Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 30E from Chapter 3 from Devore's Probability and Statistics for Engineering and Science.
Problem 30E
Chapter:
Problem:
An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is ... a. Compute E(Y). b. Suppose an individual with Y violations incurs a surcharge of $100Y2. Calculate the expected amount of the surcharge.
Step-by-Step Solution
Step 1
We are given with probability distribution function:\[p\left( y \right) = \left\{ {\begin{array}{*{20}{c}}{0.60}&0\\{0.25}&1\\{0.10}&2\\{0.05}&3\end{array}} \right.\]
Step 2: (a)
Expected value of x \[\begin{array}{l}E\left( x \right) = \Sigma xp\left( x \right)\\ = \left( {0 \times 0.60} \right) + \left( {1 \times 0.25} \right) + \left( {2 \times 0.10} \right) + \left( {3 \times 0.05} \right)\\ = 0 + 0.25 + 0.2 + 0.15\\ = 0.60\end{array}\]Therefore, \[E\left( x \right) = 0.60\]
Step 3: (b)
Expected amount of surcharge 100Y$^2$\[\begin{array}{l}E\left( {100{Y^2}} \right) = 100E\left( {{Y^2}} \right)\\ = 100\left( {\sum {{y^2}p\left( y \right)} } \right)\\ = 100\left( {0.6 \times {0^2} + 0.25 \times {1^2} + 0.1 \times {2^2} + 0.05 \times {3^2}} \right)\\ = 100\left( {0 + 0.25 + 0.4 + 0.45} \right)\\ = 100\left( {1.1} \right)\\ = 110\end{array}\]Therefore, \[E\left( {100{Y^2}} \right) = 110\]
We are given with probability distribution function:\[p\left( y \right) = \left\{ {\begin{array}{*{20}{c}}{0.60}&0\\{0.25}&1\\{0.10}&2\\{0.05}&3\end{array}} \right.\]
Step 2: (a)
Expected value of x \[\begin{array}{l}E\left( x \right) = \Sigma xp\left( x \right)\\ = \left( {0 \times 0.60} \right) + \left( {1 \times 0.25} \right) + \left( {2 \times 0.10} \right) + \left( {3 \times 0.05} \right)\\ = 0 + 0.25 + 0.2 + 0.15\\ = 0.60\end{array}\]Therefore, \[E\left( x \right) = 0.60\]
Step 3: (b)
Expected amount of surcharge 100Y$^2$\[\begin{array}{l}E\left( {100{Y^2}} \right) = 100E\left( {{Y^2}} \right)\\ = 100\left( {\sum {{y^2}p\left( y \right)} } \right)\\ = 100\left( {0.6 \times {0^2} + 0.25 \times {1^2} + 0.1 \times {2^2} + 0.05 \times {3^2}} \right)\\ = 100\left( {0 + 0.25 + 0.4 + 0.45} \right)\\ = 100\left( {1.1} \right)\\ = 110\end{array}\]Therefore, \[E\left( {100{Y^2}} \right) = 110\]