Let x1, x2, and x3 represent the times necessary to perform three successive repair tasks at a certain service facility. suppose they are independent, normal rv's with expected values m1, m2, and m3 and variances s21, s2, and s23, respec- tively.

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

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