excans:C3bbb52ac1: Difference between revisions

Latest revision as of 22:09, 6 May 2023

Solution: B

Let X and Y be the two independent losses and Z = min(X,Y). Then,

[[math]] \operatorname{P}(Z \gt z) = \operatorname{P}(X \gt z \cap Y \gt z ) = \operatorname{P}(X \gt z ) \operatorname{P}(Y \gt z) = e^{-z}e^{-z} = e^{-2z} [[/math]]

[[math]] F_Z(z) = \operatorname{P}(Z \leq z) = 1-\operatorname{P}(Z \gt z) = 1-e^{-2z} [[/math]]

which can be recognized as an exponential distribution with mean 1/2.

@@ Line 1: / Line 1: @@
 '''Solution: B'''
-Let X and Y be the miles driven by the two cars. The total cost, is then
+Let X and Y be the two independent losses and Z = min(X,Y). Then,
 <math display = "block">
-C = 3(X/15 + Y/30) = 0.2X + 0.1Y.
+\operatorname{P}(Z > z) = \operatorname{P}(X > z \cap Y > z ) = \operatorname{P}(X > z ) \operatorname{P}(Y > z) = e^{-z}e^{-z} = e^{-2z}
 </math>
-C has a normal distribution with mean 0.2(25) + 0.1(25) = 7.5 and variance 0.04(9) + 0.01(9) = 0.45. Then,
 <math display = "block">
-\operatorname{P}(C < 7) = \operatorname{P}(Z < (7-7.5)/\sqrt{0.45} = -0.7454) = 0.23.
+F_Z(z) = \operatorname{P}(Z \leq z) = 1-\operatorname{P}(Z > z) = 1-e^{-2z}
 </math>
+which can be recognized as an exponential distribution with mean 1/2.
 {{soacopyright | 2023}}