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A message of n bytes is transmitted through a communication channel.
Each byte is distorted with probability p.
What is the probability that more than k distorted bytes will be received in the transmitted message?
Use the Poisson theorem. (p = 0.01, n = 677, k = 6)
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Poisson theorem is irrelevant; so, the instruction in the post is inconsistent.
We should ignore it. Another method should be applied. See my solution below.
In this problem, the original distribution is binomial.
It has very large number of trials n=677.
Therefore, using direct formulas for binomial distribution is not recommended: it is a bad style.
Good style is to use a normal distribution as an approximation to the binomial.
This normal distribution has the mean value of m = n*p = 677*0.01 = 6.77
and the standard deviation SD = = = 2.5889.
Now we should estimate the area under this normal curve on the right of z-score value k= 6.
More precisely, we must find normalcdf(-9999, 5.5, 6.77, 2.5889).
Value 5.5 is the continuing approximation to k= 6.
z1 z2 mean SD <<<---=== formatting pattern
Thus the final probability is P = normalcdf(-9999, 5.5, 6.77, 2.5889) = 0.6881. ANSWER
Solved.