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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.1
@Zarkon can we change to \(\lambda = 900\)?
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
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.
 one year ago

satellite73Best ResponseYou've already chosen the best response.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
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
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.
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
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.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
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.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yeah each of these damned books has poisson all over the place still looking.
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
haha.... okay sure... thanks! i'm still trying too.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
use a normal approximation
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
@Zarkon so we need a table for that too right?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
My calculator can do it
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
I have tried this idea too. The probability is around 0.7 something. There is a formula for poisson.
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
But with this idea, the interval considered for the probability computed would be for each day, not 100 days any more?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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 \(\lambda =9\) \(P(x<10)=0.70599\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
sorry but every other example i can find sums up all the others and subtracts from one, as you did at first
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
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.
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
*casts doubts on the interval that the probability value is computed for.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
do you know what the answer is supposed to be?
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
Thanks a lot, @satellite73. :)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
yw, sorry it was not better
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
@Zarkon, unfortunately, I don't know the correct answer for this question. I am not given. :(
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
what don't you like about the answer Mathematica gave you?
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
Because I won't have Mathematica in the exam. :( This question suppose to be an exam question.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
will you have a normal table?
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
yea, I will have normal tables. Can I approximate poisson with normal?
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
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.
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
yes...quite accurately with continuity corrections
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
\[P(X>1000)=P(X>1000.5)\] \[=P\left(\frac{X900}{\sqrt{900}}>\frac{1000.5900}{\sqrt{900}}\right)\] \[=P\left(Z>\frac{1000.5900}{\sqrt{900}}\right)\]
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
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?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
No... here you essentially have the sum of 100 iid random variables. By the CLT it will be approx normal
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
that is how you get \(\lambda=9\cdot 100\) if \(X_1,X_2,\cdots X_n\) are poisson(\(\lambda\)) then \[\sum_{k=1}^{n}X_k\sim\text{poisson}(\lambda\cdot n)\]
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
ohhh.....yeah!! This sounds very logical. And it looks like a very workable hand method.!
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
sorry i didn't jump into the conversation earlier...I'm grading now...man I hate to grade.
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
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
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
haha....thanks! it's okay. I am very happy to learn something from you. And, grading? You mean like grading students' assignment?
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
@satellite73 Zarkon gave a pretty good explanation. Just thought you may be interested to learn about it too.
 one year ago

xEnOnnBest ResponseYou've already chosen the best response.0
oh wow... Sounds like a difficult subject. haha... thanks for your help!
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
grading is the worst. guess what i am doing?
 one year ago

ZarkonBest ResponseYou've already chosen the best response.0
it is the worst.....worst part of the job.. I hope you are not grading :)
 one year ago
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