MATH SOLVE

8 months ago

Q:
# 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.

Accepted Solution

A:

Answer:a) 0.8132 or 81.32%b) 0.1868 or 18.68%c) 0.6116 or 61.16%d) 0.7667 minutese) 0.8667 minutesf) 0.1 minutesStep-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)
[tex]\large P(X<1.31)=\int_{0}^{1.31}f(x)dx=\int_{0}^{1.31}(0.85-0.35x)dx=\\\\=0.85\int_{0}^{1.31}dx-0.35\int_{0}^{1.31}xdx=\\\\=0.85*1.31-0.35*\frac{(1.31)^2}{2}=0.8132[/tex]
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)
[tex]\large P(0.25<X<1.31)=\int_{0.25}^{1.31}f(x)dx=\int_{0}^{1.31}f(x)dx-\int_{0}^{0.25}f(x)dx=\\\\=0.8132-\int_{0}^{0.25}(0.85-0.35x)dx=0.8132-0.85\int_{0}^{0.25}dx+0.35\int_{0}^{0.25}xdx=\\\\=0.8132-0.85*0.25+0.35*\frac{(0.25)^2}{2}=0.6116[/tex]
d)
the expected time it will take a trainee to complete the task
is E(X)
[tex]\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[/tex]
e)
[tex]\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[/tex]
f)
[tex]\large Var(X)=E(X^2)-(E(X))^2=0.8667-0.7667=0.1000[/tex]