate probability that the insurance company will have claims exceeding the premiums collected if 100 policies are sold; we have :
P(X > 1100 n )
where n = numbers of premium sold
[tex]P (X> 1100n) = P (\dfrac{X - n \mu}{\sqrt{n \sigma ^2 }}> \dfrac{1100n - n \mu }{\sqrt{n \sigma^2}})[/tex]
[tex]P(X>1100n) = P(Z> \dfrac{\sqrt{n}(1100-1000}{1000})[/tex]
[tex]P(X>1100n) = P(Z> \dfrac{10*100}{1000})[/tex]
[tex]P(X>1100n) = P(Z> 1) \\ \\ P(X>1100n) = 1-P ( Z \leq 1) \\ \\ P(X>1100n) =1- 0.841345[/tex]
[tex]\mathbf{P(X>1100n) = 0.158655}[/tex]
Therefore: the approximate probability that the insurance company will have claims exceeding the premiums collected is [tex]\mathbf{P(X>1100n) = 0.158655}[/tex]
