can u find the range & the standard deviation of the data?
437434
Range: 19\nStdev: 6.27\n\nRange is just the max value minus the min value.  To learn more about calculating standard deviation, visit http://en.wikipedia.org/wiki/Standard_deviation\n\nSo, I saw someone got a different standard deviation than me, so I checked it again on Excel and I still got 6.27.  Instead of dividing by 7, I divided by 6 (n-1).\n\nIf these data are a sample of a larger population, then you divide by 6 (n-1) to estimate the stdev of the population, so the stdev is 6.27.  If you just want to know the stdev of those numbers, then you divide by 7 (n) so the stdev is 5.81.
149503
Right, so feed the numbers into your calculator and press the SD button.\n\nOr enter it into Excel and use teh standard deviation function.
802381
I read "Range Per Day"
826809
It is probably best to determine the range after you have the graph.  You will need to look at the local maximums, local minimums, and the horizontal asymptotes to determine the range.  (In other words, it is not a fast problem).
1153957
The website below seems to be what you're looking for.  The spreadsheet you can download uses the NORMDIST function to do the graph.  Check out the help feature for the function to see its equation.\n\nI assume you know how to calculate the mean and standard deviation for your data.  The AVERAGE and STDEV Excel functions are good for this.
861788
There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
1014917
You must  write your question more understandable.\n\nFor the test results, you can calculate the average, variance, standard deviation, min and max grades. You can also find the frequency of the marks.
928318
Standard deviation tells how far away a number in a set is from the mean. Like in data where the numbers are 2,4,6,8,10, the mean of this data is 6. So you have to find the difference of each number and the mean, then square each answer and add together(Sum of squared deviations). Divide this answer by the same number you divided the mean by - 1. So in this case you divided the the total of the numbers above by five, so divide the sum of squared deviation/5-1, this gives you the number they call the variance, take the square root of the answer, and you will get the standard deviation.  So, the sum of the squares for this example is 40, by taking the the square root, you get 6.32
1105474
You can get a standard deviation from any of them.  Define "good".\n\nIf you are asking which has the largest SD, it's the second set.  The order (from smallest SD to largest SD) is 3, 1, 2.  Set 3's SD is less than set 1's because if you take a set of numbers and multiply by the same number, it multiplies the SD by that number.  So if you take set 3 and multiply by 7, you get set 1, and its SD will be 7 times that of set 3.  Set 2's SD is larger than set 1 because the numbers in set 2 are more spread out than those of set 1.
1118574
Find the midpoint of each interval.\nFor 1-3, the midpoint is (3+1)/2 = 4/2 = 2\nFor 4-6, the midpoint is (6+4)/2 = 10/2 = 5\nFor 7-9, the midpoint is (9+7)/2 = 16/2 = 8\n2(1) + 5(6) + 8(3)\n= 2+30+24\n= 56\nn = number of data points = 1+6+3 = 10\nvariance = [1/(10-1)][((2^2)(1) +(5^2)(6) +(8^2)(3)) - (56^2)/10]\n= (1/9)((4+150+192) - 3136/10)\n= (1/9)(346 - 313.6)\n= (1/9)(32.4)\n= 3.6\nstandard deviation = sqrt(variance)\n= sqrt(3.6) \n= 1.9
937201
First find the mean (average).  Then subtract each data number from the mean, and square the difference.  Add up all these squared differences and find the average of that sum.  Then take the square root of that average.  That's the standard deviation.
484697
Find the inter quartile range, which is IQR = Q3 - Q1, where Q3 is the third quartile and Q1 is the first quartile.  Then find these two numbers:\n\na) Q1 - 1.5*IQR\nb) Q3 + 1.5*IQR\n\nAny observation that is below a) or above b) can be considered an outlier.\n\nedit: Chadwick, quartiles are considered robust, meaning that they are not highly effected by outliers.  This is because it takes location into account, not the values.  Let's look at your data set (sorted).\n\n2 3 6 9 13 18 21 106\n\nposition of Q1 = (8+1)/4 = 2.25\nQ1 = 0.75(3)+0.25(6) = 3.75\nposition of Q2 = (8+1)/2 = 4.5\nQ2 = (9+13)/2 = 11\nposition of Q3 = 3(8+1)/4 = 6.75\nQ3 = 0.25(18)+0.75(21) = 20.25\n\nNotice that none of these actually use the value 106.  Let's continue.\n\nSo IQR = Q3-Q1 = 20.25-3.75 = 16.5\n\nQ1-1.5*IQR = 3.75-1.5*16.5 = -21\nQ3+1.5*IQR = 20.25+1.5*16.5 = 45\n\nNo numbers are below -21, but 106 is above 45, so it can be considered an outlier.
