The Denver Post reported that, on average, a large shopping center had an incide

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). Let r be the number of shoplifting incidents caught by security in an 11-hour period during which the center is open.     
What is ? (Use 2 decimal places.)


(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.)


(c) What is the probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security? (Use 4 decimal places.)


(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.)

Explanation / Answer

Solution:

a) From the given information an incident of shoplifting caught by security 2.5 times every seven hours.

This implies that on average, 2.5/7= 0.36 incidents caught Over 11 hours, the average incidents caught is 11*0.36 = 3.96.

This problem is Poisson 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 (n-->large P-->0) with an average of 3.96 over a period of 11 hours.

b) Let X = number of incidents caught

P(X 1)=1-P(X=0) = 1- e-3.96 = 1- 0.0191 = 0.9809

c) P(X 3)=1-P(X 2)=1 - p(X=0)-P(X=1)= 1- 0.9809 - 3.96 . e-3.96 = 0.0945

d) P(X = 0) = e-3.96 = 0.0191

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