The probability distribution for the number of automobiles sold during a day (x) at Bob Iron Motors is as follows. x f(x) 0 0.001 1 0.007 2 0.034 3 0.099 4 0.188 5 6 0.220 7 0.136 8 0.055 9 0.015 10 0.001 24 Suppose the gross profit per vehicle sold is $1,820 . The standard deviation of gross profit is, a $3,105.50 b $3,029.76 c $2,955.86 d $2,883.77
The probability distribution for the number of automobiles sold during a day (x) at Bob Iron Motors is as follows. x f(x) 0 0.001 1 0.007 2 0.034 3 0.099 4 0.188 5 6 0.220 7 0.136 8 0.055 9 0.015 10 0.001 24 Suppose the gross profit per vehicle sold is $1,820 . The standard deviation of gross profit is, a $3,105.50 b $3,029.76 c $2,955.86 d $2,883.77
Managerial Economics: A Problem Solving Approach
5th Edition
ISBN:9781337106665
Author:Luke M. Froeb, Brian T. McCann, Michael R. Ward, Mike Shor
Publisher:Luke M. Froeb, Brian T. McCann, Michael R. Ward, Mike Shor
Chapter17: Making Decisions With Uncertainty
Section: Chapter Questions
Problem 8MC
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Question
The probability distribution for the number of automobiles sold during a day (x) at Bob Iron Motors is as follows. |
x | f(x) |
0 | 0.001 |
1 | 0.007 |
2 | 0.034 |
3 | 0.099 |
4 | 0.188 |
5 | |
6 | 0.220 |
7 | 0.136 |
8 | 0.055 |
9 | 0.015 |
10 | 0.001 |
24 | Suppose the gross profit per vehicle sold is | $1,820 | . The standard deviation of gross profit is, |
a | $3,105.50 |
b | $3,029.76 |
c | $2,955.86 |
d | $2,883.77 |
Expert Solution
Step 1
0.001+0.007+0.034+0.099+0.188+X1+0.220+0.136+0.055+0.015+0.001=1
X1=1-0.756=0.244
Mean=/
=(0.0010+0.0071+0.0342+0.0993+0.1884+0.2445+0.2206+0.1367+0.0558+0.0159+0.00110) /1
=5.20
variance()=
=(0-5.20)2*0.001+(1-5.20)2*0.007+(2-5.20)2*0.034+(3-5.20)2*0.099+(4-5.20)2*0.188+(5-5.20)2*0.244+(6-5.20)2*0.220+(7-5.20)2*0.136+(8-0.520)2*0.055+(9-5.20)2*0.015+(10-5.20)2*0.001
=2.4999
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