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I have approximately 50,000 product records with descriptions and costs
however in each group of similar products there are some that have outlier costs assigned. for instance one group of like items may have costs ranging from $2-$3 with one or two of the items assigned a cost of $150. Is there a way of looking for these kind of outliers? Thanks in advance. -- Jeff C Live Well .. Be Happy In All You Do |
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1. For a common group, calculate both the mean and standard deviation.
2. Then, for each member of group, calculate the absolute value of its deviation from the mean. 3. If the item deviation calculated in step 2 exceeds three times the standard deviation calculated in step 1, the item may be considered an outlier. -- Gary''s Student - gsnu200809 "Jeff C" wrote: I have approximately 50,000 product records with descriptions and costs however in each group of similar products there are some that have outlier costs assigned. for instance one group of like items may have costs ranging from $2-$3 with one or two of the items assigned a cost of $150. Is there a way of looking for these kind of outliers? Thanks in advance. -- Jeff C Live Well .. Be Happy In All You Do |
#3
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![]() -- Jeff C Live Well .. Be Happy In All You Do "Gary''s Student" wrote: 1. For a common group, calculate both the mean and standard deviation. 2. Then, for each member of group, calculate the absolute value of its deviation from the mean. 3. If the item deviation calculated in step 2 exceeds three times the standard deviation calculated in step 1, the item may be considered an outlier. -- Gary''s Student - gsnu200809 Great! Thanks - is there a way to apply these calculations over the entire record set at once? "Jeff C" wrote: I have approximately 50,000 product records with descriptions and costs however in each group of similar products there are some that have outlier costs assigned. for instance one group of like items may have costs ranging from $2-$3 with one or two of the items assigned a cost of $150. Is there a way of looking for these kind of outliers? Thanks in advance. -- Jeff C Live Well .. Be Happy In All You Do |
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