(i) Here the poission parameter
= 16
so if x is the number of application received on a given day
Pr(x>= 10) = 1 - POISSONCDF (x < = 9 ; 16) = 1 - 0.0433 = 0.9567
(ii) here expected number of application to be received in 5 days = 5 * 16 = 80
so
Pr(x > 80) = 1- POISSONCDF (x < = 80 ; 80) = 1 - 0.5297 = 0.4703
(iii) expected number ofapplication tobe received in the 2 hours from 00.00 and 02.00 am = (16/24)* 2 = 4/3
Here assumption is "the number of applications received are independent of the time they are coming"
so
Pr(x = 0 ; 4/3)= e-4/3 = 0.2636
16 pointsl The National Broadband Network (NBN) is an Australian national wholesale-only, open access data network....