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Related questions with answers

 Claims against an insurance company follow a Poisson process with rate λ>0. A total  of N claims were received over two periods of combined length t=t1+t2, with t1 and t2 being the lengths of the separate periods. \begin{array} { l } { \text { Claims against an insurance company follow a Poisson process with rate } \lambda > 0 . \text { A total } } \\ { \text { of } N \text { claims were received over two periods of combined length } t = t _ { 1 } + t _ { 2 } , \text { with } t _ { 1 } \text { and } } \\ { t _ { 2 } \text { being the lengths of the separate periods. } } \end{array}

 (a) Given this information, derive the (conditional) probability distribution of N1, the  number of claims made in the 1 st period, given N.\begin{array} { l } { \text { (a) Given this information, derive the (conditional) probability distribution of } N _ { 1 } , \text { the } } \\ { \text { number of claims made in the } 1 \text { st period, given } N . } \end{array}

 (b) The amount paid for the ith claim is Xi, with X1,X2, i.i.d. and independent  of the claims process. Let E(Xi)=μ,Var(Xi)=σ2, for i=1,,N. Given N, find  the mean and variance of the total claims paid in period 1. That is, find these two  conditional moments of the quantity \begin{array} { l } { \text { (b) The amount paid for the ith claim is } X _ { i } , \text { with } X _ { 1 } , X _ { 2 } , \ldots \text { i.i.d. and independent } } \\ { \text { of the claims process. Let } E \left( X _ { i } \right) = \mu , \operatorname { Var } \left( X _ { i } \right) = \sigma ^ { 2 } , \text { for } i = 1 , \ldots , N . \text { Given } N , \text { find } } \\ { \text { the mean and variance of the total claims paid in period } 1 . \text { That is, find these two } } \\ { \text { conditional moments of the quantity } } \end{array}

W1=i=1N1XiW _ { 1 } = \sum _ { i = 1 } ^ { N _ { 1 } } X _ { i }

Question

Based on the following table, which shows crashworthiness ratings for several categories of motor vehicles. In all of these exercises, take X as the crash-test rating of a small car, Y as the crash-test rating for a small SUV, and so on as shown in the table.

Overall Frontal Crash-Test RatingNumber Tested3 (Good)2 (Acceptable)1 (Marginal)0 (Poor)Small Cars X1611122Small SUVs Y101441Medium SUVs Z153534Passenger Vans U133037Midsize Cars V153507Large Cars W199532\begin{matrix} & \text{} & \text {} & \text{Overall Frontal Crash-Test Rating}\\ & \text{Number Tested} & \text{3 (Good)} & \text{2 (Acceptable)} & \text{1 (Marginal)} & \text{0 (Poor)}\\ \text{Small Cars X} & \text{16} & \text{1} & \text{11} & \text{2} & \text{2}\\ \text{Small SUVs Y} & \text{10} & \text{1} & \text{4} & \text{4} & \text{1}\\ \text{Medium SUVs Z} & \text{15} & \text{3} & \text{5} & \text{3} & \text{4}\\ \text{Passenger Vans U} & \text{13} & \text{3} & \text{0} & \text{3} & \text{7}\\ \text{Midsize Cars V} & \text{15} & \text{3} & \text{5} & \text{0} & \text{7}\\ \text{Large Cars W} & \text{19} & \text{9} & \text{5} & \text{3} & \text{2}\\ \end{matrix}

Compute the probability distributions and expected values of Z and V. Based on the results, which of the two types of vehicle performed better in frontal crashes?

Solution

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Answered 2 years ago
Answered 2 years ago
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Given the frequency table for a random variable, we need to calculate its expected value.

Use the given table to calculate P(Z=z)P(Z=z) and P(V=v)P(V=v), and

the expected value of both variables.

The vehicle type represented by a variable with the higher expected value has better performance.

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