Math pobability and stats question?
106983
Here is how to do a confidence interval for a proportion.  It is based on the binomial distribution.\n\n100 (1-alpha)% confidence interval for p is:\n[Y/n +or- z(alpha/2)*\nsqrt([Y/n*(1-Y/n)]/n)]\n(wouldn't fit on one line)\nY is the number of "successes," n is the number of "trials" and z(alpha) can be found on a table (Normal Distribution).\np is the proportion you want the confidence interval for.\n\nHere are some tips on how to set up your problem:\nY, the number of successes, will be 300.  n=450.\nalpha is equal to 0.05.  If memory serves me correctly, the value of z(0.025) is 1.96.  \n\nHope this helps.
1166976
As to part one, I can not recommend any acceptable, known test for such data (perhaps what I've written below might work).\n\nAs to part two, this is a horrid approximation since the midrange as the mean is a major assumption (even more far-reaching than assuming something about the populations' probability distributions).\n\nFor part 3 and also part 1:\n\nThe first problem you run into is that you do not know anything about the nature of the population distribution (normal, uniform, etc).  Thus, a nonparametric test would be required.  I know of no nonparametric tests for any of the statistics you have (minimum, maximum or range).\n\nIf it is possible to obtain the actual values for each of the samples in each set, the Kruskal-Wallis test should suffice.  It is nonparametric, so there are no problems with the requirement of assumptions about population distributions.\n\nFurther, its job is to determine if two samples are far enough apart such that the null hypothesis that the population distributions are the same can be rejected.  This would give you what you desire.\n\nAs far as your problem is concerned, the only statistics you have are minimum, maximum and range.   Perhaps one could derive a test to test the minima or maxima of each set, but one still runs into the problem that there can be no parameter placed on the population distribution.\n\nIf you can make an assumption about the distributions of the populations of interest (normal or whatnot), then it would be possible to derive such a test.  This would be a tedious task, however, as it would require thorough knowledge of order statistics and distribution functions.\n\nOne would need to use a test statistic based of off the difference in the two statistics that you use (I would use the maxima since they are further apart than the minima and probability distributions for ranges are uncharted territory unless you're a Ph.D.).  Thus, your test statistic is based off of the difference between the maxima (Y(n) - Z(n) where the (n) are subscripts, which is 2 in this case).\n\nTo perform this test, you would need to derive the probability distribution of the difference of the two maxima and use that to find the probability of the difference being two or greater.  This is quite a task for anyone and requires a large amount of knowledge in mathematical statistics.  I believe I could perform such a task, but it would likely take some time to complete.\n\nRemember though that you must make a critical assumption about the population before you can even do this.  I would recommend using the assumption of uniformity if you are unsure of the populations' distributions.  This is because the derived test would then be least likely to give a type I error, where you reject the data improperly, because the test is less likely to reject.
19818
try\nhttp://www.peterwebb.co.uk/probability.htm
554599
4.2\n\nfor a binomial distribution the mean is the number * the probability
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.
836301
Try this website for help with your statistics homework:\n\nhttp://www.statsoft.com/textbook/stathome.html
1026561
Okay, your possible outcomes number 16. They are:\n\n1+1=2\n1+2=3\n1+3=4\n1+4=5\n2+1=3\n2+2=4\n2+3=5\n2+4=6\n3+1=4\n3+2=5\n3+3=6\n3+4=7\n4+1=5\n4+2=6\n4+3=7\n4+4=8\n\nYour probabilites are:\n2: 1/16\n3: 2/16\n4: 3/16\n5: 4/16\n6: 3/16\n7: 2/16\n8: 1/16\n\n2,3,5 and 7 are prime, thus the probability of rolling a prime is (1+2+4+2)/16 or 9/16 exactly.
202632
The level of the student you are asking the question is to be taken into consideration for this....\n\n\ntry this site for your need...\n\n\nhttp://www.futureaccountant.com/probability/\n\n\n\nhttp://www.schoolingkids.com/
154380
I posted these sources earlier to someone else who was looking for math help.\n\nGood Luck!!
236313
Wouldn't the probability than 5 of the 38 thought it was exciting be\n\n38C5(.79^5)(.21^33) = 6.63x10^-18 : basically impossible?  Even if you added in the probabilities of 4, 3, 2, 1, and 0, it's still tiny.\n\nSo I would infer that either the 79% is wrong or the sample was not independently chosen.
