Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 22E from Chapter 5 from Devore's Probability and Statistics for Engineering and Science.
Problem 22E
Chapter:
Problem:
An instructor has given a short quiz consisting of two parts. For a randomly selected student, let X = the number of points earned on the first part and Y =the number of points earned on the second part. Suppose that the joint pmf of X and Y is given in the accompanying table. ... a. If the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score E(X + Y)? b. If the maximum of the two scores is recorded, what is the expected recorded score?
Step-by-Step Solution
Step 1
We are given with a joint probability distribution function
Step 2: (a)
The expected value of sum of two variables:\[\begin{array}{l}E\left[ {X + Y} \right] = \sum\limits_x {\sum\limits_y {\left( {x + y} \right)p\left( {x + y} \right)} } \\ = \left\{ {\begin{array}{*{20}{l}}{(0 + 0)0.02 + (0 + 5)0.06 + (0 + 10)0.02 + (0 + 15)0.1}\\{ + (5 + 0)0.04 + (5 + 5)0.15 + (5 + 10)0.2 + (5 + 15)(0.1)}\\{ + (10 + 0)0.01 + (10 + 5)0.15 + (10 + 10)0.14 + (10 + 15)0.01}\end{array}} \right\}\\ = \left\{ {\begin{array}{*{20}{l}}{0 + 5 \times 0.06 + 10 \times 0.02 + 15 \times 0.1 + 5 \times 0.04 + 10 \times 0.15 + 15 \times 0.2}\\{ + 20 \times 0.1 + 10 \times 0.01 + 15 \times 0.15 + 20 \times 0.14 + 25 \times 0.01}\end{array}} \right\}\\ = 14.1\end{array}\]Therefore, \[E\left[ {X + Y} \right] = 14.1\]
Step 3: (b)
The expected value of maximum of the two variables:\[\begin{array}{l}E\left[ {\max \left( {x,y} \right)} \right] = \sum\limits_x {\sum\limits_y {\max \left( {x + y} \right)p\left( {x + y} \right)} } \\ = \left\{ {\begin{array}{*{20}{l}}{\max (0,0) \times P\{ 0,0\} + \max \{ 0,5\} \times P\{ 0,5\} + \max \{ 0,10\} \times P\{ 0,10\} + \max \{ 0,15\} \times P\{ 0,15\} }\\{\max (5,0) \times P\{ 5,0\} + \max \{ 5,5\} \times P\{ 5,5\} + \max \{ 5,10\} \times P\{ 5,10\} + \max \{ 5,15\} \times P\{ 5,15\} }\\{\max (10,0) \times P\{ 10,0\} + \max \{ 10,5\} \times P\{ 10,5\} + \max \{ 10,10\} \times P\{ 10,10\} + \max \{ 15,15\} \times P\{ 05,15\} }\end{array}} \right\}\\ = \left\{ {\begin{array}{*{20}{l}}{0 \times 0.02 + 5 \times 0.06 + 10 \times 0.02 + 15 \times 0.1 + 5 \times 0.04 + 5 \times 0.15 + 10 \times 0.2}\\{ + 15 \times 0.1 + 10 \times 0.01 + 10 \times 0.15 + 10 \times 0.14 + 15 \times 0.01}\end{array}} \right\}\\ = 9.6\end{array}\]Therefore, \[E\left[ {\max \left( {x,y} \right)} \right] = 9.6\]
We are given with a joint probability distribution function
Step 2: (a)
The expected value of sum of two variables:\[\begin{array}{l}E\left[ {X + Y} \right] = \sum\limits_x {\sum\limits_y {\left( {x + y} \right)p\left( {x + y} \right)} } \\ = \left\{ {\begin{array}{*{20}{l}}{(0 + 0)0.02 + (0 + 5)0.06 + (0 + 10)0.02 + (0 + 15)0.1}\\{ + (5 + 0)0.04 + (5 + 5)0.15 + (5 + 10)0.2 + (5 + 15)(0.1)}\\{ + (10 + 0)0.01 + (10 + 5)0.15 + (10 + 10)0.14 + (10 + 15)0.01}\end{array}} \right\}\\ = \left\{ {\begin{array}{*{20}{l}}{0 + 5 \times 0.06 + 10 \times 0.02 + 15 \times 0.1 + 5 \times 0.04 + 10 \times 0.15 + 15 \times 0.2}\\{ + 20 \times 0.1 + 10 \times 0.01 + 15 \times 0.15 + 20 \times 0.14 + 25 \times 0.01}\end{array}} \right\}\\ = 14.1\end{array}\]Therefore, \[E\left[ {X + Y} \right] = 14.1\]
Step 3: (b)
The expected value of maximum of the two variables:\[\begin{array}{l}E\left[ {\max \left( {x,y} \right)} \right] = \sum\limits_x {\sum\limits_y {\max \left( {x + y} \right)p\left( {x + y} \right)} } \\ = \left\{ {\begin{array}{*{20}{l}}{\max (0,0) \times P\{ 0,0\} + \max \{ 0,5\} \times P\{ 0,5\} + \max \{ 0,10\} \times P\{ 0,10\} + \max \{ 0,15\} \times P\{ 0,15\} }\\{\max (5,0) \times P\{ 5,0\} + \max \{ 5,5\} \times P\{ 5,5\} + \max \{ 5,10\} \times P\{ 5,10\} + \max \{ 5,15\} \times P\{ 5,15\} }\\{\max (10,0) \times P\{ 10,0\} + \max \{ 10,5\} \times P\{ 10,5\} + \max \{ 10,10\} \times P\{ 10,10\} + \max \{ 15,15\} \times P\{ 05,15\} }\end{array}} \right\}\\ = \left\{ {\begin{array}{*{20}{l}}{0 \times 0.02 + 5 \times 0.06 + 10 \times 0.02 + 15 \times 0.1 + 5 \times 0.04 + 5 \times 0.15 + 10 \times 0.2}\\{ + 15 \times 0.1 + 10 \times 0.01 + 10 \times 0.15 + 10 \times 0.14 + 15 \times 0.01}\end{array}} \right\}\\ = 9.6\end{array}\]Therefore, \[E\left[ {\max \left( {x,y} \right)} \right] = 9.6\]