251probex1 12/01/06 (Open this document in 'Page Layout' view!)
Real Estate Example – Functions of a Random Variable
Problem:
We are selling real estate. x represents the number of sales each week. Our sales have the discrete uniform
distribution between 0 and 2.
x
Px 
0
1/3
1
1/3
2
1/3
1.0
a) Find the expected value and variance of our sales.
b) If we receive a combination of salary and commission of $200 a week plus $2000 for each sale find the
expected value and variance of our sales.
Solution:
a)
x
Px 
x P x 
x 2 P x 
0
1/3
0
0
1
1/3
1/3
1/3
2
1/3
2/3
4/3
1.0
1
5/3
From the table above the expected value of our sales is  x  E x  

  
 x  Px =1 and the variance is
 x2  Varx   E x   2  E x 2   x2  5 / 3  12  2 / 3 . (Note that  x  2 3  0.8165 .)
b) The hard way: We receive a combination of salary and commission of $200 a week plus $2000 for each
sale. If we represent our earnings by y , we can compute the mean and variance of our earnings the hard
way by noting that for no sales we receive only the $200 salary, for one sale we receive the $200 salary
plus $200 and for 2 sales we receive $200 salary plus $2000 commissi0on.
P y 
y P y 
y 2 P y 
1/3
200/3
40000/3
1/3
2200/3
4840000/3
1/3
4200/3
17640000/3
1.0
6600/3
22520000/3
y  P y   6600 / 3  2200 , and that E y 2  22520000/ 3  7506666.67 ,
So we can write  y  E  y  
y
200
2200
4200
so that Var y    y2 
 

 y Py  
2
2
y
 750666.67  22002  2666666.67 .
The easy way: We can do this the easy way by noting that the formula relating sales and earnings is
y  2000x  200 , which has the form y  ax  b , where a  2000 and b  200 . And that, from the
formulas in the outline  y  Eax  b  aEx   b  2000 1  200  2200 , and Var  y    y2 
 
Varax  b  a 2Varx   2000 2 2 3  2666666 .67 (  y  1632.9932 .)