Trainees must complete a specific task in less than 2 minutes. Consider the probability density function below for the time ittakes a trainee to complete the task.

f (x) = 0.67 - 0.17x 0 < x < 2

1. What is the probability a trainee will complete the task inless than 1 minutes? Give your answer to four decimalplaces.

2. What is the probability that a trainee will complete the task inmore than 1 minutes? Give your answer to four decimal places.

3. What is the probability it will take a trainee between 0.68minutes and 1 minutes to complete the task? Give your answer tofour decimal places.

4. What is the expected time it will take a trainee to complete thetask? Give your answer to four decimal places

5. If X represents the time it takes to complete the task, what isE (X2) ? Give your answer to four decimal places.

6. If X represents the time it takes to complete the task, what isVar (X) ? Give your answer to four decimal places.

1) P (X<1) = 0.5850

2) P (X>1) = 0.41500

3) P (0.68 < X < 1) = 0.0013

4) E (X) = 0.8867 minute

5) E (X²) = 1.1064

6) Var (X) = 0.3202

Step-by-step explanation:

Probability density function = f (x) = 0.67 - 0.17x (for 0 < x < 2)

1) The probability a trainee will complete the task inless than 1 minutes is given as P (X<1)

And,

P (X<1) = 1 - P (X≥1)

P (X≥1) = ∫²₁ f (x) dx = ∫²₁ (0.67 - 0.17x) dx = [0.67x - 0.085x²]²₁ = [0.67 (2) - 0.085 (2²) ] - [0.67 (1) - 0.085 (1²) ] = (1) - (0.585) = 0.415

P (X<1) = 1 - P (X≥1) = 1 - 0.415 = 0.5850

2) The probability that a trainee will complete the task inmore than 1 minutes is given as P (X>1)

And

P (X>1) = 1 - P (X≤1)

P (X≤1) = ∫¹₀ f (x) dx = ∫¹₀ (0.67 - 0.17x) dx = [0.67x - 0.085x²]¹₀ = (0.67 (1) - 0.085 (1²)) = 0.67 - 0.085 = 0.585

P (X>1) = 1 - P (X≤1) = 1 - 0.585 = 0.4150

3) The probability it will take a trainee between 0.68minutes and 1 minutes to complete the task. P (0.68 < X < 1)

P (0.68 < X < 1) = P (X0.68)

P (X<1) = 0.5850 (from question number 1)

P (X>0.68) = 1 - P (X≤0.68)

P (X≤0.68) = ∫⁰•⁶⁸₀ f (x) dx = ∫⁰•⁶⁸₀ (0.67 - 0.17x) dx = [0.67x - 0.085x²]⁰•⁶⁸₀ = (0.67 (0.68) - 0.085 (0.68²)) = 0.4556 - 0.039304 = 0.416296 = 0.4163

P (X>0.68) = 1 - P (X≤0.68) = 1 - 0.4163 = 0.5837

P (0.68 < X < 1) = P (X0.68) = 0.585 - 0.5837 = 0.001296 = 0.0013 to 4d. p

4) Expected value = E (X) = Σ xᵢpᵢ = ∫²₀ x f (x) dx (that is, a sum of all the products of possible values and their respective probabilities)

∫²₀ x f (x) = ∫²₀ x (0.67 - 0.17x) = ∫²₀ (0.67x - 0.17x²) dx = [0.335x² - 0.0567x³]²₀ = [0.335 (2²) - 0.0567 (2³) ] = 1.34 - 0.4533 = 0.8867 minute.

E (X) = ∫²₀ x f (x) = 0.8867 minute

5) E (X²) = Σ xᵢ²pᵢ = ∫²₀ x² f (x)

∫²₀ x² f (x) = ∫²₀ x² (0.67 - 0.17x) = ∫²₀ (0.67x² - 0.17x³) dx = [0.2233x³ - 0.0425x⁴]²₀ = [0.2233 (2) ³ - 0.0425 (2) ⁴] = 1.7864 - 0.68 = 1.1064

6) Variance = Var (X) = Σx²p - μ²

where μ = E (X) = 0.8867 minute

Σx²p = ∫²₀ x² f (x) = 1.1064

Var (X) = 1.1064 - 0.8867² = 0.3202