IFoA April 2012 - Qn. 4 and Qn. 10

Discussion in 'CT3' started by Rupel, Oct 1, 2016.

  1. Rupel

    Rupel Member

    4) Claim amounts arising under a particular type of insurance policy are modelled as having a normal distribution with standard deviation £35. They are also assumed to be independent from each other. Calculate the probability that two randomly selected claims differ by more than £100.

    In the above Question, D~N(0,2450) and it is solved by P(|D| > 100) = P( |Z| > 100/2450^.5) = P(|Z| > 2.020) = ?.
    Kindly help.

    10) In a portfolio of car insurance policies, the number of accident-related claims, N, made by a policyholder in a year has the following distribution:
    No. of claims, n: 0 1 2
    Probability :0.4 0.4 0.2
    The number of cars, X, involved in each accident that results in a claim is distributed as follows:
    No. of cars, x: 1 2
    Probability : 0.7 0.3
    It can be assumed that the occurrence of a claim and the number of cars involved in the accident are independent. Furthermore, claims made by a policyholder in any year are also independent of each other. Let S be the total number of cars involved in accidents related to such claims by a policyholder in a year.
    (i) (a) Determine the probability function of S.
    (b) Hence find E(S).

    In the above Question, kindly explain the procedure to obtain the Probability Function of S.
     
  2. John Lee

    John Lee ActEd Tutor Staff Member

    4) What's the problem?
    10) You're just working through the possible combinations.
     
  3. Rupel

    Rupel Member

    Dear Sir,

    In Qn No. 4, I don't know how to solve P(|z|>2.020)
     
  4. John Lee

    John Lee ActEd Tutor Staff Member

    |Z| > 2.020 is equivalent to Z < -2.020 or Z > 2.020
     
  5. Rupel

    Rupel Member

    Sir, thank you for your reply.
    I have a doubt, Kindly tell me if this is correct
    P(|Z|>2.020) = 2 * P(Z > 2.020) = 2 * ( 1 - .9783) = 0.434
     
  6. John Lee

    John Lee ActEd Tutor Staff Member

    Yes it's correct except you wrote 0.434 instead of 0.0434
     
  7. Rupel

    Rupel Member

    Thank you Sir.
     

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