How do I Find the first and third quartiles, Q1 and Q3, of the following set of numbers.?
493335
First you have to put the numbers in order:\n\n\n3, 4, 8, 8, 8, 10, 12, 13, 14, 16\n\nThere are 10 numbers. To find the median, you want to go to the middle. The middle is between 8 and 10, so you average those to find a median of 9.\n\nThere are two quartiles (like quarters) made up of the elements less than 9 and two quartiles made up of the elements more than 9.\n\nNow you need to find the median of each half. For the first five numbers, the median, middle number, is 8. For the second five numbers, the median is 13. \n\nSo the Q1 goes from 3 to 8, Q2 goes from 8 to 9, Q3 goes from 9 to 13, and Q4 goes from 13 to 16. These are all based on how the data is distributed (spread out) around the median in the middle.\n\nHope that helps.
1120383
Fine.\n\nYou need the find the average and SD of the box.\n\nAverage = (0+0+1)/3 = 1/3\nSD = (1-0)sqrt{(1/3)(2/3)} = 0.471\n(using the shortcut SD = (big-small)*sqrt(frac of big * frac of small))\n\nThe expected number of 1's is the number of draws times average of the box, so 600(1/3) = 200.\n\nHowever, we would expect to be off of the expected by a standard error or so.  The standard error of the number of 1's is \nsqrt(number of draws)*sd of the box, so sqrt(600)*0.471 = 11.55.
547907
1) 4 3/4\n\n2) 1356.66\n\n3) 760
954354
Your boxplot is based on the '5 number summary', that is... \n1. The minimum value\n2. The 1st quartile (25th percentile)\n3. The median (called the 2nd quartile)\n4. The 3rd quartile (75th percentile) and..\n5. Tthe maximum.\n\nThe 1st, 2nd and 3d quartiles make up the box, and the lines extend from the top and bottom ends of the box to the maximum and minimum.
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
419172
(0, 1)\n(1, 1.1)\n(2, 1.21)\n(3, 1.331)\n(4, 1.4641)
776941
-25 < 5(w + 3)\n-25< 5w +15\n-40< 5w\n-8<w\n\n-9q + 18 -3 > 6q\n15 > 15q\n1> q\n\n\n19 < 7 + 1/3k + 2 -2/3k\nmultiply by 3 to get rid of denominators\n57 < 21 +k +6 -2k\n30< -1k\ndivide by -1\n-30<k
769753
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
839453
1) 21.69\n2) 14.10\n3) 7.59
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
238182
p1    p2\n--- = ---\nq1    q2\n\n 5     2\n--- = ---\n 2    q2\n\nCross multiply,\n\n4/5
913451
A box and whisker plot!? No problem! \n\nThe Box is from Quartile 1(25th percentile) to Quartile 3(75th percentile),.\n\nThe Whiskers are Q1 minus 1.5 Interquartile ranges and Q3 plus 1.5 Inter Quartile ranges.\n\nIQR (Interquartile ranges) is defined as Q3-Q1\n\nWell say that Q1 is 12 and Q3 is 25 then the IQR is 13, and 1.5IQR is 19.5\n\nTherefore, Anything below -7.5 (12-19.5) and above 44.5 would be considered an outlier. \n\nAs far as drawing the damn thing, just draw a box with the outside edges being Q1 adn Q3, and then draw the whiskers 1.5IQR long.\n\nAdditional info:\n\nThe box and whisker is only 1 criterion for outliers, + or - 3 standard deviations from the mean is another common way, which can get other results.\n\nHope that 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
793308
1. sqrt 91a\n2. asqrt75\n3. sqrt15 - 3\n4.-4sqrt2-19
872177
Coefficients are:  1  1  3
202512
There is no reason for the quartile to correspond to a data point. It is simply a cutoff line to illustrate. See http://en.wikipedia.org/wiki/Quartile . 7.5 is correct.
776835
First put them in order:\n157, 225, 248, 248, 271, 310, 337, 388, 451\n\nThe first number is now 157; the last 451\n\nThe median is the middle number: 271\n\nThe lower quartile is the median of the numbers below the data median; that's 1/2 way from 225 to 248 or 236.5\n\nSimilarly, the upper quartile is 1/2 way between 337 and 388 or 362.5\n\nSo your 5 number summary is 157, 236.5, 271, 362.5, 451\n\nHope that helps