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DISCLAIMER: This is one question with multiple parts. Please do all the parts an

ID: 3303481 • Letter: D

Question

DISCLAIMER: This is one question with multiple parts. Please do all the parts and show working as I do not know how to do this. Thank you!

This is the background info of the question.

Please do 2a. I'm not sure if it's correct.

Suppose that a computer network of servers experiences failures according to a Poisson process with an average of 0.8 failures per week. Let X be the number of failures in one week, so thatX Poisson 0.8), with pmf 0.81 0.. | , x=0, 1,2, f(x) = X! 0, , otherwise Recall that the expected value and variance of a Poisson random variable both equal

Explanation / Answer

from above given formula probability P(Y=0) = P(X=0) e-0.8* 0.80/0! =0.4493

and P(X=1 or 2 ) =P(X=1)+P(X=2) =e-0.8* 0.81/1! +e-0.8* 0.82/2! = 0.5032

for P(X>2) =1-P(X=0)-P(X=1 or 2) =0.0474

2 a) from above for pmf option (ii) is correct

2 b)

E(Y) =0.60

2(c) Var(Y )=0.3352

y p(y) yP(y) y2P(y) (y-)2 (y-)2P(y) 0 0.4493 0.000 0.000 0.358 0.161 1 0.5032 0.503 0.503 0.162 0.081 2 0.0474 0.095 0.190 1.965 0.093 total 1 = 0.60 0.693 2.485 2= 0.3352 std deviation=     =    2 = 0.5790