Respuesta :
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.
