×
Get Full Access to Applied Statistics And Probability For Engineers - 6 Edition - Chapter 3.9 - Problem 199se
Get Full Access to Applied Statistics And Probability For Engineers - 6 Edition - Chapter 3.9 - Problem 199se

×

# Messages that arrive at a service center for an 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

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) = 0.1 0.7+0.3 0.3

P(X=0) = 0.16

P(X=1) = P(X=1) = 0.1 0.7+0.4 0.3

P(X=1) = 0.19

P(X=2) = P(X=2) = 0.2 0.7+0.2 0.3

P(X=2) = 0.20

P(X=3) = P(X=3) = 0.4 0.7+0.1 0.3

P(X=3) = 0.31

P(X=4) = P(X=4) = 0.2 0.7+0 0.3

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 0.2 0.31 0.14

Step 2 of 1

##### ISBN: 9781118539712

Unlock Textbook Solution