Math Probability and Stats?
107133
You have all the information you need.\n\nThe Mean is the same as the average. It's just another term for a simple average. So it's $7500\n\nThe standard error of the mean is the standard deviation divided by the square root of the number of samples. So it's just $650 divided by 10, or 65.
1071099
Here is a link to a users manual-specifically the statistics section. You can go to the full contents.\nhttp://www.infj.ulst.ac.uk/NI-Maths/Guides/TexTI-85/Ch3/THBStats.html
121605
I'm assuming that you mean introductory statistics.  This online textbook is very good, it makes good use of hyperlinks and also provides exercises and answers for each chapter.  It also has links to useful sites and books for the various topics.  \nhttp://davidmlane.com/hyperstat/\nHope this is helpful :-)
1099243
No. This is what we call probability. Take a statistics course.
1076864
Put the data of each set into bins according to their leading digit. Then according to Benford's law the distribution should be as follows:\nLeading digit Probability\n1 30.1%\n2 17.6%\n3 12.5%\n4 9.7%\n5 7.9%\n6 6.7%\n7 5.8%\n8 5.1%\n9 4.6%\n\nUsing a chi-squared or Kolmogorov test it is possible to determine which of the four data sets differs significantly from Benford's distribution.
365087
You can expect mind numblingly boring calculations for most of your summer.  The mathematics behind statistics is fairly elementary, it's the staying awake and alert part that's difficult...\nGood Luck, I had to take Statistics in college, and I have to say... It was the best one hour nap 3 times a week a person could possibly ask for...
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/
366018
try this site\n\n\nhttp://www.futureaccountant.com/probability/
552270
would calculate your chance of me calculating a statistics question for you at 1234.367:1
560810
1)  I think you're missing a bit.\n\nGiven that P(W<=4)=0.825 then P(W>4) = 1-0.825 = 0.175\n\nThis is because all probabilities sum to 1.  So since <=4 annd >4 make up all teh possibilities the value could take - then they're total probabilties must be 1.\n\n2)  Expected value\nTo do these you simply have to multiply the probability by the value and total them up.\n\n+36,000×0.7 -6,000×0.3 = $23,400\n\n3)  Binomial\nYou have a bin(6, 1/6) as there are 6 trials and the probability of a success (ie getting a four) is 1/6.\n\nYou require the probability of 2 fours so just using the binomial formula P(X = x) = nCx × p^x × (1-p)^(n-x) you get:\n\nP(X = 2) = 6C2 × (1/6)² × (5/6)^4 = 0.2009\n\n4)  Binomial\n\nthis again is a binomial Bin(10, 0.33) - as we have 10 trials (students) and the probability of each one being a success (ie member of ethnic minority) is 0.33\n\nUse the formula above to get the probability of more than 6:\n\nP(X = 7) = 10C7 × 0.33^7 × 0.67^3 = 0.01538...\nP(X = 8) = \nP(X = 9) = \nP(X = 10) = \n\ntotalling these up gives the probability of more than 6.\nThen the probability of NOT more than 6, P(X <=6) = 1 - P(X>6) using the idea from 1.\n\n5)  The mean of teh binomial is just np (ie the number of trails time the probabilty of success).\n\nSo for your example where we have bin(6, 0.7) it will be 6×0.7 = 4.2 seeds germinating.
601572
Statistics would be a good math course for you to take. It's called "Stats" by students taking the course. The course has histograms and charts and graphs you could use to show your data visually. It also had permutations and factorials good for crunching numbers in the books better. It also has normal bell curve probability.
354101
This is easy, just say there are 100 shaft holders.  So of A, 50 of them are processed there, and 3% don't work, so 1.5 out of 50.  For B, it is 30 x 4%=1.2, and C is 20 x 5%=1.  \n\nSo out of a total 100, 3.7 don't work, the the probability that this one chosen at random wont work is 3.7%.  \n\nHopefully the way i calculated the answer helps you out in the future.
