Question

The Denver Post reported that, on average, a large shopping center had an incident of shoplifting caught by security 1.4 times every eight 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.
(a) Explain why the Poisson probability distribution would be a good choice for the random variable r.
Frequency of shoplifting is a common occurrence. It is reasonable to assume the events are dependent.
Frequency of shoplifting is a rare 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 dependent.
Frequency of shoplifting is a common occurrence. It is reasonable to assume the events are independent.

What is ? (Round your answer to three 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? (Round your answer to four 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? (Round your answer to four 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? (Round your answer to four decimal places.)

Answer 1

Using the Poisson distribution, it is found that:

• a) Frequency of shoplifting is a common occurrence. It is reasonable to assume the events are independent.

• b) 0.8541 = 85.41% probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security.

• c) 0.303 = 30.3% probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security.

• d) 0.1459 = 14.59% probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

The parameters are:

• x is the number of successes

• e = 2.71828 is the Euler number

• $\mu$ is the mean in the given interval.

Item a:

In this problem, we are given the mean during an interval, which is a common event(1.4 every 8 hours), and intervals can be assumed to be independent, hence, the correct option is:

• Frequency of shoplifting is a common occurrence. It is reasonable to assume the events are independent.

Item b:

Mean of 1.4 every 8 hours, hence, every 11 hours, the mean is:

$\mu = 11 \times \frac{1.4}{8} = 1.925$

The probability is:

$P(X \geq 1) = 1 - P(X = 0)$

In which:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.925}(1.925)^{0}}{(0)!} = 0.1459$

Then

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1459 = 0.8541$

0.8541 = 85.41% probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security.

Item c:

The probability is:

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.925}(1.925)^{0}}{(0)!} = 0.1459$

$P(X = 1) = \frac{e^{-1.925}(1.925)^{1}}{(1)!} = 0.2808$

$P(X = 2) = \frac{e^{-1.925}(1.925)^{2}}{(2)!} = 0.2703$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1459 + 0.2808 + 0.2703 = 0.697$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.697 = 0.303$

0.303 = 30.3% probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security.

Item d:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.925}(1.925)^{0}}{(0)!} = 0.1459$

0.1459 = 14.59% probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security.

To learn more about the Poisson distribution, you can take a look at brainly.com/question/13971530