A shipping company handles containers in three different sizes: (1) 27 ft3 (3 × 3 × 3), (2) 125 ft3, and (3) 512 ft3. Let Xi (i = 1, 2, 3) denote the number of type i containers shipped during a given week. With μi = E(Xi) and σi2 = V(Xi), suppose that the mean values and standard deviations are as follows:

μ1 = 230 μ2 = 240 μ3 = 120
σ1 = 11 σ2 = 12 σ3 = 7

Assuming that X1, X2, X3 are independent, calculate the expected value and variance of the total volume shipped.

Expected value of the total volume shipped = 97,650 ft³

Variance of the total volume shipped = 15,183,265

Standard deviation = 3896.6 ft³

Step-by-step explanation:

The mean of number of type 1, 2 and 3 containers in a week

μ₁ = 230, μ₂ = 240, μ₃ = 120

The standard deviations for the number of type 1, 2 and 3 containers in a week

σ₁ = 11, σ₂ = 12, σ₃ = 7

When independent distributions are combined, the combined mean and combined variance are given through the relation

Combined mean = Σ λᵢμᵢ

(summing all of the distributions in the manner that they are combined)

Combined variance = Σ λᵢ²σᵢ²

(summing all of the distributions in the manner that they are combined)

Volume of each container type

λ₁ = 27 ft³

λ₂ = 125 ft³

λ₃ = 512 ft³

Distribution of total volume shipped

= 27X₁ + 125X₂ + 512X₃

Expected value = Combined Mean = 27μ₁ + 125μ₂ + 512μ₃

= (27×230) + (125×240) + (512×120) = 590

Combined Variance = 27²σ₁² + 125²σ₂² + 512²σ₃²

= (27² × 11²) + (125² × 12²) + (512² × 7²)

= 88,209 + 2,250,000 + 12,845,056

= 15,183,265

Standard deviation = √(15,183,265) = 3896.6 ft³

Hope this Helps!!!

i think it would be 42cm