What distinguishes Poisson distributions from other binomial distributions?
92833
The Poisson distribution is commonly applied to a count of events happening within some time interval or spatial area.  Examples include the number of accidents happening on a certain stretch of road in a month, the number of purchases of a given product in a store in a day, the number of flaws in a bolt of fabric, the number of weeds in a 100 square foot plot of land.\n\nThe binomial distribution describes the number of events that occur out of a fixed number of trials.  Examples include, count the number of heads in 100 coin flips, the number of correct answers when someone guesses on every question of a multiple choice test, etc.\n\nThe difference between them is that in the Poisson applications, we don't really know how many events are possible.  Practically speaking, there is an upper limit (such as, we can't sell more of a product than is on the shelf).  However, it's usually applied to cases where the events are pretty rare, in which case it's an excellent approximation to the real behavior.\n\nMeanwhile, the Binomial distribution assumes a fixed number of attempts or trials (each coin toss, or each test question).\n\nPossibly adding to the confusion is this fact.  A binomial distribution with a very large number of trials is practically indistinguishable from a Poisson distribution.  For example, the number of red-heads in a stadium full of people could either be modeled as a Poisson distribution, or as a Binomial.\n\nWhen the number of trials is very large, the Binomial distribution probabilities are difficult to compute.  In such cases, people tend to either approximate it with a Poisson distribution, or use the continuous Normal (bell curve) distribution.\n\nIf it's a count of fairly uncommon events occuring in a specific period of time or spatial area, then use the Poisson.  If it's a count of events out of a known, limited number of trials or opportunities, use the Binomial.\n\nI hope this helps.
242094
In the experiment of testing for the number of students out of 37 who consider statistics exciting\n\nNo. of trials made \n==> No. of students tested = 37\n==>n = 37\n\nLet the event of a student tested being intersted in the subject of statistics be a success\n\nProbability of success in each trial\n= probability that a student tested is interested in the subject of statistics\n==> p = 0.75\n[Given that 75% of the population of all business statistics 1 students consider statistics exciting]\n\nTherefore, \nq = 1 - p ==> q = 1 - 0.75 ==> q = 0.25\n\nLet "x" indicate the number of successes in "n" trials\nsince the number of students who are interested in statistics can be either 0, 1, 2, .... 37\nThe values "x" can carry are 0, 1, 2, .... 37\n==> "X" is a discrete random variate with range = {0, 1, 2, .... 37}\n\nThe probability distribution of "x" is a binomial distribution..\n\n[Since there a number of trials all of which are identical and independent with only two possible outcomes in each trial either a success i.e. the student being interested or a failure i.e. the student not being interested, with the probability of success in each trial remaining the same all throught]\n\nThe probability function of a binomial distribution\n==> P(X=x) = nCx * p^x * q^(n-x)\n\nTherefore, for the above distribution\nP(X=x) = 37Cx * (0.75)^x * (0.25)^(37-x)\n==>P(X=x) = 37Cx * (0.75)^x * (0.25)^(37) * (0.25)^(-x)\n==>P(X=x) = 37Cx * (0.75)^x * (0.25)^(37) * 1/(0.25)^(x)\n==>P(X=x) = 37Cx * (0.75/0.25)^x * (0.25)^(37)\n==>P(X=x) = 37Cx * (3)^x * (0.25)^(37)\n\nTherefore, \nProbability of observiing 36 or more of the students who consider statistics to be an exciting subject\n==> P(X >= 36) = P(X=36) + P(X=37)\n\n\nP(X = 36) = C(37,36) * (3)^36 * (0.25)^(37)\nP(X = 37) = C(37,37) * (3)^37 * (0.25)^(37)\n\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
631161
This is a binomial probability.\nP(success) = 1/5, P(failure) = 4/5.  There are 6 events here, so P(2 successs) = (1/5)(1/5)(4/5)(4/5)(4/5)(4/5).\nP(at least 2 successes) = P(2 successes) + P93 successes) + . . . . P(6 successes) or 1 - P(O successes) + P(1 success) + P(2 successes) which you already computed above.\nLet me know if you don't understand.
702097
Are you looking for the formula of finding the range of group data.\n\nActually, i have read two methods. \nsome books just get the difference of the highest upper limit and the lowest lower limit of the distribution.\n\nbut some get the difference of the highest upper limit and the lowest lower limit of the distribution plus one.\nThat is, it is the difference between the highest upper boundary and the lowest lower boundary.
