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(X<1) - P(X>0.68)

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(X<1) - P(X>0.68) = 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

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