1011849
a) probability of finding exacttly 2 working bulbs in first three =  3*( 9/12*8/12*3/10) = 27/55.\n\n\nb)\nprobability of not finding even one = 3/12 * 2/11 * 1/10 = 1/220\nprobability of finding exactly 1 = 3 * ( 9/12 * 3/11*2/10)  = 9/220 \n \nthus required probability = 10/220 = 1/22.
458039
In the experiment of testing for the number of errors in a chapter...\n\nLet the event an error appearing in a chapter be a success..\n\nLet "x" represent the number of success.\nThe number of errors in a chapter can be either 0,1,2,...\n==> "x" can carry the values 0,1,2,3,...\n==> "x" is a discrete random variable with range = {0,1,2,..} [Countably infinite]\n\nThe probability distribution of "x" is a poisson distribution. \n[Since there are a large number of trials all of which are identical and independent with only two possible outcomes in each trial, either a success (there being an error) or a failure (there not being an error), with the probability of success in all the trials remaining the same all throught ]\n\nThe probability function of a poisson distribution\n==> P(X = x) = {e^(-m) * m^(x)}/x!\n \nParameter of the distribution\n==> Mean of the distribution\n==> Mean number of errors per chapter =0.8 \n==> m = 0.8\n\nTherefore P(X = x) = {e^(-0.8) * (0.8)^(x)}/x!\n\nProbability that there are less than 2 errors in a chapter \n==> P(x < 2) \n= P(x=0) + P(x=1)\n= {e^(-0.8) * (0.8)^(0)}/0! + {e^(-0.8) * (0.8)^(1)}/1!\n= {e^(-0.8) * 1)/1 + {e^(-0.8) * (0.8)}/1\n= e^(-0.8) + {e^(-0.8) * (0.8)}\n= e^(-0.8) [ 1 + 0.8]\n= e^(-0.8) [1.8]\n\n\nhttp://www.futureaccountant.com/probability/
91512
20 questiosn with five options, the probabilty to choose the correct one is 1/5 per each question.\nThe probability to answer all 20 questions is (1/5)^20.\nThe probability to answer more than 10 is (1/5)^11\nThe probability to answer any eight is (1/5)^8\nThe probability to answer maximun 7 is (4/5)^13
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
1119536
1) If no one is injured, then no subs ever get to play.  The probability is (.9)^11 = .3138106 or about 31.38%.\n\n2) (.9)^9 * (.1)^2 = .003874205 or about 0.39%\n\n3) "At least" is best calculated as the complement of "no substitutes were needed."  In the first calculation we determined the probability of not needing any substitutes.  So the probability of needing at least one sub would be .6861894 or about 68.62%\n\n4) The trick here is that the two injured could both be players, or one could be a player and the other could be the sub.  (They could not both be subs, however.)  The calculation in #2 (above) is part of this--the probability that two of the players were injured.  For the other part we need to know the probability that one player is injured (while 10 are not) and that the sub is also injured.  .1 * (.9)^10 * .15 = .0052302.  So the overall result for either of these outcomes (add this result with the result from #2) is 0.00910438 or about 0.91%.
95999
Don't you think you could have written this more clearly? \n\nIf there is a uniform distribution of the weight w being between 5 and 10, then each of these probabilities is equal:\n\nP5_6=P{5<=w<6}\nP6_7=P{6<= w<7}\nP7_8=P{7<=w<8}\nP8_9=P{8<=w<9}\nP9_10=P{9<=w<10} \n\nThe sum of these is 1, so \n\n1=P5_6+P6_7+P7_8+P8_9+P9_10, and\nP5_6=P6_7=P7_8-P8_9=P9_10\n\nso 1=5xP9_10, so P9_10=0.2\n\nThe probability is 0.2.
815085
Not every problem should be solved with an elephant gun.\nJust ask yourself a different question. What is the probability that x2 is closer to x1 than to x3. since the answer to that question is the same as the question you ask and the only 2 possibilities are certain\nand mutually exclusive, the answer should be the same, 1/2.\n\nWhat about if x2 is exactly in the middle you ask?\n\nIt's like asking what is the probability that a random number in an interval is the midpoint. 1/infinity=0