Question

The Honolulu Advertiser states that in Honolulu there was an average of 834 burglaries per 100,000 households in a given year. In the Kohola Drive neighborhood there are 306 homes. Let r = number of these homes that will be burglarized in a year.(a) What is the probability that there will be no burglaries this year in the Kohola Drive neighborhood? (Use 4 decimal places.)(b) What is the probability that there will be no more than one burglary in the Kohola Drive neighborhood? (Use 4 decimal places.)(c) What is the probability that there will be two or more burglaries in the Kohola Drive neighborhood? (Use 4 decimal places.)

Answer 1

a) 0.0781 = 7.81% probability that there will be no burglaries this year in the Kohola Drive neighborhood.

b) 0.2772 = 27.72% probability that there will be no more than one burglary in the Kohola Drive neighborhood

c) 0.7228 = 72.28% probability that there will be two or more burglaries in the Kohola Drive neighborhood

Explanation:

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

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

In which

x is the number of sucesses

e = 2.71828 is the Euler number

`\mu` is the mean in the given time interval.

834 burglaries per 100,000 households in a given year. In the Kohola Drive neighborhood there are 306 homes.

This means that
`\mu = (306*834)/(100000) = 2.55`

(a) What is the probability that there will be no burglaries this year in the Kohola Drive neighborhood?

This is P(X = 0).

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-2.55)*2.55^(0))/((0)!) = 0.0781`

0.0781 = 7.81% probability that there will be no burglaries this year in the Kohola Drive neighborhood.

(b) What is the probability that there will be no more than one burglary in the Kohola Drive neighborhood?

`P(X \leq 1) = P(X = 0) + P(X = 1)`

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-2.55)*2.55^(0))/((0)!) = 0.0781`

`P(X = 1) = (e^(-2.55)*2.55^(1))/((1)!) = 0.1991`

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0781 + 0.1991 = 0.2772`

0.2772 = 27.72% probability that there will be no more than one burglary in the Kohola Drive neighborhood

(c) What is the probability that there will be two or more burglaries in the Kohola Drive neighborhood?

Either there will be no more than one burglary, or there will be at least two.

The sum of the probabilities of these events is decimal 1. So

`P(X \leq 1) + P(X \geq 2) = 1`

`0.2772 + P(X \geq 2) = 1`

`P(X \geq 2) = 0.7228`

0.7228 = 72.28% probability that there will be two or more burglaries in the Kohola Drive neighborhood