The probability distribution for the number of automobiles sold during a day (x) at Bob Iron Motors is as follows. xf(x)00.00110.00720.03430.09940.1885 60.22070.13680.05590.015100.001 24Suppose the gross profit per vehicle sold is $1,820. The standard deviation of gross profit is,a$3,105.50b$3,029.76c$2,955.86d$2,883.77

∑f(x)=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=∑x…

Answered: The probability distribution for the…

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

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

$\sum_{} f (x) = 1$

$\Rightarrow$ 0.001+0.007+0.034+0.099+0.188+X¹+0.220+0.136+0.055+0.015+0.001=1

$\Rightarrow$ X¹=1-0.756=0.244

Mean= $\sum_{} x f (x)$ / $f (x)$

=(0.001 $\times$ 0+0.007 $\times$ 1+0.034 $\times$ 2+0.099 $\times$ 3+0.188 $\times$ 4+0.244 $\times$ 5+0.220 $\times$ 6+0.136 $\times$ 7+0.055 $\times$ 8+0.015 $\times$ 9+0.001 $\times$ 10) /1

=5.20

variance( $σ^{2}$ )= $\sum_{} {(x - μ)}^{2} p (x)$

=(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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