Actually your data seems to be fairly symmetrically distributed so you could assume that the mean is quite close to the median.
regards
Assuming the data is normal, how do I find the mean and std deviation? I have the following:
10th percentile - 75,174
25th percentile - 81,627
50th percentile - 89,621
75th percentile - 98,576
90th percentile - 108,463
I can't assume the mean is the 50th percentile right?
I know this should be easy....
Actually your data seems to be fairly symmetrically distributed so you could assume that the mean is quite close to the median.
regards
Yes, you would use the sample median as the measure of central tendency. In terms of a measure of dispersion, what works best is the inter-decile range i.e., the range between the 10-th percentile and 90-th percentile. Other measures for shape (tantamount to skew and kurtosis) would be the "left-right tail-weight ratio" [ (median - 10-th percentile) / (90-th percentile - median) ] and the "tail-weight factor" [ (75-th percentile - 25-th percentile) / (inter-decile range) ].
See Karian and Dudewicz (Chapter 5, pp. 172-174) for further details in the "Handbook of Fitting Statistical Distributions with R."
Thank you for the reply. So here's my next question: If I do that and calculate the SD using a Z-score and the mean, why am I getting a different SD? I would assume it should be constant... The end goal is to take two other values and determine their Z scores so I can determine their percentile. Thanks!
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