804399
I just took Intro to Statistics at Hunter College, and got an A+.  It's not really difficult.  There's no real calculation in it, except to find the average, and range, and such.  They emphasize understanding the terminology (normal distribution, variance, etc.), understanding the advantages of using one method over another, which test to use under certain situations, and such. \n\nHope I helped!
418571
The following link has a number of Yahoo Groups on the topic of math and statistics:\nhttp://groups.yahoo.com/search?query=math.html&submit=Search\n\nThe following Yahoo search for statistics societies has a number of associations included in the results:\nhttp://search.yahoo.com/search?ei=utf-8&fr=slv1-&p=statistics%20societies\n\nAnother Yahoo search for Bayesian Statistics Society lists some other resources:\nhttp://search.yahoo.com/search?ei=utf-8&fr=slv1-&p=bayesian%20statistics%20society.\n\nA couple of specific groups or forums of interest may be:\nhttp://www.mathforum.org/\nhttp://www.bayesian.org/\nhttp://www.rss.org.uk/\n\nIf there is a local college or university, you might contact their statistics department to see if they have such a group or club.\n\nBTW, I quite like Statistics and, 20 years after graduating from university, I'm teaching myself statistics and just finished taking a class in Bayesian statistics. I'm not in the "looney bin". I've used Statistics almost every day for the past 4 years. And the best part -- I'm getting paid pretty well to do it. :^)\n\nGood luck to you!
412043
I can't believe that this question is being asked in the sixth grade.  Maybe half of my college students could answer this.\n\nThe best way to answer this question is to make a table.  Make the rows the results of the first die and the columns the results of the second die.  Then put the product of the two roll inside of the table.\n            1   2   3   4   5   6\n          ------------------\n        1| 1   2   3   4   5   6\n        2| 2   4   6   8  10  12\n        3| 3   6   9  12  15  18\n        4| 4   8  12  16  20  24\n        5| 5  10  15  20  25  30\n        6| 6  12  18  24  30  36\n\n(Sorry I can't get the columns to even out!)\n\nNotice that there are 36 possible rolls of the two dice.  To get the two probabilities, just count the number of times you get the event in the table and divide by 36.\n\nProbability of an even product = 27/36\nProbability that the product is less than or equal to six = 14/36\n\nI see a lot of incorrect answers up there.  I teach statistics at the college level.  Trust me, this is right.
461896
Wikipedia reveals all.\n\nThe third link lists online statistics textbooks.  The Statistics Homepage is especially good.
292028
For and event, odds are ususally expressed as the number of possible outcomes that qualify as a success compared to the total number of possible outcomes: 1 in 6, 3 in 10, etc.\n\nProbability is the same thing expressed as a fraction:\n1/6 = .66... 3/10 = .3\n\nWhen you select k objects from a set of n objects, the number of possible ways that you can do this is n!/(k!(n-k)!).\n\n\nIn all of your problems, your event is selecting a set of 3 objects from a set of 16. The number of possible outcomes of this is: \n16!/(3!13!) = 16x15x14/6 = 8x5x14 = 560.\n\nThe number of outcomes that are considered to be a success are:\n\n3 forks (pick 3 forks from the set of 7) = 7!/(3!4!) = 7x6x5/6 = 35\n\n2 forks 1 knife  (pick 2 forks from the set of sevan and 1 knife from the set o 5) = 7!/(2!5!) x 5 = 7x6/2 X 5 = 21x5 = 105\n\n3 spoons (Select 3 spoons from the set of 4) = 4!/(3!1!) = 4\n\n1 fork, 1 knife, 1 spoon  (Select 1 fork from the set of seven, 1 knife from the set of 5 and one spoon from the set of 4) = 7!/(1!6!) x 5!/(1!4!) x 4!/(1!3!) = 7 x 5 x 4 = 140\n\n1. P(3 forks) = 35/560\n2. P(2 forks, 1 Knife) = 105/560\n3. P (3 spoons) = 4/560\n4. P(1 fork, 1 knife, 1 spoon) = 140/560