Question

10 It is estimated that 0.44 percent of the callers to the Customer Service department of Dell Inc. will receive a busy signal. What is the probability that of today's 1,200 callers at least 5 received a busy signal? (Round your answer to 4 decimal places.) 25 Toints Probability

Answer 1

Ans We use a Poisson distribution Poisson distribution the to solve this problem probability of s'events In a happenning is given ley P(o) P(o) = -22 s = number of times the event is happenning. a= Expected number of events e = Mathematical constant - 2.71828. In the given question, it is stated that 0.44 percent of the calls to the customers recieve a busy signal sostre 22

FT 0.44) 100 Dhe throbability that the calles recieves a busy signal = 0.oouu pe fresconttin decimals =0.44 0.0044 Let p=0.0044. Jotal number of callers 2 1200-N. we need to calculate the mean or) the Expected value (1) 9 - Stample sije (n) x probability (Đ slere n=1200 620.0044. Aan p = 1200 x 0.0044.-5.28. Now we need to calculate the probability that catleast 5 recieved a busy signal P(825) It denotes the probability that atleast 5 callers recieved a busy signal. P(x2) = P(s) + P(6)+ P(7) +......... +P(199) + P(1200) They all denote the probability that minimum 5 callers and manimum 2 000 callery recieve a busy signal.

Instead of solving all the above probabilities, of we subtract from total probability & I, The probability that O callers, I calles, 2 or 4 callers recieve abury signal Then we will get the required probability. Plzzs) - Total probability – P[825) = 1-plano) + PCD +P(2) +P(3) +P6] P/s) = ct.gr P(0) - - 5.28 0 1820, de 5.28 5:2830 PCO) - 0.0051 PCI) Fe ? Até 5.28 (5.28) 0.0 0.0269 PCI) 20.0269 p(2) "" = *** (520) 0.0710 P(2) = 0.0710. P(3) = x 5.28 (5.28)0 – 0.1249 31

Plu) = ? ar [a= 5.28, 824 S-28 r - u S-28 a & S. 28 (5.2879 4) - 0.1649 plozs) - 1-Plocs) 21-[P(0) + PCI) +P(2) +P(3) +P14] substituting the values we got plozs) = 1-10-0051 +0.0269 +0.0710 + 0.1249 P(875) = 1-0.39 28 - O. 6072 The probability that out of 1200 callers atleast 5 recieved a busy signal = 10.6072) + o.16u9