Question 1050271
a. {{{lambda = (3mb)*(1/(10mb)) = 0.3}}}error.

(i) The pmf is {{{p(x) = e^(-lambda)*((lambda)^x/(x!))}}} ====> {{{p(0) = e^(-lambda)*((lambda)^0/(0!)) = e^(-0.3) = 0.74}}}, to two significant figures.

(ii) {{{p(x >= 2) = 1-p(0) - p(1) = 1- e^(-0.3) -0.3* e^(-0.3) = 0.037}}}


b. Let s = number of megabytes transmitted.
===> {{{lambda = s/10}}}
===> {{{p(x) = e^(-s/10)*((s/10)^x/(x!))}}} 
===> {{{p(0) = e^(-s/10)*((s/10)^0/(0!)) = e^(-s/10)}}}.

We want {{{e^(-s/10) = 0.95}}}.

===> s = -10ln0.95 = 0.51 megabyte of files transmitted, and the statement is proved.


c. The average is {{{lambda = (1000mb)*(1/(10mb))}}} = 100 errors.