Respuesta :
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:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
- [tex]\mu[/tex] 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:
[tex]\mu = 11 \times \frac{1.4}{8} = 1.925[/tex]
The probability is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.925}(1.925)^{0}}{(0)!} = 0.1459[/tex]
Then
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1459 = 0.8541[/tex]
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:
[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]
In which:
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
Then:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.925}(1.925)^{0}}{(0)!} = 0.1459[/tex]
[tex]P(X = 1) = \frac{e^{-1.925}(1.925)^{1}}{(1)!} = 0.2808[/tex]
[tex]P(X = 2) = \frac{e^{-1.925}(1.925)^{2}}{(2)!} = 0.2703[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1459 + 0.2808 + 0.2703 = 0.697[/tex]
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.697 = 0.303[/tex]
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:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.925}(1.925)^{0}}{(0)!} = 0.1459[/tex]
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 https://brainly.com/question/13971530