893197
Try a histogram and use a range of temperatures.
302694
A statistic used as a measure of the dispersion or variation in a distribution. It is equal to the square root of the arithmetic mean of the squares of the deviations from the arithmetic mean. Simply put, standard deviation measures how spread out the values in a data set are. More precisely, it is a measure of the average difference between the values of the data in the set. If the data points are all similar, then the standard deviation will be low (closer to zero). If the data points are highly variable, then the standard variation is high (further from zero). the greater the standard deviation, the higher the risk. it is always positive, and NEVER negative. the formula you use is....well i can't write it because i don't have the symbols on my keyboard, but i'll try:\n\n\nσ= √(Σ[D-E]^2P)\n\nwhere:\nσ=sigma=standard deviation\nΣ=capital sigma= the sum of\nD=outcome\nE=expected value\n^2= squared\nP= Probability\n√=square root\n\nhere is an example:\n\n(Assumptions)(Outcome probability)(Outcomes)\n(Pessimistic)(0.2)                     (300)\n(Successful)(0.6)                     (600)\n(Optimistic)(0.2)                     (900)\n\n(sorry, the table won't keep the spaces, so i put brackets in)\n\nok, you need the expected value, which is the sum of all of the outcomes, times their probabilities, like so:\nexpected value = 300×0.2+600×0.6+900×0.2 = 600\n\n\nthen standard deviation would be calculated like so:\nstandard deviation = {(300-600)2×0.2+(600-600)2×0.6+(900-600)2×0.2}½ = 190\n\nhope that helped! oh, and if you don't understand, the "to the power of 1/2" part is the same as taking the square root.
898911
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
47438
Take each of your values, and then subtract the mean from them.  E.g., 70 - 56 = 14, 60 - 56 = 4, etc.\n\nNow take each of these answers and square it.  Half of the starting numbers will be negative (eg 30 - 56 = -26) but once you square all of these numbers, they should all be positive.\n\nNow you'll have 7 positive numbers. Add them all together and divide the answer by 7.  Finally, take the square root of that value, and that's the standard deviation.\n\nSo 14, 4, -6, 4, 4, 9, and -26\nbecome 196, 16, 36, 16, 16, 81, and 676\nSummed together gives 1037\ndivided by 7 is ~148.14\nSquare-rooted is ~12.2\n\nThe standard deviation is 12.2.\n\nNOTE: the rt11guru answer that follows is based on a technique used in some larger sample sets where 1 is subtracted from the number of entries, ie the sum is divided by 7-1=6 rather than 7.  That is called the *sample standard deviation*, as shown in the link.
278272
You'll use Z-scores and the Z-table. Z-scores convert deviations to standard deviation units. \n\nIn this case, you're looking for the range of scores that makes up the middle 88% of the distribution - 44% on either side of the mean. \n\nThe z-score is 1.56 (look in the table for the z-score that cuts off 44% of the distribution.) This tells you that 44% of the scores fall between the mean and 1.56 standard deviation units above the mean - and 44% of all scores fall between the mean and 1.56 standard deviation units below the mean. \n\nNow, since each standard deviation unit in this particular distribution is 4ml (as per your question), 1.56 standard deviation units is equal to 1.56 x 4, which is equal to 6.24ml.\n\nBack to the problem - your range is 6.24 ml above and below the mean - which is 250ml - therefore the range of volumes is from 6.24ml less than 250 to 6.24ml more than 250 - or in other words - 243.76ml to 256.24 ml.\n\nHope that helped!