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How do you find the 10th percentile of a set of data?
How would you find the 10th percentile given
4.9 8.8 10 10 10.2 10.9 11 11.1 11.1 11.3 11.4 11.4 11.4 11.6 11.9 12 12 12.3 12.4 12.4 12.5 12.5 12.5 12.6 12.6 12.6 12.6 13 13.1 13.3 13.3 13.4 13.4 13.5 13.6 13.7 13.7 13.9 13.9 13.9 14.2 14.3 14.4 14.5 14.5 15.2 15.3 15.7 15.9 18.6
There are 50 numbers.
You don't have to actually solve it but if you could explain it, it would be very helpful.
Also how would you find the 90th percentile?
2 Answers
- UniversalistLv 78 years agoFavorite Answer
The kth percentile is the value that is greater than k percent of the data values after ranking them. The kth percentile is denoted Pk, and for us in this problem k = 10.
Procedure for finding
1. Rank the data from the lowest to highest (already done)
2. Multiply the sample size by k/100 to find the depth of the kth percentile (10/100 x 50 = 5)
3. If the depth is a whole number, add 0.5. If the depth is not a whole number, round up to the next higher whole number. (5 + 0.5 = 5.5)
4. The kth percentile is the value in the depth position. If the depth ends in 0.5, then you need to average the two values on either side. For example, if the depth is 19.5, then the kth percentile will be the average of the 19th and 20th values. ((10.2 + 10.9) / 2 = 21.1 / 2 = 10.55)
Answer = 10.55
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Another method specifies that the kth percentile is found by first calculating the index i=(k/100)(n+1) [n is the number of data]. Then if i is an integer, the kth percentile is the ith datum from the bottom. If i is not an integer, the kth percentile is the mean of the data in the positions obtained by rounding i up and rounding i down. Using this method we get i=5.1 so the relevant positions are 5 and 6 corresponding to 10.2 and 10.9 data points. Again we get P10 = (10.2 + 10.9) / 2 = 10.55
The first method however is the one I usually see taught in the statistics textbooks.
I hope this response is helpful. ^--^
JJ, would you please vote on a tie-breaker here: http://answers.yahoo.com/question/index;_ylt=AquF4...
Source(s): http://people.richland.edu/james/ictcm/2001/descri... http://www.cs.uni.edu/~campbell/stat/relstat.html