poisson distribution

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maths_arghh234

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okay so the question is :-

the number of accidents per week at a factory is a poisson random variable with the parametre of 2
a. find the probability that in any week chosen at random exactly 1 accident occurs
b. the factory is observed for 100 weeks. determine the expected number of week in which 5 or more accidents occuroks

okay so what i'm stuck on is ... 'with the parametre of 2' ?? .. is that the X value if not is is p(x is less or equal to 2) if not then i don't what to do with that information. also i know that equation and sir gave us tables ..(there's a chance i've lost 1 of them :\) but in this question would i use the table or equations?? i'm not sure ... thanks for your help :)
 
The poisson distribution has a parameter that determines the "shape" of the distribution.
Here on this page you can see a graph of the poisson distribution with different parameters.
If X is a random variable representing the number of accidents per week, then
X ~ Pois(?)
in this case ? = 2, that's what the parameter means.

If you look at the probability mass function definition for the poisson distribution, you'll see that you need the parameter to calculate the probability:
\(\displaystyle P[X=k] = f(k; \lambda)=\frac{\lambda^k e^{-\lambda}}{k!},\,\!\)
You can use this equation to solve the first problem.
 
maths_arghh234 said:
b. the factory is observed for 100 weeks. determine the expected number of weeks in which 5 or more accidents occur
Click to expand...
I hope my correction (in red) is what the question is supposed to be, otherwise I couldn't make sense of it.
If it is, then try calculating the probability that in any week the number of accidents is more than 5 (let's call the event A).
You can do it easily if you calculate the complement event instead (probability of the event that there will be four or less accidents, B) and use the formula for complementary events: P(A) = 1-P(B).
You should be able to finish from here, just multiply by 100...
 
your correction was correct :D... thanks so much ... that seemed okay after you explained it .. i tried it did what u said and come out with 85.71 looked in the back of the question book (where the answers are) and it says 5.3 :s
 
In order to calculate the probability that there will be four or less accidents I need to take the sum of
  • the probability that there are exactly four accidents P[X=4][/*:m:dr9ajwmq]
  • the probability that there are exactly three accidents P[X=3][/*:m:dr9ajwmq]
  • the probability that there are exactly two accidents P[X=2][/*:m:dr9ajwmq]
  • the probability that there is exactly one accident P[X=1][/*:m:dr9ajwmq]
  • the probability that there are no accidents P[X=0][/*:m:dr9ajwmq][/list:eek::dr9ajwmq]
    That is \(\displaystyle P(\textbf{B}) = \sum_{i = 0}^4{P[X=i]}\)
    You can get each of these from the equation:
    \(\displaystyle P[X=k] = f(k; \lambda)=\frac{\lambda^k e^{-\lambda}}{k!},\,\!\)
    where \(\displaystyle \lambda = 2\)

    P[X=4] = 0.090223522157742
    P[X=3] = 0.18044704431548
    P[X=2] =0.27067056647323
    P[X=1] = 0.27067056647323
    P[X=0] = 0.13533528323661
    The sum is: 0.94734698265629202

    So P(B) = 0.94734698265629202, now to get P(A) I use the complement rule:
    P(A) = 1-P(B)
    P(A) = 0.052653017343707975
    This is the probability that one week will have five or more accidents.
    The expected value for 100 weeks is 100 times that, that is
    5.265301734370798, or 5.3 when rounded.
 
phhew i get it now .. thank you :) ... i have poisson distribution tables that the techer gave us .. and i used them. to be honest i'm not sure when to use them and when to use that equation
 
Using the tables would actually be better...
Your tables probably show the cumulative probability function, that is \(\displaystyle P[X\le k]\), and \(\displaystyle P[X\le 4] = P(\textbf{B})\) is exactly what we need to calculate.

The equation I gave was the probability mass function, the tables show cumulative probability function, which is basically a sum of values of the other.

For \(\displaystyle \lambda = 2\) the tables show that \(\displaystyle P[X\le 4] = 0.947\), and \(\displaystyle P(A) = 1 - 0.947 = 0.053\), hundred times that is 5.3 ...
 
the tables were what i looked at first but on mine it say 4 is 0.1429 ... i'm probably looking at the wrong place.. at the top of the collum i'm looking at it says 2 .. and go down 4 and it say 0.1429 (just realized i basically wrote the same thing twice lol)
 
Oh the table you're looking at must be showing this:
\(\displaystyle P[X\ge r]\) and \(\displaystyle P[X\ge 5] = 0.0527\) (right below the number you were looking at), which right away is P(A), the probability that five or more accidents happen in any week, so you can just multiply that by 100 and get the right answer... I'm sorry I explained it in an unnecessarily compilacated way, but hopefully you'll understand it better...
 
doesn't matter i would still be starring at the question if you didn't explain it evan if it was the long lol . ..so chhers for that :) ..and wonders of wonders there's the number right below the one i was looking at ....wow i think this mean i am pretty dense ...:\
 
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