What is the Probability and Risk Assessment for Motor Insurance Claims?

In summary, the conversation discusses a survey on 50 claims for a particular class of customers of a motor insurance company, finding that the average cost for car damage is £700 with a standard deviation of 400. The insurance manager believes that the average cost is actually £80 higher. The conversation then delves into hypothesis testing at a 5% significance level to determine if the manager is correct. In the second part, the conversation explores the use of normal approximation to the Poisson distribution to compute the probability of a group of 400 insured cars producing less than 10 claims per year and more than 20 claims per year. Finally, the conversation concludes with calculating the risk premium and value at risk for the insurer losses, taking into
  • #1
Samantha24
8
0

Homework Statement



(a) A survey on 50 claims for a particular class of customers of a motor insurance company found out that the average cost for car damage is £700 with a standard deviation of 400. The insurance manager believes instead that the average cost is £80 bigger.

(i) Does the sample data suggest that the manager is right? Test at a 5% significance level the hypothesis that the average cost is £780. Specify: null hypothesis, alternative hypothesis, critical region and comment on the result of the test.

(b) For the same type of accidents, the number of claims per vehicle insured is known to follow a Poisson distribution with mean 0.03 per year. Using the normal approximation to the Poisson distribution, compute the probability that a group of 400 insured cars produces :

(i) Less than 10 claims per year.

(ii) More than 20 claims per year.

(c) Assuming the cost per claim is 780 with standard deviation of 400, claims’ frequency and portfolio’s size are as specified above:

(i) Compute the risk premium.

(ii) Compute the value at risk at 99.5% level for the Insurer losses.


The Attempt at a Solution



For part (a) this is what I did;

H0 : m = 780
H1 : m = 700

From my test statistic I got 1.41. This is less than 1.96. So you don't reject null hypothesis. Also we can't conclude anything is statistically significant. Therefore, we can't tell if the manger is right or not.

For (b) I don't get what it means by Normal Approximation to the Poisson Distribution but this is what I did. Also I don't know if I should use the 5% significance level from before (If I did I could use the Poisson Table?).

(i)

P(X < 10) = P(0) + P(1) + .. + P(9)

To calculate each value I use: P(X)= (e^-m)*(m^x)/ x!

I don't know if I should be doing it like that or using the poisson table.. It doesn't seem to justify 4 marks..

(ii)

P(X > 20) = 1 - P(X < 20)

From here I use the same method as before but have many more values..? I'm sure there must be a short cut as for 4 marks each.. Seems like a lot of work.

Furthermore, what do I do with the 400? Does it just change my notation on my probabilities? I.e. P(X < 10, 400)?

For part (c)

(i)

RP = average cost per claim * probability of claims. So RP= 780*[the answer I get from (b)(ii)]

However, I don't know if my method of calculating in part (b)(ii) is correct, let alone (b)(i)..

Finally,

(ii) I'm completely lost..
 
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  • #2
Your expression for P(X < 10) is correct, but you need to say what "m" is in that expression (it is not necessarily the same as the previous "m"). By normal approximation to the Poisson we mean that under certain conditions, you can get quite good approximations to quantities like P(X < 10) by replacing the exact Poisson distribution by the normal distribution having the same mean and variance.

RGV
 
  • #3
Yeah the m would be 0.03 there.

Does my part (a) seem correct? Since there are other ways to write the hypothesises?
 
  • #4
No, m is NOT 0.03 in this question. You are not reading the question properly.

RGV
 
  • #5
Oh.. m is 12..
 

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Statistics is the study of collecting, analyzing, and interpreting data. It focuses on understanding and describing patterns and relationships in data. Probability, on the other hand, is the study of the likelihood of events occurring. It uses mathematical principles to assign numerical values to the likelihood of different outcomes.

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