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A message consists of a string of bits (0s and1s). Due to

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Problem 21E Chapter 4.2

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 21E

A message consists of a string of bits (0s and1s). Due to noise in the communications channel, each bit has probability 0.3 of being reversed (i.e., a 1 will be changed to a 0 or a 0 to a 1). To improve the accuracy of the communication, each bit is sent five times, so, for example, 0 is sent as 00000. The receiverassigns the value 0 if three or more of the bits are decoded as 0. and 1 if three or more of the bits are decoded as 1. Assume that errors occur independently.

a. A 0 is sent (as 00000). What is the probability that the receiver assigns the correct value of 0?

b. Assume that each bit is sent n times, where n is an odd number, and that the receiver assigns the value decoded in the majority of the bits. What is the minimum value of n necessary so that the probability that the correct value is assigned is at least 0.90?

Step-by-Step Solution:
Step 1 of 3

Solution 21E

Step1 of 3:

Let us consider a random variable X it presents the number of bits that are reserved.

Then X follows binomial distribution with parameters “n and p” that is X B(n, p),

The probability mass function of binomial distribution is given by

, x = 0,1,2,...,n.

Where,

n = sample size

   = 5

x = random variable

p = probability of success

   = 0.3

q = 1 - p (probability of failure)

   = 1 - 0.30

   = 0.70

Here our goal is:

a).We need to find the probability that the receiver assigns the correct value of 0.

b).We need to find the minimum value of n necessary so that the probability that the correct value is assigned is at least 0.90.

Step2 of 3:

a).

We have n = 5, p = 0.3 and correct value is assigned if

Then  

 Here can be obtained from Excel by using the function  “=binomdist(X,n,p,false)”

X

P(X2)

0

0.16807

1

0.36015

2

0.3087

Total

0.83692

Therefore,

Step3 of 3:

b).

We have p = 0.3 and we need to find the smallest odd value of n.

X B(n, 0.3)

, x = 1, 3, 5, 7, 9.                              . . . . (1)              

1). Put n = 3 in above equation(1) we get

   

     is obtained from Excel by using the function  “=binomdist(X,n,p,false)”

...

X

Step 2 of 3

Chapter 4.2, Problem 21E is Solved
Step 3 of 3

Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

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