Refer to the Excel “Insurance” data posted in eLearning. (12) What is the expect
ID: 3364628 • Letter: R
Question
Refer to the Excel “Insurance” data posted in eLearning. (12)
What is the expected claim payment made per customer in a year?
What is the probability that a randomly selected customer receives exactly the expected claim payment you calculated in “a”?
What is the (sample) standard deviation of the claim payments made per customer in a year?
If the insurance company charges a $750 premium to each customer, each year, then what amount does the insurance company make/lose per customer, each year? In a sentence or two, explain why the customers might be willing to pay the $750 premium, given your calculations.
The critical value (or “z value”) for statistical significance at the 10% level is 1.645 (remember?!). Thus, I’d like you to create a 90% confidence interval by taking:
Expected Claim Payment + 1.645(standard deviation of claim payment)
Now that I have the interval, I’ll know that a randomly selected insurance customer will have a 90% chance of receiving claim payments within this confidence interval.
Do you agree with the above, italicized, analysis? In a sentence, explain why or why not.
Claim Payment Probability This probability distribution describes the probabilities that 0 0.80 customers of the insurance company will receive various 1000 0.08 claim payment amounts in a given year. 2000 0.06 5000 0.05 25000 0.01Explanation / Answer
What is the expected claim payment made per customer in a year?
Answer : Expected claim payment made per customer = 0 * 0.80 + 1000 * 0.08 + 2000 * 0.06 + 5000 * 0.05 + 25000 * 0.01 = $ 700
What is the probability that a randomly selected customer receives exactly the expected claim payment you calculated in “a”?
Answer : Pr(X= $ 700) = 0 as there is no 700 payment here.
What is the (sample) standard deviation of the claim payments made per customer in a year?
ANswer : Variance = 0.80 * (0 - 700)2 + 0.08 * (1000 - 700)2 + 0.06 * (2000 - 700)2 + 0.05 * (5000 - 700)2 + 0.01 * (25000 - 700)2 = 73,30,000
Standard deviation = sqrt(variance) = $ 2707.40
If the insurance company charges a $750 premium to each customer, each year, then what amount does the insurance company make/lose per customer, each year? In a sentence or two, explain why the customers might be willing to pay the $750 premium, given your calculations.
If the insurance company charges a $ 750 premium to each customer, each year.
THe amount does the insurance company will makeper customer , each year = $ 750 - $ 700 = $ 50
Here, the customere may be willing topay the $ 750 premium even if is more than the expected value is because the large amount of standard deviation of claim payments made per customer in a year. As this have the value of around $ 2700 so that put an serious effect if there is an emergency occur on any given person.
The critical value (or “z value”) for statistical significance at the 10% level is 1.645 (remember?!). Thus, I’d like you to create a 90% confidence interval by taking:
Expected Claim Payment + 1.645(standard deviation of claim payment)
Now that I have the interval, I’ll know that a randomly selected insurance customer will have a 90% chance of receiving claim payments within this confidence interval.
Do you agree with the above, italicized, analysis? In a sentence, explain why or why not.
Answer : here as the given distribution is not a normal distribution, even not a symmetric distribution towards mean. So we will not use this type of analysis.