Question

The Honolulu Advertiser stated that in Honolulu there was an average of 661 burglaries per 100,000 households in a given year. In the Kohola Drive neighborhood there are 323 homes. Let r be the number of homes that will be burglarized in a year. Compute the probability for . Round your answer to the nearest ten thousandth. 0.3969 0.2001 0.0613 0.1743 0.1890

Answer 1

Solution:

given,

average number of burglaries in honolulu = 661

total number householdsin Honolulu = 100000

probability of burglary in Honolulu = 661 / 100000 = 0.00661

total number of homes in Kohola drive = 323

expected value of = 0.00661 * 323 = 2.13503

number of homes that will be burglarized in a year in Kohola = r

probability of burglary can be found with

checking r = {0 , 1, 2, 3} //assuming the missing data

probability of having 0 burglaries = P(0, 2.13503) = 0.1182 //rounded values

probability of having 1 burglary = 0.2524

probability of having 2 burglaries = 0.2695

probability of having 3 burglaries = 0.1918

probability of having not more than 1 burglary = P(0, ) + P(1, ) = 0.3707

the question isn't having complete information, please proceed as shown above with the missing data.