Question

The Denver post reported that on average a large shopping center had an incident of shoplifting caught by security 1.9 times every 4 hours. The shopping center is open from 10:00 a.m. to 9:00 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.

b) What is the probability that from 10:00 a.m. to 9:00 p.m. there will be at least one shoplifting incident caught by security ?(use 4 decimal places)

c) what is the probability that from 10:00 a.m. to 9:00 p.m. there will be at least 3 shoplifting incidents caught by security? (use 4 decimal places)

D) what is the probability that from 10:00 a.m. to 9:00 p.m. there will be no shoplifting incidents caught by security? (use 4 decimal places)

Answer 1

Answer:

a)

Given,

$\lambda$ = 1.9

For 4 hours , $\lambda$ = 1.9

For 11 hours , $\lambda$ = 1.9*11 / 6

$\lambda$ = 3.48

b)

To determine the probability that from 10:00 a.m. to 9:00 p.m. there will be at least one shoplifting incident caught by security

P(X >= 1) = 1 - P(0)

= 1 - [e^-3.48*3.48^0 / 0!]

= 1 - 0.0308

= 0.9692

c)

To determine the probability that from 10:00 a.m. to 9:00 p.m. there will be at least 3 shoplifting incidents caught by security

P(X >= 3) = 1 - P(X < 3)

= 1 - [P(0) + P(1) + P(2)]

= 1 - [e^-3.48*3.48^0/0! + e^-3.48*3.48^1/1! + e^-3.48*3.48^2/2!]

= 1 - [0.0308 + 0.1072 + 0.1865]

= 1 - 0.3245

= 0.6755

d)

To determine the probability that from 10:00 a.m. to 9:00 p.m. there will be no shoplifting incidents caught by security

P(no incidents) = P(X = 0)

= e^-3.48*3.48^0/0!

= 0.0308