Answer the following questions below.

The numbers of claims made last year by individual motor insurance policyholders were: Number of claims 0 2 3 4+ Number of policyholders 2,962 382 47 25 4 Carry out a chi square test to determine whether these frequencies can be considered to conform to a Poisson distribution.Vehicles in a certain country are required to be assessed every year for road-worthiness. At one vehicle assessment centre, drivers wait for an average of 15 minutes before the road-worthiness assessment of their vehicle commences. The assessment takes on average 20 minutes to complete. Following the assessment, 80% of vehicles are passed as road-worthy allowing the driver to drive home. A further 15% of vehicles are categorised as a "minor fail"; these vehicles require on average 30 minutes of repair work before the driver is allowed to drive home. The remaining 5% of vehicles are categorised as a "significant fail"; these vehicles require on average three hours of repair work before the driver can go home. A continuous-time Markov model is to be used to model the operation of the vehicle assessment centre, with states W (waiting for assessment), A (assessment taking place), M (minor repair taking place), S (significant repair taking place) and H (travelling home). (i) Explain what assumption must be made about the distribution of the time spent in each state. [1] (ii) Write down the generator matrix for this process. [2] (iii) (a) Use Kolmogorov's Forward Equations to write down differential equations satisfied by Pww () and by PWA (!). (b) Verify that PWA(1) =4e-/20 -4e-1/15 for 1 20, where I is measured in minutes. (c) Derive an expression for PWM (1) for : 20. [7] (iv) Let 7; be the expected length of time (in minutes) until the vehicle can be driven home given that the assessment process is currently in state i . (a) Explain why Ty =15+TA. (b) Derive corresponding equations for TA, Ty and Is- (c) Calculate Tw . [4] [Total 14]