Question 625986: Currently an average of 25 residents of the village of Westport (Population 1260) dies each year.
A)Find the mean number of deaths per day?
B)Find the probability that on a given day there are no deaths.
C)Find the probability that on a given day there is one death.
D)Find the probability that on a given day there is more than 1 death.
E) Based on the results, should westport have a contingency plan to handle more than 1 death per day. Why or why not?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! averaqe number of deaths in a given year is 25 out of 1260.
that's a probability of 25/1260 = .01984127.
in one year's time, you will have .01984125 * 1260 = 25 deaths.
the mean number of deaths per day can be calculated as 25/365 = .068493151 deaths per day.
that's a probability of .068493151 / 1260 = .00005436
in one day's time, you will have .00005436 * 1260 = .068493151 deaths.
that's actually 1 death every 14.6 days on the average.
if we assume a binomial distribution (discrete random variable), then we get:
p = .00005436 probability of death on a given day.
q = 1-p = .99994564
p(0 deaths on a given day) = C(1260,0) * p^0 * q^1260 = .933798118
p(1 death on a given day) = C(1260,1) * p^1 * q^1259 = .063962252
p(more than 1 death on a given day) = 1 minus p(0 deaths on a given day) - p(1 death on a given day) = 1 - .933798118 - .063962252 = .00223963.
that's a probability of less than 2 tenths of a percent.
even though it can happen based on random probability, it's probably a good idea to have a plan in place, but not a very large amount of money should be spent on it because the likelihood the plan will need to be used is very remote, and even if required, it would probably not be required more than once in a very long time.
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