This is a math question in my statistics class?
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r decreases becuase the ellipse is becoming thinner i think that would happen\n\nhope u find out if my answer is wrong GOOD LUCK!!!!!!!!
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.
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P(A or B) = P(A) + P(B) - P(A and B)\n\nP(A) = probability of trying out for basketball\nP(B) = probability of trying out for baseball\n\nNow, P (A and B) = 0, since Jamie will not try out for basketball and baseball\n\nP(A or B) = 1 (or 100%), since Jamie will definitely try out for basketball or baseball\n\nSo, P(A) + P(B) = 100%\nP(B) is more than 42% or more (since you have to consider that Jamie will try out for other positions for baseball).\n\n100% - 42% = 58%\nP(A) has to be 58% or lesser. The only answer that is less than 58% is option A (40)
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Here are two sites that look into the problem:\n\nhttp://www.kw.igs.net/~jackord/bp/f6.html\nhttp://math.fullerton.edu/mathews/N310/projects/e6.htm
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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.
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There are 20 students in the class. There are 4 possibilities when you choose the 3 random students. I've included the probabilities of these 4 possibilities. 1) and 3) are the answers to your questions.\n\n1)   3 boys = (4/20)(3/19)(2/18) = 0.0035\n\n2)    2 b, 1 g = (4/20)(3/19)(16/18)x3(=3C2) = 0.0842\n\n3)    2 g, 1 b = (16/20)(15/19)(4/18)x3(=3C2) = 0.4211\n\n4)     3 g = (16/20)(15/19)(14/18) =  0.4912\n\nFor the probabilities to be correct they must sum to 1:\n\n0.0035+0.0842+0.4211+0.4912 = 1
219821
This cannot be ascertained without knowing how many are being selected in all...\n\n\nhttp://www.futureaccountant.com/probability/\nhttp://www.schoolingkids.com/
158411
Start from www.statistics.com or search "statistics books" in Yahoo Answers. Greetings.
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Your going to compare your mean with the population mean and see if your class really better than the average of the population.\nHo: u = xbar\nHa: u < xbar\n\nalpha = 0.10\none tailed test (of course)\ntherefore: z(table) = -1.28 (i'm not sure)\n\nuse z test since you know the population standard deviation.\n\nz = (x - u)/o/sqrt of n\n\nthat's it.\n\ngood luck
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I've just answered this in your previous question: "Statistics: Variability & SRS Samples of Population?"\n \nIf you look there, you'll find the formula for the s.d. of p_hat. So that increasind n by multiplying it by "m", reduces (devides) sd (or se) by sqrt(m).\n \nFor example, lets say the sd is 0.2 and the sample size is 100. If you increase the sample to 400 (multiply it by 4), the sd is devided by sqrt(4) = 2, i.e., sd = 0.1.
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order the data: 7,9,9,11,12,13,13,13,15,19\n\nfind the Median (middle point, as there are 10 data points it's the average of the 5th and 6th): 12.5\nLower Quartile: 9 (25th percentile, median of lower half, i.e. here 3rd observation)\nUpper Quartile: 13 (75th percentile, median of upper half, i.e. here 8th observation)\nmin. observation: 7\nmax observation: 19\n\nso it's:\n\n    12.5\n9          13\n7          19
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.
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sample space {1,2,3,4,5,6}
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If the population has a mean of 58 and a sd of 8, then the sample means of size 100 will have a mean of 58 and a sd of 8/sqrt(100) = 0.8\n\na) Within 3 of 58 means that the sample mean will be between 58-3 = 55 and 58+3 = 61\n\nz = (55-58)/0.8 = -3.75\nz = (61-58)/0.8 = 3.75\n\nFrom the table, the area between the two is almost 1.\n\nb) The mean weight is less than 2.5 of 58 or more means less than 58-2.5 = 55.5\n\nz = (55.5-58)/0.8 = -3.13\n\nFrom the table, the are greater than that is almost 0.\n\nc) z = (57.5-58)/0.8 = -0.63\nz = (58.7-58)/0.8 = 0.88\n\nFrom the table, the area between is 0.811 - 0.264 = 0.547.\n\nd) 56 +- (1.96)(0.8)\n56 +- 1.568\n(54.432,57.568)
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Let us call x the minutes hand on the clock.\nIf 0<x<=5, then it is train B. (5 minutes)\nIf 5<x<=12, then it is train A. (7 minutes)\nIf 12<x<= 24, then it is train A. (12 minutes)\nIf 24<x<= 25, then it is train B. (1 minute)\nIf 25<x<= 45, then it is train B. (20 minutes)\nIf 45<x<=48, then it is train A. (3 minutes)\nIf 48<x<=60, then it is train A. (12 minutes)\n\nWe add it up to get that for 34 of the minutes, he gets on A, and that for of the 26 minutes he gets on B. There are 60 minutes in an hour, so the probability is 34/60, which reduces to 17/30.
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Joel, you need some help.  This is one of the easiest questions I have ever heard!  You should know this, man!  You're gonna fail your test if you keep cheating on here and don't actually figure out how it works.  Good luck, brotha'\n\nL = -2 or L(x) = -2
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If you have an odd number of numbers, it's true.\nE.g., given the set:   1,   2,   3\nthe median is 2.\n\nIf you have an even number of numbers, it's false.\nE.g., given the set:  1,   2,   3,   4\nthe median is 2 1/2  (the average of 2 and 3)