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?

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

@Zarkon can we change to \(\lambda = 900\)?

- anonymous

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.

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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.

- anonymous

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.

- anonymous

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

- anonymous

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.

- anonymous

yeah each of these damned books has poisson all over the place
still looking.

- anonymous

haha.... okay sure... thanks! i'm still trying too.

- Zarkon

use a normal approximation

- anonymous

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

- anonymous

@Zarkon so we need a table for that too right?

- Zarkon

My calculator can do it

- anonymous

I have tried this idea too. The probability is around 0.7 something. There is a formula for poisson.

- anonymous

But with this idea, the interval considered for the probability computed would be for each day, not 100 days any more?

- anonymous

i found this
http://www.wku.edu/~david.neal/statistics/discrete/poisson.html
which tells you how to do it with a calculator, but if my idea is right, then i found a table that says
for \(\lambda =9\) \(P(x<10)=0.70599\)

- anonymous

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

- anonymous

sorry but every other example i can find sums up all the others and subtracts from one, as you did at first

- anonymous

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.

- anonymous

*casts doubts on the interval that the probability value is computed for.

- anonymous

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

- Zarkon

do you know what the answer is supposed to be?

- anonymous

Thanks a lot, @satellite73. :)

- anonymous

yw, sorry it was not better

- anonymous

@Zarkon, unfortunately, I don't know the correct answer for this question. I am not given. :(

- Zarkon

what don't you like about the answer Mathematica gave you?

- anonymous

Because I won't have Mathematica in the exam. :(
This question suppose to be an exam question.

- Zarkon

will you have tables?

- Zarkon

will you have a normal table?

- anonymous

yea, I will have normal tables. Can I approximate poisson with normal?

- anonymous

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.

- Zarkon

yes...quite accurately with continuity corrections

- Zarkon

\[P(X>1000)=P(X>1000.5)\]
\[=P\left(\frac{X-900}{\sqrt{900}}>\frac{1000.5-900}{\sqrt{900}}\right)\]
\[=P\left(Z>\frac{1000.5-900}{\sqrt{900}}\right)\]

- anonymous

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?

- Zarkon

No...
here you essentially have the sum of 100 iid random variables. By the CLT it will be approx normal

- Zarkon

that is how you get \(\lambda=9\cdot 100\)
if \(X_1,X_2,\cdots X_n\) are poisson(\(\lambda\)) then
\[\sum_{k=1}^{n}X_k\sim\text{poisson}(\lambda\cdot n)\]

- anonymous

ohhh.....yeah!! This sounds very logical. And it looks like a very workable hand method.!

- Zarkon

good

- Zarkon

sorry i didn't jump into the conversation earlier...I'm grading now...man I hate to grade.

- anonymous

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

- anonymous

haha....thanks! it's okay. I am very happy to learn something from you.
And, grading? You mean like grading students' assignment?

- Zarkon

Complex analysis exams

- anonymous

@satellite73 Zarkon gave a pretty good explanation. Just thought you may be interested to learn about it too.

- anonymous

oh wow... Sounds like a difficult subject. haha... thanks for your help!

- anonymous

grading is the worst. guess what i am doing?

- Zarkon

it is the worst.....worst part of the job..
I hope you are not grading :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.