How do you find outliers? i have all my quartiles calculated.?
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.
775593
Load the Analysis Toolpak add-in and then use this formula, it's shorter:\n\n=A1/GCD(A1, B1)&":"&B1/GCD(A1, B1)
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...
1136885
so your scores will look like\n11 12 13 13 15 15 15 15 16 16 17 17 17 18 18 18 19 19 19 19 20 20 20 20 20 20 21 21 21 22 22 22 22 23 23 23 23 24 25 25\n\nthere are 40 observations, so Q1 is the (41/4=10.25)th observation, which is somewhere between 16 and 17\nusing linear interpolation, Q1 is 16.25
889683
There is, it is called Standarad Deviation, if you have a decent Scientific Calculator it shouyld show how to use it.
824818
OK, here goes:\n\nThe data could be\n7 9 10 12 12 13 14\n\nThe median (center number) is 12\nThe mode (most repeated number) is 12\nThe mean (average) is 11.\n\nThe point is that you need to look at all three measures to get a good idea of the data.\n\nHere's another set:\n1 3 10 12 12 19 20\nThis has the same measurements.\n\nIt is only when we look at things like the variance and std deviation that we can really see the difference between those two data sets.\n\nGood luck!
479611
just use SPSS 13.0  ;)
72964
thats easy, just chaeck your points and click on the link. thaen look at the second chart and look at the 1st and 4th colums
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.
267652
use ur score x 0.25 and the other score x 0.75\n\nthen add them together \n\nthat would be ur new score!!
142207
An outlier is a value that is more than 1.5 times the interquartile range (Q3 - Q1) away from either Q1below it) or Q3 (above it).
932261
hey there is a good website to check your work \nits www.webmath.com i use it all the time
333334
data xi = 4, 13, 5, 6, 9\nnumber of data: N = 5\nmeans:\nxm = sum(xi)/N\nxm = (4 + 13 + 5 + 6 + 9)/5\nxm = 7.4\n\nstandard deviation:\nsd = sqrt(sum(xm - xi)²)/N\nsd = sqrt{(4 - 7.4)² + (13 - 7.4)² + (5 - 7.4)² + (6 - 7.4)² + (9 - 7.4)²}/5\nsd = sqrt{11.56 + 31.36 + 5.76 + 1.96 + 2.56}/5\nsd = sqrt{53.2}/5\nsd = 1.46
470556
generally it depends on the data, but to correct a previous answer, median is NOT the average, the MEAN is the average. THe average includes all, if any, outliers (data that is noticeably different from the curve. If your data is a large sample, and a normal curve (reasonably bell shaped) the mean may better represent the data. If the data is skewed, the Median may be a better choice. But the best way to summarize the data is the "5 number summary" which includes the low number, high number, median (middle number) and the Q1 (25th percentile) and the Q3 (75th percentile). There are specific methods to calculate the Q1 and Q3
979515
the formula for sd.dev of binomial distribution is "\nsqrt(n*p*q).\ncan't understand the table so use the fomula urself.hope this helps
780864
arrange the numbers in ascending sequence then the first 25% make the first quartile and so on\n\nso...\n\n1,1,1,2,2, 3,4,5,9,9, 10,12,12,15,20 21,21,21,23,28\n\nthe first quartile is 1-2,\nsecond 3-9\nthird10-20\nfourth 21-28\n\nIf you want to find the 95th percentile it's 95% of the way up.\nIn the sequence above the 23rd value (21) is 92% and the 24th value (23) is 96%.  So the 95% percentile lies 75% of the way between these values.  (23-21) * 0.75 = 1.5 so the 95th percentile is 22.5\n\nThe median is the 50th percentile (9.5)\n\nIf the median > mean then the distribution is positively skewed\nIf the median < mean then the distribution is negatively skewed\nIf the median = mean then the distribution is normally distributed
979610
Well, you've got the averages. You can now find the total points my multiplying the mean with 10. You can also determine the mean deviation, standard deviation, variance, co-variance etc of the data of each group and determine the relative levels of consistancy of each group and compare the spread of the data.\nFor further stuff...look it up in wikipedia.