Question

Trainees 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) 0.8132 or 81.32%

b) 0.1868 or 18.68%

c) 0.6116 or 61.16%

d) 0.7667 minutes

e) 0.8667 minutes

f) 0.1 minutes

Step-by-step explanation:

a)

If f(x) = 0.85-0.35x (0<x<2) is the PDF and X is the random variable that measures the time it takes for a trainee to complete the task, the probability a trainee will complete the task in less than 1.31 minutes is P(X<1.31)

$\large P(X$

b)

The probability that a trainee will complete the task in more than 1.31 minutes is  

P(X>1.31) = 1 - P(X<1.31) = 1 - 0.8132 = 0.1868

c)

The probability it will take a trainee between 0.25 minutes and 1.31 minutes to complete the task is P(0.25<X<1.31)  

$\large P(0.25$

d)

the expected time it will take a trainee to complete the task

is E(X)

$\large E(X)=\int_{0}^{2}xf(x)dx=\int_{0}^{2}x(0.85-0.35x)dx=0.85\int_{0}^{2}xdx-0.35\int_{0}^{2}x^2dx=\\\\=0.85*\frac{2^2}{2}-0.35*\frac{2^3}{3}=0.7667\;minutes$

e)

$\large E(X^2)=\int_{0}^{2}x^2f(x)dx=\int_{0}^{2}x^2(0.85-0.35x)dx=0.85\int_{0}^{2}x^2dx-0.35\int_{0}^{2}x^3dx=\\\\=0.85*\frac{2^3}{3}-0.35*\frac{2^4}{4}=0.8667$

f)

$\large Var(X)=E(X^2)-(E(X))^2=0.8667-0.7667=0.1000$