poisson distribution?
922757
a) Can't sell more than 4, so 1*p(selling 1) + 2*p(selling 2) + 3p(selling 3) + 4p(selling4).\n\nb) Find the number n so that p(x>n) is less than 0.05 for a Poisson "vif" mean 3.
165850
Poisson distribution is a discrete distribution whereas the normal (Gaussian) distribution is continuous.  The Gaussian distribution is symmetrical and goes to infinity on each side, whereas Poisson is infinite on one side and finite on the other.  Per source, “[Poisson] expresses the probability of a number of events occurring in a fixed time if these events occur with a known average rate, and are independent of the time since the last event."  Thus, Poisson is often the best model of a queue.\n\nThe phenomenon of people arriving in groups is not explicitly modeled by Poisson.  It may be partially accounted for by samples with a very small inter-arrival interval.  If you need to model a queue that typically services people arriving in groups of varying sizes, Poisson may not be the best model.
324717
what type of statistics?
549133
substitute in the equation for the poisson probability function\n\nP(n)=(e^-V)(v^n)/n!\n\nv=mean\nn=number\n\nThen eliminate variables until you are left with\n\nv^4+3v^2-4=0\n\nfactorise\n(v^2+4)(v^2-1)=0\n\ntherefore v=1\n\nso mean =1
764469
X follows a Binomial distribution with n = 10 and probability of success p = 1/6
286964
If a random experiment has two possible outcomes, success with proability  p and failure with probability q=1-p. And the experiment is repeated until a success happens. The number of trials before the success is a random variable  with density function\nf(x)=q^(x-1) * p     is called the geometric distribution.\n\n\nhttp://library.thinkquest.org/10030/6dpdgd.htm\n\nhttp://en.wikipedia.org/wiki/Geometric_distribution
165255
Take a look at the error function, and modify it so that you are dealing with distributions in 2 dimensions, using polar coordinates.\n\nFor example, the counterpart of the error function in 2 dimensions, in polar coordinates can actually be integrated, to produce the comulative distribution function\n\nF(R,s)=1-e^[-(R^2)/(2*s^2)]\n\nwhere s is the standard deviation, and R is the radius of a circle centered on the origin. \n\nAs R--> infinitity, then F(R,s)----> 1\nF(0,s)=0\n\nSuppose that the target has an edge R0 such that\n\nF(R0,s)=.5 for a given value of s. \n\nThen, you have to find the value of S for which F(R0,S)=0.9.
1180637
I was answering another similar question, and I came across a website that has a downloadable sample spreadsheet for doing statistical distributions.  Play with this and see if it helps you out.
1028806
I'll assume you already know what permutations and combinations are and how they work.  Now lets say you're conducting an experiment where the result of the experiment can either be a success or a failure (this is called a Bernoulli trial in probability).  To ensure the accuracy of your results, you are going to repeat the exact same experiment three times, and you will look at the total number of successes in those three Bernoulli trials.  If we know the probability that each individual trial will be a success, we can calculate how many successes we can expect to get in three trials.  This is called a binomial distribution.  Let's look at all possible outcomes:\n\nYou could have all successes:\ns s s\nYou could have all failures:\nf f f\nThe first trail may be a success and the other two failures:\ns f f\nAnd so on... here's the list of all possible outcomes:\ns s s\ns f f\ns s f\ns f s\nf s s\nf s f\nf f s\nf f f\n\nLet's say you want to know what's the probability that you will get exactly two successes and one failure from your three trials.  The number of possible outcomes from this list that have exactly two successes in it is (3 choose 2)=3 out of the eight in the list.  Now if "p" is the probability that each individual trial is a success, and if "q" is the probability that it is a failure (meaning q = 1 - p), then the density function for this binomial distribution is:\nf(x) = (3 choose x) * p^x * q^(3-x)\nSince we want to know the probability of getting exactly two successes, we plug in two for x:\nf(2) = (3 choose 2) * p^2 * q^(3-2)\nf(2) = 3 * p^2 * q\nDepending on what p and q are, the probability of getting two successes in your three experiments can change.\n\nThe binomial distribution is one example of many discrete distributions that use combinations in its formulas.  Permutations have other applications, but this answer is getting long, so I'll leave it at that.
199925
Probability function of a poission distribution\n==> P(X=x) = e^(-m)*m^(x)/x! \n[x= 0,1,2,.... ]  countably infinite\n\nMean = 4 ==> m = 4\n\nProbability that x = 2\n==> P(X=2)= e^(-4)*4^(2)/2!\n          = \n\nProbability that x <= 2\n==> P(X<=2)= P(X=0) + P(X=1) + P(X=2)\n==> P(X<=2)\n= e^(-4)*4^(0)/0! + e^(-4)*4^(1)/1! + e^(-4)*4^(2)/2!\n= e^(-4) {1/1 + 4/1 + 16/2}\n= e^(-4) {1 + 4 + 8}\n= e^(-4) {13}\n\nProbability that x > 2\n==> P(X > 2)\n= P(X=3) + P(X=4) + ...... \n(Or) = 1 - {P(X<=2)}\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/probability/
475811
Here's a step by step explanation of how the probability is assessed.  Hope this will make things clearer for you.\n\nhttp://www.mste.uiuc.edu/java/java/birthday/explanation2.html
907438
I could tell you the answer or I could give you an idea of what to consider when you do it yourself.\n\na) Binomial numbers and p = 0.7, q= 0.3 might help here.  You do not know the size of the population of the mailing list so you might need to make an assumption that it is large enough to assume a Normal distribution.  I think the central limit theorem might help here.\n\nb) Binomial theory, and conditional probabilities.  Look it up in your text book!  Again if the size of the population is large enough you might be able to simplify the problem.
