P = {x:x is a multiple of 4, x ≤ 20} and Q = (x:x is a multiple of 5, x ≤ 20 }
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Answer:
The probability that the intersection will come under the emergency program is 0.1587.
Step-by-step explanation:
Lets divide the problem in months rather than in years, because it is more suitable to divide the period to make a better approximation. If there were 36 accidents in average per year, then there should be 3 accidents per month in average. We can give for the amount of accidents each month a Possion distribution with mean 3 and variance 3.
Since we want to observe what happen in a period of one year, we will use a sample of 12 months and we will take its mean. We need, in average, more than 45/12 = 3.75 accidents per month to confirm that the intersection will come under the emergency program.
For the central Limit theorem, the sample mean will have a distribution Normal with mean 3 and variance 3/12 = 0.25; thus its standard deviation is √0.25 = 1/2.
Lets call the sample mean distribution X. We can standarize X obtaining a standard Normal random variable W with distribution N(0,1).
The values of , the cummulative distribution function of W, can be found in the attached file. We are now ready to compute the probability of X being greater than 3.75, or equivalently, the probability than in a given year the amount of accidents is greater than 45, leading the intersection into an emergency program