Question 1201237
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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.



<pre>
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 = {{{sqrt(n*p*(1-p))}}} = {{{sqrt(677*0.01*(1-0.01))}}} = 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.    <U>ANSWER</U>
</pre>

Solved.