316964
For n = 5? Seriously?\n\nIt is not clear what you are asking.\n\nSuppose that:\n\n1. Your model is of a sequence of independent Bernoulli trials each of which succeeds with probability p.\n\n2. Your null hypothesis H0 is p = 0.3\n\n3. Your alternative hypothesis H1 is p is not 0.3.\n\n4. You plan to observe 5 trials with the intent of accepting or rejecting the null hypothesis with 95% confidence.\n\nThen from p + q = 1 (i.e., q = 1 - p) and 1^5 = (p+q)^5 = p^5 + 5p^4q + 10p^3q^2 + 10p^2q^3 + 5pq^4 + q^5 you know that the probabilities are\n\n5 successes in 5 trials: p^5 = 0.3^5 = 0.00243\n4 succeses in 5 trials: 0.02835\n3 successes: 0.13230\n2 successes: 0.30870\n1 success: 0.36015\n0 successes: 0.16807\n\ni.e., 5 or 4 successes occur <= 5% of the time so you accept H0 on 3 or fewer successes and reject H0 on 4 or 5 successes.\n\nAnother interpretation of your question:\n\nSuppose your model is of independent identical Bernoulli trials each with unknown probability p of success and you would like to estimate p from observing the outcomes of 5 independent trials.  Given k out of 5 successes as the observed outcome for which p can we say that "k" or "k or fewer" or "k or more" successes out of 5 has probability <= 5% of being observed (that is three different questions by the way).  For each of k = 0, 1, ..., 5 you plot the probability of that outcome given p and exclude those p for which the plot drops to 5% or less.
319924
It depends on the distribution.  Some probability distributions have multiple parameters (such as the mean and standard deviation for the Normal distribution).  Phi2 would be one parameter of the particular distribution used in the problem.
268119
To answer that question you would have to know if the differences between "having a variation" and "not having the variation" are normally distributed.  If so, you could make an assumption on the number to look at based on normal distribution and 3 standard deviations.\n\nIf there is no normal distribution to the data, then you would have to determine your own "statistics" for the odds that the deviation would occur, and calculate the number based on the possible number of "variations" that could occur, versus the odds that any one could occur.
4160
Random discrete variables are specific numbers.  E.g., all integers X such that 1<= X <= 6.\nX can be 1, 2, 3, 4, 5, or 6\nThe probability that X=1 is 1/6.\n\nRandom continuous variables can be any of an infinite set of numbers in a range.  E.g., all X such that  1 < X < 6\nX can be any of an infinite set of numbers between 1 and 6.\nThe probability that X is any specific number is infinitesimal.
76447
p1v1t1=p2v2t2\n\nSee Bernoulli's theorem and Pascals Law
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Probability theory is the branch of mathematics that helps assess the chances that something will or will not occur. One of the tools is using probability distributions to model processes in real life.  If the distribution "fits" the process well, predictions can be made about how often events might occur.\n\nThe process might be a coin toss (Bernoulli), the rate at which people arrive at the food line in McDonalds (Poisson), the rate at which new products fail (Weibull).  Once you have an appropriate model, you can say "what are the chances that I'll get 5 heads in a row", if I average 5 people an hour in my food line, what might be the "peak" rate at which I can expect people to arrive.".\n\nNow, to answer your question -- this is difficult.  You haven't provided enough information about the situation or the process to be able to give you any specific guidance.
561865
A permutation. P(n,r)=(n!)/((n-r)!). C(n,r)=(n!)/(r!(n-r)!)\nor (P(n,r))/(r!). x/something is smaller than x itself,(unless there's negatives or decimals, but we don't have to worry about that because r is positive) so the permutation is biggger than the combination.
270990
Descriptive Statistics do no intend to "fit" a distribution on the data or to predict new observations given the data.\nInstead, it pretends to "describe" the data in terms of its central tendency measures (mean, median, etc.) as well as dispersion statistics (variance,  rank, inter-quartile rank, etc.), bias statistics, etc.\nSummaries like histograms are also a good example of descriptive statistics.
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I have a different point of view to the post above. If the probability of the given event occuring is 0, the the probability P(A|B) is a nonsense sentence and undefined, in the same manner as division by 0. There's nothing wrong with the existence of meaningless statements.\nThe aforementioned probability being zero is only enforced if you change the probability space to that of P(A AND B) which can easily be argued to be zero.
984407
In most cases you just report them both.  This would be called a "bimodal" distribution.  (You absolutely DON'T average them.)  A related question is what do do when it's a 3-way or more tie for most common.  Usually in that case you say there is "no mode" because nothing is more frequent than anything else.
508493
I've found that the Wikipedia articles on probability distributions are top notch and contain enough information for you to solve this problem on your own.\n\nHint: solve the algebraic expressions that relate mean and variance so you can express p or n as algebraic relationship of one another and then do some substitions.\n\nAt first I thought you needed to know p or n up-front however I realize that you have all the info you need to solve this problem algebraically.\n\nGood luck with your homework.
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You can do this two ways...\n\n1. Find P(x=1) + P(x=2) + P(x=3).\n\n2. An easier way: find P(x >= 1) by taking P(x = 0) and subtracting that from 1.\n\nEither way, from the table, you should get the same thing.\n\nedit: That rule of thumb will work for the binomial distribution because it is discrete.  It would not for, say, continuous distributions.  A better rule of thumb might be this...\n\nP(X >= x) = 1 - P(X < x).\n\nThis will always work.  Hopefully you understand the reason behind it; the probability of an event is equal to one minus the probability of the opposite of that event.
90357
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.