Answer: (a) 8 users
(b) 51.5%
Explanation:
(a) Bandwidth = 1000 Mbps = 1000*100 = 100000 Kbps
Employee requirement = 12 Mbps = 12*1000 = 12000 Kbps
Total users = 100000/12000 = 8 Users approx
(b) P(x>2) = 1-(P(x=0)+P(x=1)+P(x=2))
= (8CO * 0.80^8) + (8C1 * 0.80^7*0.20) + (8C2 * 0.80^6*0.20^2)
=51.5%