Answer:
P(T₀ < 200) = 0.99856
P(150 < T₀ < 200) = 0.99856
Step-by-step explanation:
The expected values for each of the tasks is μ₁ = 60, μ₂ = 60, μ₃ = 60
The variances for each of the 3 tasks
σ₁² = 15, σ₂² = 15, σ₃² = 15
calculate P(T₀ < 200) and P(150 < T₀ < 200)
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)
Distribution of total time taken for the 3 successive tasks
= X₁ + X₂ + X₃
Expected value = Combined Mean = μ₁ + μ₂ + μ₃ = 60 + 60 + 60 = 180
Combined Variance = 1²σ₁² + 1²σ₂² + 1²σ₃²
= (1² × 15) + (1² × 15) + (1² × 15)
= 45
standard deviation of the combined distribution = √(variance) = √45 = 6.708
Since each of the distributions are said to be normal, the combined distribution too, is normal.
P(T₀ < 200)
We first standardize 200
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (200 - 180)/6.708 = 2.98
The required probability
= P(T₀ < 200) = P(z < 2.98)
We'll use data from the normal probability table for these probabilities
P(T₀ < 200) = P(z < 2.98) = 0.99856
b) P(150 < T₀ < 200)
We first standardize 150 and 200
For 150
z = (x - μ)/σ = (150 - 180)/6.708 = -4.47
For 200
z = (x - μ)/σ = (200 - 180)/6.708 = 2.98
The required probability
= P(150 < T₀ < 200) = P(-4.47 < T₀ < 2.98)
We'll use data from the normal probability table for these probabilities
P(150 < T₀ < 200) = P(-4.47 < T₀ < 2.98)
= P(z < 2.98) - P(z < -4.47)
= 0.99856 - 0.0000 = 0.99856
Hope this Helps!!!
