Question

The Denver Post reported that, on average, a large shopping center had an incident of shoplifting caught by security 2.5 times every seven hours. The shopping center is open from 10 A.M. to 9 P.M. (11 hours). Letr be the number of shoplifting incidents caught by security in an 11-hour period during which the center is open. (a) Explain why the Poisson probability distribution would be a good choice for the random variable Frequency of shoplifting is a common occurrence. It is reasonable to assume the events are dependent. O Frequency of shoplifting is a rare occurrence. It is reasonable to assume the events are dependent. Frequency of shoplifting is a common occurrence. It is reasonable to assume the events are independent. Frequency of shoplifting is a rare occurrence. It is reasonable to assume the events are independent. What is A? (Use 2 decimal places.) 3.96 (b) What is the probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security? (Use 4 decimal places.) 0909 (c) What is the probability that from 10 AM to 9 P.M. there will be at least three shoplifting incidents caught by security? (Use 4 decimal places.) 0.9054 (d) What is the probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security? (Use 4 decimal places.) 0.0191 x Show All

Answer 1

Answer 1.5 times every shoplifting caught by security 2.5 tim Given : An incident of 7 hours That implies that on average 25/9 = 0.3571 indicates caught over 1 hours, the average incidents caught is 11 X0.3571 = 3.9281 This problem is poison because there are shoplifting incidents or there are not and each incident of being caught is very small while the potential for shoplifiting incidents can be large (now larg p20) with an average of 3.9281 over a period of 1 hours 2 frequency of shoplifting is a rare occurence. It is measured reasonable to assume the events are independent - 3.92 b) plxz) = 1- P(x21) = 1-P(x=0) PLX=0) 6-3.928) X 3.9281° 0.01968) x 1 O! -0.01968) P(x21) = 1-0.01968) - 0.980319

e p(x 3) = P f - P(x2) = 1-p(x=0) - P(x=1) - P(x=2) plx=0) - 0-3.928) X 3.9281 = 0.01968) 0! 0.019681 X 3.928) Plx=1) - e-369281x 13.9281)' 11 = 0.977308 p(x=2) = (-3.9281 x (3.9281², 0.019681 x 15.42996 2 = 0.15 1838 PCX23) = 1-0.019681-0:077308 – 0.151838 = 0.751173 P(x=0) = 0.019681