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# A binary message m, where m is equal either to 0 or 1, is ISBN: 9780073401331 38

## Solution for problem 23E Chapter 4.5

Statistics for Engineers and Scientists | 4th Edition

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Problem 23E

A binary message m, where m is equal either to 0 or 1, is sent over an information channel. Because of  noise in the channel, the message received is X, where X = m + E, and E is a random variable representing the channel noise. Assume that if X ≤ 0.5 then the receiver, concludes that m = 0 and that if X > 0.5 then the receiver concludes that m = 1. Assume that E ~ N (0,0.25).

a. If the true message is m = 0, what is the probability of an error, that is, what is the probability that the receiver concludes that m = 1?

b. Let σ2 denote the variance of E. What must be the value of σ2 so that the probability of error when m =0 is 0.01?

Step-by-Step Solution:
Step 1 of 3

Solution 23E

Step1 of 3:

We have a random variable X it presents the number of messages received.

Let X = m + E

Where,

E = a random variable representing the channel noise.

E N(0, 0.25)

m = Binary message.

If X 0.5 then the receiver concludes that m = 0

If X > 0.5 then the receiver concludes that m = 1

Here our goal is:

a).We need to find the probability of an error when true message is m = 0 and also find the probability that the receiver concludes that m = 1.

b).We need to find the value of so that the probability of error when m = 0 is 0.01 where denote the variance of E.

Step2 of 3:

a).

If m = 0 then E N(0, 0.25).

The probability of an error is given by

P(Error) = P(X > 0.5)

Let,

P(X > 0.5) =1 -  P(Z < )

=1 -  P(Z < )

= 1 - P(Z < 1)

Here P(Z < 1) is obtained from standard normal table(area under normal curve) = 0.8413 (In statistical table we have to see row 1.0 under column 0.00)

Now,

P(X > 0.5) = 1 - P(Z < 1)

= 1 - 0.8413

= 0.1587

Hence, P(Error) = 0.1587.

Step3 of 3:

b).

If m = 0 then X= E N(0,...

Step 2 of 3

Step 3 of 3

##### ISBN: 9780073401331

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