Here are some statistic problems, please help if you can....?
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Try doing a search on yahoo/google about the peoples' careers. Using the search terms "Tiger Woods' finicial information" or something simular.
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1)..................\n2)..................\n\nIf unable to read Visit me at http://www.i-m-possible.co.nr
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This Electronic Statistics Textbook offers training in the understanding and application of statistics. \n\nExploratory Data Analysis is also a good site\n\n\navailable at the site below
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plz complete the question. i like to solve the questions on probability, but yur question is incomplete.
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No problem \n\ngo to google and search under "Z table" +statistics. \n\nYou'll find a lot.
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You might try DAP. You can find it here along with a lot of other free statistical software.\nhttp://members.aol.com/johnp71/javasta2.html
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1.) A - 0.6666666\n2.) C - $12.26\n3.) B - 0.308\n4.) C - 19.87\n5.) A - $.65
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By reading your question, it appears that you are first and foremost seeking to "define" mathmatically what it is that you are attempting to do to reach your results-- presumably so that you can study that area of statistics further and apply it to your experiment.\n\nWith that in mind, what you are dealing with in your experiment is called "Event Probability."  In Event Probability, you are dealing with concepts of "conditional probability", "frequency", and "relative frequency."  For instance, what you are doing is very similar to the flow charts that appear in many family's medical books with a "Decision Tree" that allows you to answer "Yes" and "No" questions along a Flow Chart to reach an ultimate probability that you have a particular illness.  In other words, you are creating an algorithm by measuring the "relative frequency" of events and compound events and drawing a conclusion from that information.  People who perform these experiments are called "frequentists."  For all intents and purposes, this is what actuaries do at insurance companies and financial firms.  I'd read about those areas of statistics first which should prove very helpful in providing you with a structure for reaching the most helpful and accurate conclusions from your experiment.
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Is there any statistics to support what U said???\n\nWeirdo..
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You can't apply statistics (to anything) unless you have a sufficient number of values to work with.  For example, if you took a poll of who people were going to vote for, if you only asked 5 people, you might not get an accurate representation - but if you ask 5000 then the margin of error gets smaller.  With molecules, there are millions of them.
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There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
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From checking statistics not so well.....
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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.
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I can tell that nobody wants to just do this problem.  There's nothing interesting about it.  It's a typical problem that a stat student does to show understanding.  Get out your textbook and study up how to solve it.
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1) Find the probability of getting a red ball when a bag contains 2 red balls and one blue ball.\n\n2) There is a red ball, a blue ball and a green ball in a bag. Find the probability of getting either a red ball or a blue ball.\n\n3) There are 10 yellow chips and 20 purple chips in a sack. When one chip is selected at random, find the probability of NOT getting yellow chips.
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Do a confidence interval calculation and see if the confidence level for the data you've collected is above that required for 98%.\n\nI can show you how but given your other questions, it's clear you haven't really done any work on this yet and you're just looking to get someone to do your homework for you.\n\nLet me know if you still can't do it after you've tried a while, and I might help you out.
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The problem is incompletely or improperly described.\n\nEach member of the population has some characteristic value.  The mean is the average of this value taken over the entire population, which, assuming each member is an equal contributor, is M = sum(values)/N, where N is the population.  If each member has the same value, say 515, then mean would be 515 and the standard deviation would be zero.  If each even-numbered member has value 1030, and each odd-number member has value 0, the mean is 515 and the standard deviation is (approximately) 515.\n\nSo, you have provided enough information to solve the problem satisfactorily.