Question

1Y, the number of accidents per year at a given intersection, is assumed to have a Poisson distribution. Over the past few years, an average of 36 accidents per year have occurred at this intersection. If the number of accidents per year is at least 45, an intersection can qualify to be redesigned under an emergency program set up by the state. Approximate the probability that the intersection in question will come under the emergency program at the end of the next year.I'm thinking of using the Central Limit Theorem but I'm not exactly sure what that'll look like and the parameters used.

Answer 1

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).

$W = \frac{X-\mu}{\sigma} = \frac{X-3}{1/2} = 2x-6$

The values of $\phi$ , 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

$P(X > 3.75) = P(2X-6 > 2*3.75-6) = P(W > 1) = 1-\phi(1) = 1-0.8413 \\= 0.1587$

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