How can the Binomial and Poisson probability distributions be utilized in solving business and marketing-relat
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The Binomial and Poisson probability distributions are both discrete probability distributions for modelling events.\n\nThe Poisson is useful for modeling arrivals per unit time. If you are running a customer support call center, a fast-food restaurant, an amusement park, or the toilet facilities at a venue, you might try to use the Poisson to estimate the rate at which people arrive in 10 minutes, or in an hour.  The Poisson tells you that while the average rate might be 20 per hour, the peak arrival rate for the 99th percentile might actually be as high as 31 per hour.  For capacity planning purposes, you might actually add more servers or more toilets so that people have to wait less time.\n\nThe Binomial can be used to model probability that someone does or does not do something.  Let's say 3 out of 10 people in one trial respond yes in a consumer survey.  Now let's say another survey has 4 out of 10 people say yes.  Does this mean that people were more likely to say use in the second survey?  The answer is probably not.  The Binomial allows you to model the the probability that a larger sample or population will say yes given the mean behavior you've measured.  This will allow you to access whether there is a true difference or if the difference is in the expected random variation.\n\nThe use of these distributions and others is often called Operations Research (OR) in business and military settings.\nDoing a Yahoo Search for operations research binomial may help give you more applications:\nhttp://search.yahoo.com/search?ei=utf-8&fr=slv1-&p=binomial%20operations%20research
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You need to know the rate at which the birds poop (i.e. how often do poops fall per hour).  You also need to know the number of people who get crapped on per hour.  Once you know this, you could probably use the Poisson distribution to provide a probability.
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1st Problem:\n\nLet's see, if we sell 198 thousand candy bars, we are giving them away, which is terrific for volume, but bad for profit.    \n\nRevenue = x P(x), = 9x-(x^2/22)  differentiate to get 9-(2x/22) so x/11=9 x=99\n\nPrice would be  4.50
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Binomial Distribution, \nPoisson Distribution, \nNormal Distribution\n\nhttp://www.futureaccountant.com/probability/
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No.  There are a lot of different types of distributions to model the many phenomena in the universe and in society.\n\nCheck out this list:\nhttp://en.wikipedia.org/wiki/Probability_distribution\nThe normal distribution is way down in the list somewhere!
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production, distribution, and consumption of goods and services
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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/
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X follows a Binomial distribution with n = 10 and probability of success p = 1/6
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Check out Markov Chain's.  You will find many uses for matrices.
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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.
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While there are many techniques you can use to find the right target audience for a company, the most common technique is based on identifying the highest potential customers of its products/services.   \n\nLet's pretend we are trying to complete this exercise for a fictional company called XYZ electronics.\n\nWe would take a list of XYZ's customers and order them from most valueable to least valueable based on their history of purchase.  The number of products a customer has bought, the length they have been a customer and the profit margin of the products purchased would help us calculate the value of the company's customers.   Assuming an 80/20 rule (the pareto principle- 20% of customer drive 80% of value), we would be able to segment out the top 20% of XYZ electronic's customers.   \n\nWe would then analyze what was common for the top 20% relative to the non-20%.  Are they men?  Do they live nearby stores?   How old are they?  Based on this analysis, we may have identified its target market.\n\nIf XYZ electronics is a new store, we may want to do a general industry analysis and do the same exercise for entire electronics category.   Who buys the most electronics?  Who are the most valueable targets based on purchase needs and purchase ability.
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One cool way that I have seen it derived is based on Jayne's Formalism.\n\nAssume we sample process, and measure the mean and standard devation of the samples. There are many possible distributions that will produce those exact mean and standard deviation. \n\nJayne's Formalism chooses the distribution which maximizes the entropy of the distribuiton. This is viewed mathematically as the distribution that makes the minimum number of assumptions regarding the distribution.\n\nI always thought that was kinda cool.
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In solving problems in probability, you must determine first if it is binomial (with replacement) or hypergeometric (without replacement).\n\nsince you did not specify that the card you'd drawn was not replaced in the deck of cards, the distribution is hypergeometric.\n\nTo solve this problem, you can use any of the ff method.\n\nfirst method\n\nfor 2 spades:\nP(spade on the first draw)=13/52=1/4\nP(spade on the second draw, provided you picked a spade on the first draw)=12/51\n\nTherefore\nP(2 spades)=1/4 * 12/51 = 1/17 = 0.058823529...\n\nfor 2 aces:\n\nP(ace on the first draw)=4/52=1/13\nP(ace on the second draw, provided you picked an ace on the first draw)=3/51\n\nTherefore\nP(2 aces)=1/13 * 3/51 = 1/221 = 0.004524886...\n\n\nSECOND METHOD:\n\nUsing combination\n \nfor 2 spades;\nThere are 13C2 ways to pick two spades\nThere are 52C2 ways to pick two cards\n\nP(2 spades) = 13C2 / 52C2 = 1/17 = 0.058823529...\n\n\nfor 2 aces\n\nThere are 4C2 ways to pick two aces\nThere are 52C2 ways to pick two cards\n\nP(2 aces) = 4C2 / 52C2 = 1/221 = 0.004524886...\n\n\nNOTE: Both solutions I have shown is applicable only for hypergeometric. That is, there was no replacement done during the experiment.
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I answered this one for you earlier...\n\nYou have X ~ Bin(6, 1/6)\n\nas you have 6 trials, with the proability of success (a four) being 1/6\n\nYou then simply need to stick this in the formula:\n\nP(X = x) = nCx × p^x × (1-p)^(n-x)\n\nIn this case you want the probability of 2 fours:\n\nP(X = 2) = 6C2 × (1/6)² × (5/6)^4 = 0.2009
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Binomial distribution, (a) p=0.5, n=12; (b) p=0.5, n=13.  You can do it!
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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/
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P(selecting a person at random and he has the symptom and is detected by the procedure)\n\n= P(patient having it) and P(detecting it using the procedure)\n\n= (0.5/100)*(98/100)\n= 0.0049\n\nBut if the person is not randomly selected and you are sure s/he has the problem and its just a matter of detecting it with the procedure, then Probability = (98/100) = 0.98