A binary message m, where m is equal either to 0 or to 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 
denote the variance of E. What must be the value of ?2 so that the probability of error when m = 0 is 0.01?
(a) If m= 0, then X = E, so X ~ N(0, 0.25).P(error) = P(X>0.5). The z-score of 0.5 is 
.
The area to the right of z= 1.00 is 1 ? 0.8413 = 0.1587.
Therefore P(error) = 0.1587.
Therefore P(error) = 0.1587.
(b) If m= 0, then X = E, so X ~ N(0, ?2).
P(error) = P(X >0.5) = 0.01. The z-score of 0.5 is (0.5 ? 0)/?.
Since P(X >0.5) = 0.01, the z-score of 0.5 is 2.33.
Therefore (0.5 ? 0)/?= 2.33. Solving for ?yields ?= 0.2146, and 
= 0.04605.
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