×
Log in to StudySoup
Get Full Access to Applied Statistics And Probability For Engineers - 6 Edition - Chapter 3.9 - Problem 199se
Join StudySoup for FREE
Get Full Access to Applied Statistics And Probability For Engineers - 6 Edition - Chapter 3.9 - Problem 199se

Already have an account? Login here
×
Reset your password

Messages that arrive at a service center for an

Applied Statistics and Probability for Engineers | 6th Edition | ISBN: 9781118539712 | Authors: Douglas C. Montgomery, George C. Runger ISBN: 9781118539712 55

Solution for problem 199SE Chapter 3.9

Applied Statistics and Probability for Engineers | 6th Edition

  • Textbook Solutions
  • 2901 Step-by-step solutions solved by professors and subject experts
  • Get 24/7 help from StudySoup virtual teaching assistants
Applied Statistics and Probability for Engineers | 6th Edition | ISBN: 9781118539712 | Authors: Douglas C. Montgomery, George C. Runger

Applied Statistics and Probability for Engineers | 6th Edition

4 5 1 246 Reviews
12
0
Problem 199SE

Problem 199SE

Messages that arrive at a service center for an information systems manufacturer have been classified on the basis of the number of keywords (used to help route messages) and the type of message, either e-mail or voice. Also, 70% of the messages arrive via e-mail and the rest are voice.

Determine the probability mass function of the number of keywords in a message.

Step-by-Step Solution:

Solution :

Step 1 of 1:

We assume that,

X is the number of keywords in the message.

E is the email message and

V is the voice mail message.

Then the table is given below.

Number of keywords

0

1

2

3

4

E-mail

0.1

0.1

0.2

0.4

0.2

Voice

0.3

0.4

0.2

0.1

0

Our goal is:

We need to find the probability mass function of the number of keywords in a message.

 

We know that 70% of the messages arrive via e-mail and the rest are voice.

So P(V) = 0.3 and

P(E) = 0.7.

Range of x = [0, 1, 2, 3, 4].

And the probability of keywords in a mail either email or voice is given in the above table.

Using total probability theorem,

P(A) =  

Then,

P(X=0) =

P(X=0) = 0.10.7+0.30.3

P(X=0) = 0.16

P(X=1) =

P(X=1) = 0.10.7+0.40.3

P(X=1) = 0.19

P(X=2) =

P(X=2) = 0.20.7+0.20.3

P(X=2) = 0.20

P(X=3) =

P(X=3) = 0.40.7+0.10.3

P(X=3) = 0.31

P(X=4) =

P(X=4) = 0.20.7+00.3

P(X=4) = 0.14

Then the probability mass function of the number of keywords in a message is

X

0

1

2

3

4

P(X=x)

0.16

0.19

0.20

0.31

0.14



Step 2 of 1

Chapter 3.9, Problem 199SE is Solved
Textbook: Applied Statistics and Probability for Engineers
Edition: 6
Author: Douglas C. Montgomery, George C. Runger
ISBN: 9781118539712

Other solutions

People also purchased

Related chapters

Unlock Textbook Solution

Enter your email below to unlock your verified solution to:

Messages that arrive at a service center for an