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.

Respuesta :

Answer:

Step-by-step explanation:

Given that 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.

[tex]f(x) = 0.67 - 0.17x, 0 < x < 2[/tex]

Hence F(x) = [tex]\int\limits^x_0 {f(x)} \, dx[/tex]= cumulative distributive function for x between 0 and 2.

1)  the probability a trainee will complete the task inless than 1 minutes

=F(1) = 0.67-0.085 = 0.585

2. the probability that a trainee will complete the task inmore than 1 minutes

=1-F(1) = 0.415

3. the probability it will take a trainee between 0.68 minutes and 1 minutes to complete the task

=F(1)-F(0.68) =0.168704

=0.1687

4.  the expected time it will take a trainee to complete thetask

=[tex]\int\limits^2_0 {x(0.67-0.17x)} \, dx \\=0.886667[/tex]

=0.8867

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

E(x^2) =

[tex]\int\limits^2_0 {x^2(0.67-0.17x)} \, dx \\=1.106667[/tex]

=1.1067

6.If X represents the time it takes to complete the task, what isVar(X)

Var(x) = E(x^2)-Mean^2

= 1.106667-0.886667^2

= 0.320489

=0.3205