Answer:
There is 0.00022 probability that the yearly income of the insurance claims will exceed 8.3 millions.
Step-by-step explanation:
Given data:
Sample Size = n = 25,000
Mean of yearly claim = Mu = 320
Standard Deviation of yearly claim = SD = 540
What we have to find out?
Probability of total yearly claims exceeding 8.3 million?
As the total no. of policy holders is 25,000 therefore the targeted value of claim will be calculated as follows:
Targeted value of policy claim = x = Total Yearly claim / No. of Policy holders
x = (8.3 * 1.000.000) / 25,000
x = 332
Central Limit Theorem (CLT):
From Central theorem we know that: z(x) = (x - Mu)/(SD/√n) ; Equation 1
By putting values in equation 1 we have:
z(332) = (332-320)/(540 / √25,000)
z(332) = 12 / (540/158.114)
z(332) = 3.51
By using the normal distribution tables we get:
P(x <= 332) = Phi(z) = Phi(3.51)
P(x <= 332) = 0.99978
By using the probability unity method we will get:
P(x > 332) = 1 - P(x <= 332)
P(x > 332) = 1 - 0.99978
P( x > 332) = 0.00022
Thus we get the probability of total yearly claim exceeding 8.3 million equals to 0.00022.
