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In a city, the daily number of traffic accidents follows a Poisson distribution with \[\lambda=9\]. What is the probability that the total number of traffic accidents exceeds 1000 in 100 days?

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@Zarkon can we change to \(\lambda = 900\)?
One of my solutions is to have: \[\lambda=9*100\] Because the interval considered in this case is 100 days. Then I calculate for: \[P(X>1000)\]. But the problem with having Lambda as 900 is that it is too huge. P(X>1000)=1.
how do you calculate \(P(x\geq 1000)\)? using poisson? usually "greater than ..." means calculate everything up to it and subtract from one. must be another formula for this. i will try to look it up

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Other answers:

Precisely. Here's what I have done: \[P(X>1000)=1-\sum _{ i=0 }^{ 1000 }{ poisson(i,\quad \lambda =900) } \] However, I can't calculate possion up to 1000. It is too huge. Not even my calculator can do it. I used an online calculator to do it but the problem is I believe this question shouldn't require me to do such calculations that require a powerful calculator like a computer.
And the value returned from Mathematica for calculating the sum from 0 to 1000 of the poisson function is 0.9995093672673, which is almost 1. So, I think my answer is still wrong.
yeah i think there is another way of computing \(P(x\geq k)\) for poisson using variance, but it might require some table i am trying to look it up in a third book
ohh thanks! I didn't know there is another way to compute poisson using variance. But if we need a table, would a table list up to 900? The variance for poisson is the same as its expectation, which is Lambda.
yeah each of these damned books has poisson all over the place still looking.
haha.... okay sure... thanks! i'm still trying too.
use a normal approximation
i have an idea, probably a bad one you expect 900 accidents, and want to know what is the probability you get over 1000 reduce all by a factor of 100 and change to you expect 9 accidents, what is the probability you get more than 10 this seems to be done by tables, one of which i found on line, where it will give you this
@Zarkon so we need a table for that too right?
My calculator can do it
I have tried this idea too. The probability is around 0.7 something. There is a formula for poisson.
But with this idea, the interval considered for the probability computed would be for each day, not 100 days any more?
i found this which tells you how to do it with a calculator, but if my idea is right, then i found a table that says for \(\lambda =9\) \(P(x<10)=0.70599\)
if it is valid to reduce all by 100, and it seems like it might be, then your answer would be \(1-.70599\) but again this is relying on a table
sorry but every other example i can find sums up all the others and subtracts from one, as you did at first
yea, it seems pretty valid. thanks! :) My only concern is if this would be computing for an interval of 1 day instead of 100 days? Because usually, I remember I would adjust the value of Lambda according to the required interval specified in the problem. But in this case, adjusting the Lambda seems too huge. Adjusting the interval value gives a more believable value but casts doubts on the interval that probability value computed for.
*casts doubts on the interval that the probability value is computed for.
i wish i knew. my only answer is that this would be an approximation, and perhaps a bad one. sorry i cannot be more help, stumped on this one, but i would go with reducing and computing the approximation
do you know what the answer is supposed to be?
Thanks a lot, @satellite73. :)
yw, sorry it was not better
@Zarkon, unfortunately, I don't know the correct answer for this question. I am not given. :(
what don't you like about the answer Mathematica gave you?
Because I won't have Mathematica in the exam. :( This question suppose to be an exam question.
will you have tables?
will you have a normal table?
yea, I will have normal tables. Can I approximate poisson with normal?
I'm only aware that I could approximate binomial with normal and poisson. But I don't know if I could approx for poisson with normal.
yes...quite accurately with continuity corrections
\[P(X>1000)=P(X>1000.5)\] \[=P\left(\frac{X-900}{\sqrt{900}}>\frac{1000.5-900}{\sqrt{900}}\right)\] \[=P\left(Z>\frac{1000.5-900}{\sqrt{900}}\right)\]
But with this, I have to make an assumption that the original data follows a normal distribution? Otherwise, how can I be sure that the data is close to a normal distribution?
No... here you essentially have the sum of 100 iid random variables. By the CLT it will be approx normal
that is how you get \(\lambda=9\cdot 100\) if \(X_1,X_2,\cdots X_n\) are poisson(\(\lambda\)) then \[\sum_{k=1}^{n}X_k\sim\text{poisson}(\lambda\cdot n)\]
ohhh.....yeah!! This sounds very logical. And it looks like a very workable hand method.!
sorry i didn't jump into the conversation earlier...I'm grading I hate to grade.
It is through CLT, that this whole poisson becomes close to normal. And therefore, the approximation can take place. It is not exact but close because of the CLT, n more than 30. Thank you so much for your help! :D
haha....thanks! it's okay. I am very happy to learn something from you. And, grading? You mean like grading students' assignment?
Complex analysis exams
@satellite73 Zarkon gave a pretty good explanation. Just thought you may be interested to learn about it too.
oh wow... Sounds like a difficult subject. haha... thanks for your help!
grading is the worst. guess what i am doing?
it is the worst.....worst part of the job.. I hope you are not grading :)

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