Math pobability and stats question 2?
113517
There will be a table in your statistics textbook that gives probabilities for a normal (Gaussian) distribution.  Usually it is for a "standard normal" where the mean is 0, and the standard deviation is 1.  The table will show upper tail probabilities for values of x ranging from 0. to 3. (or higher), where the it gives the probability that random sample xi > x.  For general samples with a mean different than 0 and a standard deviation different than 1, you have to apply a conversion to use this table, as follows:\n\nLet m = the mean of a normal distribution\nq = the standard deviation (usually called by the greek letter sigma), yi = boundary of upper tail distribution.\nIn this case, m = 92500, q = 8400, yi = 95000.  The value of x to look up in the table is \n(1) x = (yi-m)/q.  \nThis will tell you the probability that ONE random element, xi > x.\n\nHowever, if you take a sample of N elements, if the standard deviation of the entire sample is q, the standard deviation on the MEAN of the sample is \n(2) q'=q/sqrt(N).\nSo for the sample of 64 families, the standard deviation on the mean is 8400/sqrt(64) = 1050.  This is the value to use for q in equation (1).  Equation (1) then gives x = (95000-92500)/1050 = 2.38.\n\n  So the probability is the same as the probability that x > 2.38 for a standard normal distribution.  From the table (or from an online calculator like the one at\nhttp://bayes.math.montana.edu/cgi-bin/Rweb/buildModules.cgi\nI find that the probability = 0.008656319 , less than 1%
577657
since it is "at least", use 1-(probability of no aces or no kings), so:\n\n1-((44/52)*(43/51))\n\nfor the second problem, I would use the idea of (probability of no orange + probability of one orange), so:\n\nprob of no orange:\n((5/8)*(4/7)*(3/6)*(2/5))\n\nprob of one orange, assuming you've learned combinations and permutations:\n(3C1*5C3)/8C4\n\nadd the two prob from above.\n\nhope this helps...probabilty can be quite ambiguous.
933392
Probability in statistics would agree with your claim to a point of certainty
834182
sample space=8\nnumber of outcomes with at\nleast 2 heads=4\n\nP(at least 2 heads)=4/8=1/2\n\nodds:even money\n\ni hope that this helps
836301
Try this website for help with your statistics homework:\n\nhttp://www.statsoft.com/textbook/stathome.html
783821
5 out of the 8 items in Box A are non-defective.\n3 out of the 5 items in Box B are non-defective.\n\nThe probability that an item from Box A and an item from Box B are drawn and are both non-defective is:\n\n5/8 * 3/5 = 3/8\n\nThe probability that an item from Box A and an item drawn from Box B and both defective is:\n\n3/8 * 2/5 = 3/20.\n\nThis means that the probability that one of the two items drawn is defective is:\n\n1 - p(both are not defective) - p(both are defective)\n=1 - 3/8 - 3/20\n= 19/40\n\nWe can see this by showing all possible outcomes. The items in Box A are shown on the top edge, and the items in Box B are shown on the right edge\n\nN  N  N  N  N  D  D D\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nA  A  A   A  A  B  B  B  N\nB  B  B  B  B  C  C  C D\nB  B  B  B  B  C  C  C D\n\nThe events marked "A" represent 2 non-defective items. The events marked "B" represent 1 defective and 1 non-defective item. The events marked "C" represent 2 defective items.\n\nOut of 19 events where there is 1 defective and 1 non-defective item, the defective item came from box A 9 times.\n\nSo if 1 item is defective and 1 is not, the probability is 9/19 that it came from box A
822095
a) Both: .1(.02) = .002\n\nb) Exactly one:  .02(1-.1) = .018 + .1(1-.02) = .018 + .098 = .116\n\nc) (1-.02)(1-.1) = (.98)(.9) = .882\n\nThey add up to one.  I'm not a probability person, but I think this is correct.
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
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.
154380
I posted these sources earlier to someone else who was looking for math help.\n\nGood Luck!!
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/
696155
Whats the probability that the problem is *not* solved? Tom doesn't solve it (0.75) and Alice doesn't (0.5) so its 0.75*0.5 = 0.375. So the probability it is solved is 1-0.375 = 0.625.\n\nNow, we are given that it is solved. What is the probability that Tom solves it? 0.25. \nSo:\nP(Tom | solved) = P(Tom and solved) / P(solved) = P(Tom) / P(solved) = 0.25/0.625 = 0.4.\n\n(edit - note that the reason the below answer is not quite correct is because it is possible for *both* of them to solve it - so the probability Tom solves it when solved + probability Alice solves it when solved should *not* equal 100% - it should be more.)
359986
Student's t-test can be used to determines the probability that 2 random variables have the same mean. This can then be used to test whether two sets of data come from the same source.
522796
a) Sensitivity = P(Test + | Has disease), so out of the people who have the disease, 93.4% would test positive for it.\n\nSpecificity = P(Test - | Doesn't have disease), so out of the people who don't have the disease, 98.6% would test negative for it.\n\nb) This is again a Bayes' Rule problem\nP(T+|D) = 0.934\nP(T-|ND) = 0.968\nP(D) = 0.002\n\nP(D|T+) = P(D)P(T+|D)/(P(D)P(T+|D) + P(ND)P(T-|ND))\n= (0.002)(0.934)/(0.002(0.934) + 0.998(0.968))
1179837
a) Try counting the possibilities.\n     There is only one way to roll a 12...(6, 6)\n     and there are 36 different possible rolls.\n     so the chances are 1/36\n\nb) There are 2 possibilities for this one (3, 6) and (6, 3)\n     so it's 2/36 or 1/18\n\nThe difference between the book the you is notation.\n1:35 means that for every one success there are 35 failures.\n1/36 means that there is one success for every 36 tries.
894091
None. Two odds make an even
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.