Question

just solve c) and d)

42 To o data communica-libo Silent several metodes that oude o mode ave bendeed into a packet before they ove transmitted over The network. Assume that the mesonges anive at the node Decordis Do Dois process 12go mesnesiminFise messages are used to form a packet a what she mean time until a pocket is formed, that is, until I live messages have arrived at the node? b) What is the standard deviation of the time until a packet is formed? o what is the probability that a packet is formed in less than 10 secs? d what is the probability that a Loacket is formed in less than 5 seconds? n e

Answer 1

5 messages are used to form a packet. According to the problem, the messages arrive according to a Poisson process as 30 messages / min. 0 Let, x be a random variable that denotes the no. of messages arriving at the nodex follows poisson distribution with parameter x and pdf, fullwys en o , X=011; 21 - -- 30 (€ Here, A packet is formed in less than 10 seconds.i.e, at least 5 masseges has - come in 10 seconds. here, a 0 2 5 Tas, 1 min = 60 see Now, the probability that a packet will be formed in less than 10 seconds is cs P(x>5) - 1-P (*<5)

= 1 - P(x = 0) – P(0-1) -P(p = 2) – P(423) . . -P(+24) tes[l+5 tap ] 21-23T1+5 +12.5 + 20.83 + 26.04) 21-le-s *65.37) 21-0.4404 > 0.5595 (offenex) -. The required probability is 0.5595. -5 X 65.37) @ Here, a packet is formed less than 5 seconds.ie, at least 5 messages has come in less than 5 seconds. here, so 30x5 • 2.5 Now, the probability that a packet will be formed in less than 5 seconds is - CamScanner

=1 - P(x<5) -1- [p(=0) + f(*=1) +F(*22) + P(0-3)+ P(304)] -1-6-25Tit 2:5 + 2 to 7 = 1- 625 53.5+3.125 + 2.604 + 1.63] 21-12-215 210.86 ) 31-0.891 = 0.109 (apperox ) |:. The required probability is 0.109. 000 rs Scanned with CamScanner