Please help me with this Statistic question..!?
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For intents and purposes the above 2 are correct, but just\nto be absolutely correct you would have to know the population of the town. Because after the  first pick of a person, the\npercent of people having insurance is slightly less than 60%, And\nafter the second person is picked the probability is even less than that of the second pick.  This is called a dependent distribution because each subsequent pick has a probability\nsomewhat altered by the previous pick.\n  Now suppose the population of adults is n. The first pick has a probability of .6, the next is (.6n-1)/(n-1) and finally  (.6n-2)/(n-2).\n\n   Total  probability : (.6)(.6n-1)(.6n-2)/{(n-1)(n-2)}   Everyone in your class will give (.6)^3 and you may also. But just to be a smart a__ you can show off with this. Remember me well.
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Answer is 10.2% of the sample(25) have a mean score within 3 points of the population mean score of 220.\n\nSD=10 --> 34% (normal distribution)\n   3 from mean = 3/10 x 34% = 10.2%.
940569
you can start here, this may not be the exact answer but has some interesting data with relevance to your question, good luck
49011
http://www.windows.ucar.edu/tour/link=/earth/statistics.html
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.
584200
1) the pdf is actually a binomial.\nIf you consider only 1 man, what is the probability that he will be paired to his wife out of the 10 women? \nIt is 1/10\n\nLet X be the number of paired couples, paired meaning they are paired to their own spouse.\nTherefore X~B(10,1/10)\n\nSo if we want to find the probability that there is only 1 pair, using the formula of a binomial dist,\n\nP(X=1)= [10 choose 1] (1/10) (9/10)^9\n= 0.3874\n--------------------\n\n2) To find the mean and variance, using the Binomial parameters formula, we compare what we got in part 1 with the general formula,\nX~B(n,p)\nX~B(10,1/10)\n\nSo our mean = (n) multiply (p)\n= 10 x (1/10)\n= 1\n\nVariance = (n) multiply (p) multiply (1-p)\n= 10 x (1/10) x (9/10)\n9/10\n----------------------\n3) P(X >= 3) = 1 - P(X <= 3) \n= 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) ]\nreferring to the above eg in part 1, its easy to derive\n= 1 - [ 0.0349 + 0.3874 + 0.1937 + 0.0574]\n= 0.3266
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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.
17841
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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go to this site, its very good,\n\nit gives clear explanations and statistics
586725
The answer would be C.\n\nIn a two tailed test, you're looking to see if the mean is either significantly greater or less than k which is why it is called two-tailed.  The null hypothesis would be that mu = k and the alternate, not equal.  Thus, c
582964
You are going to have to directly ask an actuary or professor of actuarial science.  You may redirect this question to the mathematics section.  I have a great statistics background and I don't know where to find the answer to that.  I can tell you the history of Sommer's D, which is an obscure but unbelievably useful statistic, but I cannot tell you the origin of the name for that.
774817
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.
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.
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Let me tell you a story about anecdotal evidence.  I had this friend who always supported his arguments with anecdotal evidence, and one day I took him to task for it, by presenting him with statistical data that completely demolished his hunches.  :^)
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There would are no statistics for that.\n\nAny other answer would be an out of the hat guess.
49083
This may not be exactly what you are looking for but check it out: http://www.whois.sc/internet-statistics/
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Go to http://office.microsoft.com/en-us/results.aspx?Scope=TC&Query=statistic.\nThis is a Microsoft website for templates.  You could find any template you could need.\n\nI hope this would assist u.