Question

Probability Experts Needed! Topic: Convergence in Distributions.

Answer 1

me!

Answer 2

Problem:

The number of accidents per year X at given intersection is assumed to have a Poisson distribution. Over the past few years, an average number of 32 accidents per year has occurred at this intersection. (My notes: this is, lambda = mu = expected value = 32.) If the number of accidents per year is at least 40 (this is my X value), an intersection can qualify to be rebuilt under an emergency program set up by the state. Approximate the probability that the intersection in question will qualify under the emergency program at the end of next year.

My attempt:

I know that to approximate the probability distribution of X for large values of lambda I can use this:

\[P(Y = \frac{X-\lambda}{\sqrt{\lambda}})\], which converges in distribution toward a standard normal random variable as \(\lambda\) tends toward infinity.

I believe I can simply plug in X = 40 and \(\lambda\) = 32.

Making:

\[Y > \frac{40-32}{\sqrt{\lambda}} = \frac{8}{\sqrt{32}} = \frac{8}{4\times\sqrt{2}} = \frac{2}{\sqrt{2}} = \frac{2\times\sqrt{2}}{2} =\sqrt{2} \rightarrow Y > \sqrt{2}\].

Now, referencing my standard normal Z table, finding the probability that \(Z < \sqrt{2} = P(Z < 1.414) \approx 92.07% \approx 0.9207\). Now, since I am looking for Y greater than this value, I find the difference \(1-0.9207 = 0.0793\).

Making the probability that there will be greater than 40 accidents approximately 7.93%.

Is this correct?

Answer 3

@HenrietePurina Thanks for volunteering to help. I have posted the problem. :)

Answer 4

ok... gimme some time

Answer 5

No problemo. I will move on to the next problem for now.

Answer 6

so far so good :)

Answer 7

why is it 1 - 0.9207, why 1

Answer 8

yes, good job :)

Answer 9

Thanks very much! How confident are you (not doubting, but.. I want to be sure).

Answer 10

?

Answer 11

90% sure, im not einstein ;)

Answer 12

well bye bud and keep it up :)