Answer:
a) E(X) = 16.09 ft³
E(X²) = 262.22 ft⁶
Var(X) = 3.27 ft⁶
b) E(22X) = 354 dollars
c) Var(22X) = 1,581 dollars
d) E(X - 0.01X²) = 13.470 ft³
Explanation:
The complete Correct Question is presented in the attached image to this solution.
a) Compute E(X), E(X2), and V(X).
The expected value of a probability distribution is given as
E(X) = Σxᵢpᵢ
xᵢ = Each variable in the distribution
pᵢ = Probability of each distribution
Σxᵢpᵢ = (13.5×0.20) + (15.9×0.59) + (19.1×0.21)
= 2.70 + 9.381 + 4.011
= 16.092 = 16.09 ft³
E(X²) = Σxᵢ²pᵢ
Σxᵢ²pᵢ = (13.5²×0.20) + (15.9²×0.59) + (19.1²×0.21)
= 36.45 + 149.1579 + 76.6101
= 262.218 = 262.22 ft⁶
Var(X) = Σxᵢ²pᵢ - μ²
where μ = E(X) = 16.092
Σxᵢ²pᵢ = E(X²) = 262.218
Var(X) = 262.218 - 16.092²
= 3.265536 = 3.27 ft⁶
b) E(22X) = 22E(X) = 22 × 16.092 = 354.024 = 354 dollars to the nearest whole number.
c) Var(22X) = 22² × Var(X) = 22² × 3.265536 = 1,580.519424 = 1,581 dollars
d) E(X - 0.01X²) = E(X) - 0.01E(X²)
= 16.092 - (0.01×262.218)
= 16.0926- 2.62218
= 13.46982 = 13.470 ft³
Hope this helps!!!