can u help me to proof where is the mean is always grater than the median in statistics?
374019
Sorry your answer was wrong.\nMean can be less than median also as median is not affected by extreme numbers at its both ends while mean is.\n\ne.g. find mean and median of following data.\n     1, 3, 5, 9, 18, 180\n\nhere your answer for mean will come to\n               36\n\nwhile if you find median for it then it will come to\n            7\n\nIn this example mean was greater than median.\n\ne.g. -180, 18, 5, 3, 1, 9\nhere your mean is -24\n\nwhile your median is 4.\n\nSo from both example you can say that\n\nnone of them is always greater than other one nor it is always smaller than other one.
472192
To elaborate further, your mode does tell you what the most common value in your data is, but, like the median, it's a somewhat arbitrary value.  For example, take the data set 1,2,2,3,7.  Both the mode and the median are 2, and the mean is 3.  Now take the data set 1,2,2,25,50.  Again, the mode and the median are both 2, but the mean is now 16.  Now, say that second data set represents the number of shark attacks that have happened in the past 5 years off the coast of a beachfront property you're thinking about buying.  I'd much rather know the mean number of shark attacks, not the mode or the median, wouldn't you?
96188
I am truly wondering where you have found this statistic? I have not seen that this is true in any sense
911531
Arrange the numbers in increasing order.\nFind the middle number in the sequence of increasing numbers.\nThat's the median.\n\nIf there are an odd number of values (such as 31), then it's easy to find the middle number (it's the 16th one, since there are 15 lower numbers and 15 higher numbers).\n\nIf there are an even number of values (such as 30), then find the TWO middle numbers (the 15th and 16th numbers in this case, with 14 lower and 14 higher). The median is the AVERAGE of these two middle numbers (i.e., add them up and divide by 2).
580071
1,3,5,8,8\n\nUsing these numbers, the mean is 5 (1+3+5+8+8=25  25/5=5), the median (middle number) is 5.  The mode is 8 since it appears the most.  \n\nTherefore, the difference between the largest and smallest number is 8-1=7.
682546
Mean = 2.59\nMedian = 3.1\n\nAnswer: 3.1 - 2.59 = 0.51
791151
Median is the "resultin the middle" Mean is the average of all results\n\neg:  7 results:\n\n\n1  2  1  9   9   9   9\n\nThe Median is 9\nThe mean is 5.7\n\nIncidentally the modeal is the numberthat occurs most - eg:\n\n1  2  5  5  3498  2323\n\nThe modal of this is 5 - the only number in fact which appears more than once. As you can see, except in VERY odd circumstances, the modal result is grotesquely misleading as an "average"\n\nHaving explained some of your question, I will let others answer the other bits, you multiple-askers you :)
991331
If there were extreme data points, for example out of a sample of ten, 5 data points had the value "5," two data points had the value "6," two had the value "4,"  and 1 had the value "10," the median value would be a value between the 5th and 6th data point, the mode would simply be the one with the most frequency, 5. Since the outlier 10 data point skews the mean upward, but the trend is lopsided towards higher numbers, the mean, median, and mode are very close.
880673
hint: mean value theorem
43772
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.
56254
since it was defined, i suppose.  i'm not positive what you mean by "the point of greatest distribution" but if i interpret it as meaning the point at which the probability distribution is the greatest, then that refers to the *mode* of the distribution.\n\noften the mean, the median and the mode are the same - if you have a symmetrical distribution like a gaussian or normal distribution, then the mean, median and mode all line up at the same value.  \n\nhowever, many real-life distributions are not symmetric and are skewed towards one side.  a good example is household income.  here is a chart of weekly household income in australia:  http://www.abs.gov.au/Ausstats/abs@.nsf/0/d62255a4a6b59475ca256cbf001721a1?OpenDocument\n\nas you can see, the statistical mean and median are much higher than the most probable value.  this is because the high wage-earners skew these value upwards.\n\nas for WHEN these terms came into use, i can't tell you that for sure, but this page seems to indicate that the mean goes back at least as far as aristotle:  http://www.amstat.org/publications/jse/v11n1/bakker.html\n\ncheers!
842178
Start with the whole array as set Q.\nUse algorithm X to get the median M of Q.\nSet a = [size of Q]/2.\nIf a = k, stop; M is your answer.\nIf a > k, set Q = the values less than M, otherwise set Q = the values greater than M.\nRepeat.\n\nThis algorithm is a bit off because when [a > k], the comparison [a <=> k] is subsequently invalid.  But you get the gist of it.  Use X in a binary search mode.
391744
This is a simple two tailed test of the equality of means.\n\nLook this up in a stats book, but the Hypothesis that you are testing is Mean1=Mean2.
905903
There are three medians in a triangle, not just one.\n\na) is your answer.
817419
Proof.\n\nsin : [a, b] → ℝ is a continuous differentiable function, so by the mean value theorem, ∃ c ∈ (a, b) such that\n\nsin'(c) = (sin(b) - sin(a)) / (b - a), i.e.\ncos(c) * (b - a) = sin(a) - sin(b).\n\nNow since ∀ x ∈ ℝ, -1 ≤ cos(x) ≤ 1, it follows that\n\n| sin(a) - sin(b) |\n= | sin(b) - sin(a) |\n= | cos(c) * (b - a) |\n= | cos(c) | * | b - a |\n≤ 1 * | b - a |\n= | a - b |. ∎
773196
Mean: 113\nMedian: 108-116\nMode: 108-116\n\nThe mean was the hard one, I actually took the middle score from each range (eg 90-98=94) and multipled it by the number of times it occured (eg 94x6=564) and added all 5 (564+2266+4816+3388+1170= 12204). Then divided it by the frequency (6+22+43+28+9=108). Total (12204/108= 113). It's impossible to find a mean of a set of ranges, you need exact figures.\n\nThe mode is the most commonly occuring range, the median is the central range. These figures can't be accurate without the full actual IQ scores. But based on the figures provided, the mean, median and mode are as I stated above. If you provide the exact figures I'll work out the exact mean, mode and median.
942658
If by "resistant" you mean "unaffected", you are correct.  You can multiply the highest number in a list of data by a million and it would have no effect on the median but the mean would shoot way up, and not represent the numbers very well.