Statistics: An Avg of an Avg is incorrect, it should be an average of the data?
411133
Eh?  Not sure what you're saying here.\n\nDo you mean "what is the correct way of talking about an average"?\n\nIn which case it is the average of the data (where we add up the results and divide by how many we had)\n\nAn average of an average would only make sense in the case where we are looking at the distribution of an average and then want to find the average result of that distribution.  If you want to go that way then....\n\nSuppose we have a population (eg people's heights) with mean 170cm and standard deviation 12cm.\n\nThe we take a sample of say, 50 values from that population (ie we measure the heights of 50 individuals)\nThen the we take an average (posh word: MEAN) of this sample.\n\nWe then take another sample of 50 values and get the mean of that sample.\n\nWe then take another sample of 50 values and get the mean of this sample.\n\netc.\n\nWe cacn then plot the frequencies of these means and get the distribution of these means.\n\nIt can be shown that it will have a normal distribution with mean 170cm (same as the population), but with a standard deviation of 12/root(50) = 1.697cm.\n\nSo the average of all these averages is 170cm - the same as the original population.  But guess you didn't want all that.\n\nCheers
997805
==================\n#include <iostream.h>\n\nint main(void)\n{\n\n   int count; // holds the count\n   float sum, number, avg;\n\n   sum = 0.0;\n   count = 0;\n\n   cout << endl << endl   //  Make some space\n        << endl << "Enter a number.  Enter a zero (0) when you have"\n        << endl << "finished entering numbers and want to compute the"\n        << endl << "average of the inputted numbers." ;\n\n   do {\n         cout << endl << endl << "Data element #" << count + 1\n              << ": ";\n\n         cin >> number;\n\n         if (number != 0.)\n           sum = sum + number;\n           count = count + 1;\n\n      } while (number != 0.0);\n\n      averg = sum / (count-1);\n      cout << endl << "The average is: " << avg\n           << endl << endl ;\n      return 0;\n}\n\n\n==================\n\nThis program computes the average of a set of data values. Commented a bit for you to understand\n\nEnjoy
580641
If the data is roughly symmetric with no outliers (values that are unlike the rest of the data), then yes, SD should be fine.  However, if the data is skewed (having a tail to the left or right) or there are outliers, then I would use IQR to describe the spread of the data instead.
900254
They are different types of averages.  Mean is the actual average, median is the middle number in a set of data, and the mode is the most repeated number in a set of data(there can be more than one).
929935
variations matter greatly in interpreting data.  Something might seem like it is returning significantly different results than the control but stat evaluation may show that the variations are minimal at best
189376
Hi,\n\nA fitted value is the mean (Average)\n\nOf a statistics, I know it's confusing sometimes!!\n\nCheers!!\nHope that helps!!
383531
========\n\nApparently purity, calculation or measurement errors have been there.\n\nYou repeated the thrice from the same solution to get an average.\n\nTry another solution.
8570
that is the fit of a statistical model...it says that the results varry 56.46% becuase of the variables and data itself
551009
If you can't answer this, you have no idea what the empirical rule is.\n\nAbout 68% of all values fall within 1 standard deviation of the mean for the normal distribution.
972191
The two most common methods of normalizing data are:\n\n1. Subtract the mean, then divide by the standard deviation (or variance)\n\nor...\n\n2. Subtract the minimum, then divide by the range.\n\nThe first method above yields a distribution with a mean of zero and a standard deviation (or variance) of 1.\n\nThe second method yields a distribution which ranges from 0 to 1.\n\nOf course, other methods are possible, such as subtracting the median then dividing by the inter-quartile range.
354866
The average value of a function is given by\n\n(1/(b-a)) S f(x) dx, evaluated from a to b (I'm using S as the integration symbol).\n\nSo for this question, the average will be given by\n\nVavg = (1/6-0) S e^t + te^(2t) dt\n= (1/6){e^t + [(e^2t)/2][t - (1/2)]}\n= (e^t)/6 + te^(2t)/12 - e^(2t)/24\n\nEvaluating from 0 to 6,\n= (e^6)/6 + 6e^(2(6))/12 - e^(2(6))/24 - {(e^0)/6 + 0e^(2(0))/12 - e^(2(0))/24}\n= (e^6)/6 + e^(12)/2 - e^(12)/24 - (1/6 - 1/24)\n= 74663\n\nI've been looking at your answers and the equation and either there's something missing in your equation or there's an answer missing somewhere.
417556
the average would be considered ... well average...
935926
It means that there is a large gap in the middle of the data set with one or a small ammount of entries being grossly larger than the majority of the set:\n   the set 1,2,3,4,30 has a mean of 8 and a median of 3\n   the set 20,22,28,32,33,34,64,68,71 has a mean of 49 and \n   a median of 33
332290
Generally it is the standard deviation, or range of typical answers.  The mean usually is the center of the bars, but not always.
270417
Succinctly, for the normal ditribution, the three averages are equal, that is the mean, median and mode are all the same. Not so in other cases.
735064
That sounds familiar... Think I'm working on it now! But I don't remember q. It is n in my book. In which case, q (if I'm thinking right) is the number of samples. Then again what's up with the ^'s? If np was > 10, then it meant that the distribution was approx. normal. You might be ahead of me right now...
111022
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