Question

rainees must complete a specific task in less than 2 minutes. Consider the probability density function below for the time it takes a trainee to complete the task.f(x) = 0.85 - 0.35x0 < x < 2a) What is the probability a trainee will complete the task in less than 1.31 minutes? Give your answer to four decimal places.b) What is the probability that a trainee will complete the task in more than 1.31 minutes? Give your answer to four decimal places.c) What is the probability it will take a trainee between 0.25 minutes and 1.31 minutes to complete the task? Give your answer to four decimal places.d) What is the expected time it will take a trainee to complete the task? Give your answer to four decimal places.e) If X represents the time it takes to complete the task, what is E(X2)? Give your answer to four decimal places.f) If X represents the time it takes to complete the task, what is Var(X)? Give your answer to four decimal places.

Answer 1

Answer:

a) P(0 < x < 1.31) = 0.8132

b) P(1.31 < x < 2) = 0.1868

c) P(0.25 < x < 1.31) = 0.6116

d) E(X) = 0.7667

e) E(X²) = 0.8667

f) Var(X) = 0.2789

Step-by-step explanation:

Probability density function = f(x) = 0.85 - 0.35x for 0 < x < 2

Note that the values obtained here are the closest mathematical estimates. A greater/less than or equal to sign would make it all more accurate.

a) probability a trainee will complete the task in less than 1.31 minutes

P(0 < x < 1.31) = ∫¹•³¹₀ f(x) dx

= ∫¹•³¹₀ (0.85 - 0.35x) dx

= [0.85x - 0.175x²]¹•³¹₀

= [(0.85)(1.31) - (0.175)(1.31²)]

= 1.1135 - 0.3003175

= 0.8131825 = 0.8132 to 4 d.p.

b) probability that a trainee will complete the task in more than 1.31 minutes

P(1.31 < x < 2) = ∫²₁.₃₁ f(x) dx

= ∫²₁.₃₁ (0.85 - 0.35x) dx

= [0.85x - 0.175x²]²₁.₃₁

= [(0.85)(2) - (0.175)(2²)] - [(0.85)(1.31) - (0.175)(1.31²)]

= (1.7 - 0.7) - (1.1135 - 0.3003175) = 0.1868175 = 0.1868

c) probability it will take a trainee between 0.25 minutes and 1.31 minutes to complete the task

P(0.25 < x < 1.31) = ∫¹•³¹₀.₂₅ f(x) dx

= ∫¹•³¹₀.₂₅ (0.85 - 0.35x) dx

= [0.85x - 0.175x²]¹•³¹₀.₂₅

= [(0.85)(1.31) - (0.175)(1.31²)] - [(0.85)(0.25) - (0.175)(0.25²)]

= 0.8131825 - 0.2015625 = 0.61162 = 0.6116

d) Expected value = ∫ xf(x) dx [with the integral done all over the sample spaces, for this question, it is done over the interval (0,2)]

E(X) = ∫²₀ x(0.85 - 0.35x) dx

= ∫²₀ (0.85x - 0.35x²) dx

= [0.425x² - 0.1167x³]²₀

= [(0.425)(2²) - (0.1167)(2³)]

= [1.7 - 0.9333333] = 0.7667

e) X represents the time it takes to complete the task, what is E(X²)

E(X²) = ∫ x²f(x) dx [with the integral done all over the sample spaces, for this question, it is done over the interval (0,2)]

E(X) = ∫²₀ x²(0.85 - 0.35x) dx

= ∫²₀ (0.85x² - 0.35x³) dx

= [0.2833x³ - 0.0875x⁴]²₀

= [(0.2833)(2³) - (0.0875)(2⁴)]

= [2.2667 - 1.4] = 0.8667

f) Variance of continuous function is given as

Var(X) = E(X²) - [E(X)]²

Var(X) = 0.8667 - 0.7667² = 0.2789

Hope this Helps!!!

Answer 2

Answer:

a. P(x<1.31) = 0.8132

b. P(x>1.31) = 0.1868

c. P(0.25<x<1.31) = 0.6116

d. E(x) = 0.7667

e. E(x²) = 0.8667

f. Var(x) = 0.2789

Step-by-step explanation:

Given

f(x) = 0.85 - 0.35x 0 < x < 2

a. P(x<1.31).

This means we're working with interval of 0 to 1.31 minutes translating to Rainees completing the task between the interval 0 to 1.31 minutes

This is calculated as follows;

P(x<1.31) = ∫ f(x)dx {0,1.31}

P(x<1.31) = ∫ (0.85 - 0.35x)dx {0,1.31}

P(x<1.31) = (0.85x - 0.35x²/2){0,1.31}

P(x<1.31) = (0.85(1.31) - 0.35(1.31)²/2)

P(x<1.31) = 0.8131825

P(x<1.31) = 0.8132 ---- Approximated

b) P(x>1.31)

This means we're working with interval of 1.31 to 2 minutes translating to Rainees completing the task between the interval 1.31 to 2 minutes

This is calculated as follows;

P(x>1.31) = ∫ f(x)dx {1.31,2}

P(x>1.31) = ∫ (0.85 - 0.35x)dx {1.31,2}

P(x>1.31) = (0.85x - 0.35x²/2) {1.31,2}

P(x>1.31) = (0.85(2) - 0.35(2)²/2) - (0.85(1.31) - 0.35(1.31)²/2)

P(x>1.31) = 0.1868175

P(x>.1.31) = 0.1868 ---- Approximated

Alternatively, it can be calculated as;

P(x>1.31) = 1 - P(x<1.31)

P(x>1.31) = 1 - 0.8132

P(x>1.31) = 0.1868

c) This means we're working with interval of 0.25 to 1.31 minutes translating to Rainees completing the task between the interval 0.25 to 1.31 minutes.

This is calculated as follows

P(0.25<x<1.31) = ∫ f(x)dx {0.25,1.31}

P(0.25<x<1.31) = ∫ (0.85 - 0.35x)dx {0.25,1.31}

P(0.25<x<1.31) = (0.85x - 0.35x²/2) {0.25,1.31}

P(0.25<x<1.31) = (0.85(1.31) - 0.35(1.31)²/2) - (0.85(0.25) - 0.35(0.25)²/2)

P(0.25<x<1.31) = 0.61162

P(0.25<x<1.31) = 0.6116 ---- Approximated

d) Expected time, E(x) is calculated as

E(x) = ∫xf(x) dx

E(x) =∫ x(0.85 - 0.35x)dx {0,2}

E(x) =∫ (0.85x - 0.35x²)dx {0,2}

E(x) =(0.85x²/2 - 0.35x³/3) {0,2}

E(x) =(0.85(2)²/2 - 0.35(2)³/3)

E(x) = 0.766666666666666

E(x) = 0.7667 --- Approximated

e)

E(x²) is calculated as;

E(x²) = ∫x²f(x) dx

E(x²) =∫ x²(0.85 - 0.35x)dx {0,2}

E(x²) =∫ (0.85x² - 0.35x³)dx {0,2}

E(x²) =(0.85x³/3 - 0.35x⁴/4) {0,2}

E(x²) = (0.85(2)³/3 - 0.35(2)⁴/4)

E(x²) = 0.866666666666666

E(x²) = 0.8667 ---- Approximated

f) Var(x) = E(x²) - (E(x))²

Var(x) = 0.8667 - (0.7667)²

Var(x) = 0.27887111

Var(x) = 0.2789 --- Approximated