Consider a small ferry that can accommodate cars and buses. The toll for cars is $3, and the toll for buses is $10. Let X and Y denote the number of cars and buses, respectively, carried on a single trip. Suppose the joint distribution of X and Y is as given in the table below.
y | ||||
p(x,y) | 0 | 1 | 2 | |
x | 0 | 0.025 | 0.015 | 0.010 |
1 | 0.050 | 0.030 | 0.020 | |
2 | 0.105 | 0.075 | 0.050 | |
3 | 0.150 | 0.090 | 0.060 | |
4 | 0.100 | 0.060 | 0.040 | |
5 | 0.050 | 0.030 | 0.040 |
Compute the expected revenue from a single trip. (Round your
answer to two decimal places.)
$
X | P(X) | X.P(X) |
0 | 0.025+0.015+0.010 = 0.05 | 0.00 |
1 | 0.050+0.030+0.020 = 0.10 | 0.10 |
2 | 0.105+0.075+0.050 = 0.23 | 0.46 |
3 | 0.150+0.090+0.060 = 0.30 | 0.90 |
4 | 0.100+0.060+0.040 = 0.20 | 0.80 |
5 | 0.050+0.030+0.040 = 0.12 | 0.60 |
Total | 1.00 | 2.86 |
E(x) = Ʃ x*p(x) = 2.86
Y | P(Y) | Y.P(Y) |
0 | 0.025+0.050+0.105+0.150+0.100+0.050 = 0.48 | 0.00 |
1 | 0.015+0.030+0.075+0.090+0.060+0.030 = 0.30 | 0.30 |
2 | 0.010+0.020+0.050+0.060+0.040+0.040 = 0.22 | 0.44 |
Total | 1.00 | 0.74 |
E(y) = Ʃ y*p(y) = 0.74
Expected revenue for a single trip = $3* 2.86 + $10*0.74 =
$15.98
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