Statistics homework help!!!!!!!!!!?
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I am making the assumpition that the points you are given form the probability density distribution curve p(x) for the random variable.   This means that P(a<x<b) is ∫[a to b)p(x)dx.  In this type of distribution  ∫[-∞ to +∞]p(x)dx = 1\n\nStart by plotting the distribution from the points given.  The probability of a point being within a range is the area under the curve within that range .  Use geometry to compute these areas: break up the figure into rectangles and triangles.  First get the full area, call that A (this should come out 1).  The width of the area you want is the limit of the values given; in case i, that would be the distance from x = 1 to the right-hand limit of the curve at x = 3, or ∆x = 2.\n\nTo show you how, here is the way to get the full curve area A:\n\nRectangle from (0,.3 to (3,.3): Base 3 height .3, area .9\nTriangle from (1,.3) to (1.5,.5) to (2,.3): Base 1, altitude (.5-.3) = .2, area = .5*1*.2 = .1\n\nTotal area = 1 (as it should).\n\nArea for P(x>1) rectangle base 2, height .3, area .6; triangle base 1 altitude .2, area .1, then P(x>1) = .7, the sum of the areas.\n\nFrom a probabiliy distribution, you cannot get the probability of a single value, only a range.\n\nYou will have to interpolate altitudes for points that do not coincide with the ones given.\n\nEDIT: I have made a drawing of the distribution if that will help.  Find it at http://img218.imageshack.us/img218/8930/distributionhg6.png
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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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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%.
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Take the average of the absolute value of each of the numbers minus the mean...\n\n( |2.44-2.23| + |3.54-2.23| + |1.37 - 2.23|  + |1.5 - 2.23| + |2.3-2.23| )/5
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Hi! I'm a statistician and I use http://www.statsoft.com as well as cheap textbooks from Half Priced Books in my area. Both seem to help. If you have a specific question, feel free to email me. Don't know how much good I'll be because I use fancy software for my analysis, but I'll try.
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This link has statistical calculators:\n\nhttp://calculators.stat.ucla.edu/\n\nA webpage that will lead you to more statistic links.\n\nhttp://math.about.com/od/statistics/\n\nI hope this helps.
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yea, multiplying the single event probabilities of two independent events to get the prob of the independent events conjunction is correct. In other words (to avoid you going to grab an Advil already): p(A&B)=p(A)*P(B) assuming A and B two independent events. That's BASIC stats.\n\nIn excel, use the function POWER. First argument the probability, second argument the number of occurence in a row.
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Scroll to the bottom for Homework Help: http://answerpool.com/6/ubb.x?a=cfrm&cdra=Y&s=669605981
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Do your own homework!!!!
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Is this your idea of Homework Help??? Obviously, no one was interested in helping you plagerize someone else's effort.  I certainly am grateful for that.\n\nDO YOUR OWN HOMEWORK!!!
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try this site\n\n\nhttp://www.futureaccountant.com/probability/
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7\n\norder them; if you have odd number of entries, take the middle one; if even - take average of two middle ones.\n\n\n\nand it's a statistics question
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1-i) P(h>200)=P(S>(200-178)/10)=P(S>2.2)=0.5 - 0.4861 = 0.0139 = 1.39 %   (Where S denotes Standard normal distribution value, h the non standard value).\n1-ii) P(h<158)=P(S<(158-178)/10)=p(S<-2)=0.5 - 0.4772 = 0.0228 = 2.28 %\n2)\n  i)P(39.5 < h <40.5)= P((39.5-37)/3.5 < S < (40.5-37)/3.5)\n                   = P( 0.71 < S < 1)\n                   = P( 0 < S < 1 ) - P( 0 < S < 0.71)\n                   = 0.3413 - 0.2611 = 0.0802 = 8.02 %\n  ii)The limits are 38.5 : 39.5\n iii)P(38.5 < h <39.5)= P((38.5-37)/3.5 < S < (39.5-37)/3.5)\n                   = P( 0.43 < S < 0.71)\n                   = P( 0 < S < 0.71 ) - P( 0 < S < 0.43)\n                   = 0.2611 - 0.1664 = 0.0947 = 9.47 %\nNote : S = (h - mean)/sd.
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This is a proportioal problem\n\nMultiply the means\n\nmultiply the extremes\n\n- - - - - - -\n\nx = the amount of men\n\n4 / 7 = x / 28\n\nthe means = 7x\n\nThe extremes = 4 x 28\n\n- - - - - - - - - - - - - - -\n\n7x = 4 x 28\n\n7x = 112\n\n7x / 7 = 112 / 7\n\nx = 16\n\nThe answer there are 16 men\n\n- - - - - - s-
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Really Bob y .. .cant you do your own homework?\n\nTry searching on google and stop being lazy!
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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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check out awstats or http://google.com/analytics