1. An automobile insurer has found that repair claims for costs of small acciden

Question

1. An automobile insurer has found that repair claims for costs of small accidents are at an average of $1200 with a standard deviation of $300. Suppose that the next 36 small claims can be regarded as a random sample from the small claim process. 7.1) Find the probability that the total amount of the next 36 small claims cost less than $20,000. 7.2) How much money should the insurer reserve for the next 25 small claims in order to keep the coverage probability at 99%? Is $60,000 enough? How about $30,000?

Explanation / Answer

1) expected mean of total amount =36*1200=43200

ans std deviation of total of 36 cars =300*(36)1/2 =1800

7.1) probability that the total amount of the next 36 small claims cost less than $20,000 =P(X<20000)

=P(Z<(20000-43200)/1800) =P(Z<-12.8889) =0.0000

7.12)for 25 claims mean =1200*25=30000

=300*(25)1/2 =1500

as for 99 percentile, zscore =2.3263

therefore reserve for the next 25 small claims in order to keep the coverage probability at 99%

=mean +z*std deviation=30000+2.3263*1500 =33489.52

hence 60000 would be enough and 3000 will not be enough,

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