220083
I am pondering this question a bit. I might have a better answer for you, but here is a start:\n- - - - - - - - - - - -  - \n\nYour data is not exactly amenable to standard quantitative analysis. You can't really form an average, for example. \n\nYou can find the mode. The most common answer is, diet. So the mode of your data is "diet". \n\nThe data categories, as they stand, are not represented by ordinal numbers. So it is a bit hard to even describe the distribution, let alone the dispersion of the distribution. What you could do, is say that 35% preferred diet, and 65% did not. Then you could try to estimate the error bounds on these figures, given some assumptions. Or do a robust estimate using a jackknife or bootstrap.\n\n- - - - - -  - - \n\nHere is another idea. You might want to know the average number of people that are in a given category. That is, you might want to know how many people like a given type of soda. You have 5 categories, so the average is \n\n(20)/5=4\n\nThere are an average of 4 people in each category.\n\nNow you can find the standard deviation.\n\nFirst find an estimate of the variance V using the standard formula:\n\nV=(1/4)*[(1-4)^2 + (3-4)^2 +(3-4)^2 +(7-4)^2 +(6-4)^2 ]\nV=(1/4)*[9 + 1 + 1 + 9 +4 ]\nV=(1/4)*[24]\nV=6\n\nThen take the square root\n\nS=sqrt(V)= sqrt(6)\nS=sqrt(6) = 2.44948974\n\nThe average soft drink is liked by 4 people, but there is a standard deviation of about 2.4.\n\nSo one can say a soft drink is liked by 4 += 2.4 people (if we set the error bounds to be 1 standard deviation, but some people set it at 2).\n\nYou could even convert this to percentages. Then you could say that a soft drink is liked by 20% of the people, +- 12%.\n\nThe standard deviation assumes that the distribution is normal, that is a bell-shaped curve (aka Gaussian). However, it probably is not. A robust estimate of the dispersion of the dispersion can be formed just resampling methods, like the bootstrap or the jackknife. I have placed a link to these if you are interested, but I think it might be too advanced for your needs.\n\nAnother thing to do is to look at the median, and the median average deviation, or the median average deviation away from the median. A reference discusses these and other measures of dispersion if you are interested.\n\nI still feel that one might be able to do more with this. I will think some more and if I have any good ideas I will post them. I am not a statistician, so it is likely that one of them might have a much better suggestion right away.
402972
http://www.sunnybrook.on.ca/~yaffe/research/density/density.html
251771
Suppose P(k) is the probability that k of the N pots are occupied. Considering the process to begin on the second throw, there are N possible states (k=1,2,3...N). The initial probability distribution (after the first rock is thrown) is:\n\nP(1)=1, P(2)=P(3)=...=P(N)=0\n\nEach time a rock is thrown at random, the probability distribution undergoes a constant linear transformation (this is a Markov process):\n\nP --> P'(k) = (k/N)*P(k) + ((k-1)/N)*P(k-1)\n\nTo answer the question it's necessary to calculate P(N) after r transitions, equate it to .95 and solve for r. That can be done in principle for various specific values of N. For example for N=2, it's not to hard to see that:\n\nP(1) = (1/2)^r\nP(2) = 1 - (1/2)^r\n\nSolving (1/2)^r = 1-.95 = .05 gives about r = 4 (i.e. 5 actual rocks).\n\nI can also solve the case N=3 exactly. In this case, the transition P -> P' caused by each throw is: \n\nP'(1) = (1/3)P(1)\nP'(2) = (2/3)P(1) + (2/3)P(2)\nP'(3) = P(3) + (1/3)P(2)\n\nThis has solution (P(k) after r transitions):\n\nP(1) = (1/3)^r\nP(2) = (2^(r+1) - 2)/3^r\nP(3) = 1-P(1)-P(2)\n\nSo the exact answer in this case is the solution of:\n\nP(1) + P(2) = (2^(r+1)-1)/3^r = (2^(r+1)-1)/3^r = .05\n\n==> r is between 7 and 8 (i.e. between 8 and 9 actual throws).\n\nIf there's an expression for P(N) for general values of N, it's probably relatively complicated. \n\nHowever for large N it might be a pretty good approximation to treat the occupation states of each pot as independent (even though they're not really). Taking r as the actual number of rocks thrown, the probability that any given pot is occupied is exactly:\n\nP = 1 - ((N-1)/N)^r\n\nIf these events were independent, the probability that all pots are occupied would be:\n\nP^N = [1-((N-1)/N)^r]^N = .95\n\n==>\n\nr = ln(1- .95^(1/N))/(ln(N-1)-ln(N))\n\nFor N=3 this gives r = 10 which is a bit larger than the exact value, but it's probably more accurate for larger N.
418035
Abby's answer is for getting exactly \nTTTTTTTTHH\n\nBut the same probability applies for\nHHTTTTTTTT\n\nand\n\nHTHTTTTTTT\n\nThere are C(10,2) ways the 8 tails and 2 heads can occur, so the correct answer is to multiply .5^10 * C(10,2)\n\nThis is the same as c(10,8)/ 2^10 as in the last answer.
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/