A)
P(0 Bottles) = 0.61 + 0.07 = 0.68
P(1 Bottle) = 0.17 + 0.15 = 0.32
B)
P(0 Cartons) = 0.61 + 0.17 = 0.78
P(1 Carton) = 0.07 + 0.15 = 0.22
C)
For Bottle of liquor:
X | P(X) | X.P(X) | X².P(X) |
0 | 0.68 | 0 | 0 |
1 | 0.32 | 0.32 | 0.32 |
Total | 1 | 0.32 | 0.32 |
Mean, μ = Ʃ[X.P(X)] = 0.32
Variance, σ² = Ʃ(X².P(X)) - μ² = 0.32 - 0.32² = 0.2176
D)
For cigarette cartons:
X | P(X) | X.P(X) | X².P(X) |
0 | 0.78 | 0 | 0 |
1 | 0.22 | 0.22 | 0.22 |
Total | 1 | 0.22 | 0.22 |
Mean, μ = Ʃ[X.P(X)] = 0.22
Variance, σ² = Ʃ(X².P(X)) - μ² = 0.22 - 0.22² = 0.1716
Canadians who visit the United States often buy liquor and cigarettes, which are much cheaper in...
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