excans:13d8516dc1: Difference between revisions

Revision as of 21:52, 8 May 2023

Solution: D

[[math]] \begin{align*} F(x) &= \int_0^{x}2e^{-2y} dy = -e^{-2y} \Big |_0^x = 1-e^{-2x} \\ P[X \leq 0.5 | X \leq 1.0] &= \frac{P[X \leq 0.5]}{P[X \leq 1.0]} = \frac{F(0.5)}{F(1.0)} = \frac{1-e^{-1}}{1-e^{-2}} = 0.731. \end{align*} [[/math]]

@@ Line 1: / Line 1: @@
-'''Solution: E'''
+'''Solution: D'''
-The given mean of 5 years corresponds to the pdf <math>f(t) = 0.2 e^{-0.2t}</math> and the cumulative distribution function
-<math display = "block">
-F(t) = 1 - e^{-0.2t}.
-</math>
-The conditional pdf is
-<math display = "block">
-g(t) = \frac{f(t)}{F(10)} = \frac{0.2e^{-0.2t}}{1-e^{-2}}, 0 < t < 10.
-</math>
-The conditional mean is (using integration by parts)
 <math display = "block">
 \begin{align*}
-\operatorname{E}(T | T < 10) = \int_0^{10}tg(t) dt &= \int_0^{10}t \frac{0.2e^{-0.2t}}{1-e^{-2}} dt \\
+F(x) &= \int_0^{x}2e^{-2y} dy = -e^{-2y} \Big |_0^x = 1-e^{-2x} \\
-&= 0.2313 \int_0^{10}te^{-0.2t} dt \\
+P[X \leq 0.5 | X \leq 1.0] &= \frac{P[X \leq 0.5]}{P[X \leq 1.0]} = \frac{F(0.5)}{F(1.0)} = \frac{1-e^{-1}}{1-e^{-2}} = 0.731.
-&= 0.2313 \left [ t(-5e^{-0.2t} \Big |_0^{10}  - \int_0^{10}-5e^{-0.2t} dt \right ] \\
-&= 0.2323 \left [ −6.7668 + 0 - 25e^{-0.2t} \Big |_0^{10}\right] \\
-&=0.2313[−6.7668 − 3.3834 + 25] \\
-&= 3.435.
 \end{align*}
 </math>
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