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.

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

walter needs to make at least 5 more cookies but no more than 40. 5 ≤ x ≤ 40

step-by-step explanation:

20 ≤ x + 15 ≤ 55

subtract 15 from all parts

20-15 ≤ x + 15-15 ≤ 55-15

5 ≤ x ≤ 40

this means how many more cookies he has to make

walter needs to make at least 5 more cookies, but no more than 40 cookies; 5 ≤ x ≤ 40

need more info

step-by-step explanation: