Amelie,
I have seen you post this same question several times now, and you must feel frustrated that nobody has answered it.
If I don't have enough confidence that I can give you a satisfactory and correct answer I often stay away from a reply.
This is the case for me for this question.
The last time I studied Poisson distribution was about 38 years ago, and at this moment I cannot help you with this question.
This appears to be a bit beyond highschool level math.
I am expressing my own opinion, and it does not reflect anybody else's view.
I wish I could help you.
MY MAIN PROBLEM IS FIGURING OUT WHAT DISCRETE DISTRIBUTION TO USE, BERNOULLI, BINOMIAL, DISCRETE UNIFORM, GEOMETRIC NEGATIVE BINOMIAL, OR POISSON. Every time I choose one, it's the incorrect one. Is there some way I can easily find out which one to use.
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3.40
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Let Y denote a random variable that has a geometric distribution, with a probability of success on any trial denoted by p.
b) Find P(Y>4 | Y>2) for general p. Compare this result with the unconditional probability P(Y>=2).[This property is referred to as "lack of memory"]
3.58
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In the article cited in Exercise 3.57, the projected fatality rate for 1975 if the NMSL had not been in effect was 25 per 10^9 vehicle miles. Assume that these conditions had prevailed.
a) Find the probability that at most 15 fatalities occurred in a given block of 10^9 vehicle miles.
b) Find the probability that at least 20 fatalities occurred in a given block of 10^9 vehicle miles.
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3.66
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The number of bacteria colonies of a certain type in samples of polluted water has a Poisson distribution with a mean of two per cubic centimeter.
a)If four 1-cubic-centimeter samples of this water are independently selected, find the probability that at least one sample will contain one or more bacteria colonies.
b)How many 1-cubic-centimeter samples should be selected to establish a probability of approximately 0.95 of containing at least one bacteria colony?
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4 answers
Here is how I would do problem numbers 3.58 and 3.66
Use Poisson statistics whenever you know the AVERAGE rate per volume or time of a particular event happening, and you want the probability P(n) of a specific INTEGER number of events in a random sample.
In problem 3.58, for the first part, you have to add the probabilities P(0), P(1).. up to P(15) in a one-billion-mile "block", and (for the second part), add the P(20), P(21), .. etc until the numbers become negligible.
As examples of typical calculations, with a = 25 (average number per sample),
P(n) = a^n*e^-a/n!
P(0) = 25^0*e^-25/0! = e^-25
P(10) = 25^10*e^-25/10! = 0.000365
P(15) = 25^15*e^-25/15! = 0.009891
P(20) = 25^20*e^-25/20! = 0.051917
P(25) = 25^25*e^-25/25! = 0.079523
P(30) = 25^30*e^-25/30! = 0.045413
P(40) = 25^40*e^-25/40! = 0.001408
P(50) = 25^50*e^-25/50! = 0.000004
Adding upo all the Poisson P(n) numbers in his case is a tedious exercise.
You can save a lot of effort by using the Gaussian approximation whenever average number in a sample greatly exceeds 1, as is the case here. In that case, the most likely number of deaths in a "block" is 25, and the standard deviation is sqrt(25)= 5 . Then use a normal distribution function table, integrating from 0 to 15 and then from 20 to infinity for the two parts of your question.
For your problem 3.66, the Poisson probability of getting ZERO cultures in a 1 cm^3 sample is, with a = 2 as the average,
P(0) = 2^0/e^-2/0! = e^-2 = 0.1353
Now if you want to have a 0.95 probability of getting at least one sample with one colony, you must reduce the probability of NOT getting one to be <or = 0.05. That requires N independent samples, there
0.1353^N <= 0.05
N = 2 is enough to provide this assurance.
Use Poisson statistics whenever you know the AVERAGE rate per volume or time of a particular event happening, and you want the probability P(n) of a specific INTEGER number of events in a random sample.
In problem 3.58, for the first part, you have to add the probabilities P(0), P(1).. up to P(15) in a one-billion-mile "block", and (for the second part), add the P(20), P(21), .. etc until the numbers become negligible.
As examples of typical calculations, with a = 25 (average number per sample),
P(n) = a^n*e^-a/n!
P(0) = 25^0*e^-25/0! = e^-25
P(10) = 25^10*e^-25/10! = 0.000365
P(15) = 25^15*e^-25/15! = 0.009891
P(20) = 25^20*e^-25/20! = 0.051917
P(25) = 25^25*e^-25/25! = 0.079523
P(30) = 25^30*e^-25/30! = 0.045413
P(40) = 25^40*e^-25/40! = 0.001408
P(50) = 25^50*e^-25/50! = 0.000004
Adding upo all the Poisson P(n) numbers in his case is a tedious exercise.
You can save a lot of effort by using the Gaussian approximation whenever average number in a sample greatly exceeds 1, as is the case here. In that case, the most likely number of deaths in a "block" is 25, and the standard deviation is sqrt(25)= 5 . Then use a normal distribution function table, integrating from 0 to 15 and then from 20 to infinity for the two parts of your question.
For your problem 3.66, the Poisson probability of getting ZERO cultures in a 1 cm^3 sample is, with a = 2 as the average,
P(0) = 2^0/e^-2/0! = e^-2 = 0.1353
Now if you want to have a 0.95 probability of getting at least one sample with one colony, you must reduce the probability of NOT getting one to be <or = 0.05. That requires N independent samples, there
0.1353^N <= 0.05
N = 2 is enough to provide this assurance.
Another hingis that you should not think like: "Every time I choose one, it's the incorrect one". Because you should learn the stuff by deriving everything from first principles (unless you are not a math student but, say, a sociology or psychology student, who only needs to use foirmulas without understanding them).
So, take your time to derive the Poison and other distributions from first principles and do not attempt to get the answer of problems correct by guesing which formula to use.
So, take your time to derive the Poison and other distributions from first principles and do not attempt to get the answer of problems correct by guesing which formula to use.
3. A used car salesperson estimates the probabilities for the number the number of cars sold in a week as follow-
number—0, 1, 2 ,3,4,5
Probability-.05,.20,.35,.25,.10,.05
a) What is the probability for a given week that four or more cars will be sold?
b) Find the expected number of cars sold in a week
c) What is the probability that the salesperson will sell more than 2 cars in a given week?
d) For a period of 3 weeks, what is the probability that salesperson will sell more than 2 cars every week?
number—0, 1, 2 ,3,4,5
Probability-.05,.20,.35,.25,.10,.05
a) What is the probability for a given week that four or more cars will be sold?
b) Find the expected number of cars sold in a week
c) What is the probability that the salesperson will sell more than 2 cars in a given week?
d) For a period of 3 weeks, what is the probability that salesperson will sell more than 2 cars every week?
0 answers