Answer:
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
