excans:237dbe9dad: Difference between revisions

Latest revision as of 22:39, 2 May 2023

Solution: D

The cumulative distribution function for the exponential distribution is

[[math]] F(x) = 1-e^{-\lambda x} = 1-e^{-x/\mu} = 1-e^{-x/100}, x \gt 0. [[/math]]

From the given probability data,

[[math]] \begin{align*} F(50) - F(40) &= F(r) - F(60) \\ 1-e^{-50/100} - (1-e^{-40/100}) &= 1-e^{-r/100} - (1-e^{-60/100}) \\ e^{-40/100} - e^{-50/100} &= e^{-60/100} - e^{-r/100} \\ e^{-r/100} &= e^{-60/100} - e^{-40/100} + e^{-50/100} = 0.4850 \\ -r/100 &= \ln(0.4850) = -0.7236 \\ r &= 72.36. \end{align*} [[/math]]

@@ Line 1: / Line 1: @@
-'''Solution: B'''
+'''Solution: D'''
-Let <math>X</math> be the number of burglaries. Then,
+The cumulative distribution function for the exponential distribution is
+<math display =  "block">
+F(x) = 1-e^{-\lambda x} = 1-e^{-x/\mu} = 1-e^{-x/100}, x > 0.
+</math>
+From the given probability data,
 <math display = "block">
 \begin{align*}
-\operatorname{E}(X | X \geq 2) = \frac{\sum_{x=2}^{\infty}xp(x)}{1-p(0)-p(1)} &= \frac{\sum_{x=0}^{\infty}xp(x)-(0)p(0)-(1)p(1)}{1-p(0)-p(1)} \\
+F(50) - F(40) &= F(r) - F(60) \\
-&=  \frac{1-p(1)}{1-p(0)-p(1)} = \frac{1-e^{-1}}{1-e^{-1}-e^{-1}} = 2.39.
+-e^{-50/100} - (1-e^{-40/100}) &= 1-e^{-r/100} - (1-e^{-60/100}) \\
+e^{-40/100} - e^{-50/100} &= e^{-60/100} - e^{-r/100} \\
+e^{-r/100} &= e^{-60/100} - e^{-40/100} + e^{-50/100} = 0.4850 \\
+-r/100 &= \ln(0.4850) = -0.7236 \\
+r &= 72.36.
 \end{align*}
 </math>
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