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A company offers collision insurance and liability insurance. The claim amount under collision insurance is normally distributed with mean 10, 000 and standard deviation 2, 000; the claim amount under liability insurance is normally distributed with mean 9, 000 and standard deviation 2, 000. Assuming independence, find the probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.

Answer :

Answer:

36.32% probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Two variables:

X1 and X2, probability X1 exceeds X2

[tex]\mu(X_{1} - X_{2}) = \mu_{X_{1}} - \mu_{X_{2}}[/tex]

[tex]\sigma^{2}(X_{1} - X_{2}) = \sigma_{X_{1}}^{2} +  \sigma_{X_{2}}^{2}[/tex]

In this problem, we have:

X1: Liability insurance:

So [tex]\mu_{X_{1}} = 9000, \sigma_{X_{1}} = 2000[/tex]

X2: Collusion insurance:

So [tex]\mu_{X_{2}} = 10000, \sigma_{X_{2}} = 2000[/tex]

Then

[tex]\mu(X_{1} - X_{2}) = \mu_{X_{1}} - \mu_{X_{2}} = 9000 - 10000 = -1000[/tex]

[tex]\sigma^{2}(X_{1} - X_{2}) = 2000^{2} + 2000^{2} = 8000000[/tex]

[tex]\sigma = \sqrt{8000000} = 2828.43[/tex]

Assuming independence, find the probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.

This is [tex]P(X_{1} - X_{2} > 0)[/tex], which is 1 subtracted by the pvalue of Z when X = 0. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0 - (-1000)}{2828.43}[/tex]

[tex]Z = 0.35[/tex]

[tex]Z = 0.35[/tex] has a pvalue of 0.6368

1 - 0.6368 = 0.3632

36.32% probability that the claim amount under liability insurance exceeds the claim amount under collision insurance.

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