Should I use some sort of Chi Square Distribution or another method... Any assistance is greatly appreciated.

I am asking for assistance or guidance to point me in the right direction for this problem I am trying to find a solution.

I need to come up with a percentage to show a Severity of Defects for monthly software releases at my company. Each month our customers inform us of bugs/issues with the software and we create a "Report Card" to rank these defects.

There are 4 severity levels and each one has a different weight
SEV 1 has a weight of 1.00
SEV 2 has a weight of 0.75
SEV 3 has a weight of 0.50
SEV 4 has a weight of 0.25

My first attempt at it was a simple weighted average in excel. I used a dot product calculating the # of each SEV by its weight. The issue with this method is that it accounts for the outliers and this could skew the results if there is a high # of a particular SEV.

I think I need to steer towards some sort of Chi-Square Distribution. I also think I need some bounding built in to take care of cases with more defects than opportunities for defects. In other words, if you are counting only number of tickets entered vs. number of changes made and 10 customers enter a ticket for the same issue, does that count as 10 separate defects?

Any help would be greatly appreciated.


Well-Known Member
Hi Matoosky,

There are no outliers since outlier is an extreme value, like a person height of 2.5 meters. in your case there are only 4 can't get an outlier of SEV -8 ...

I'm not sure if the weighted approach is the correct one. I don't know your exact definitions for the severities but if for example, SEV1 is "system down and nobody can use the software" and SEV4 is a minor issue (like an internal report with no margin) definitely you prefer to have four SEV4 over one SEV1. You can give an exponential weight like (1,10,100,1000) but I would present it separately or 3 groups (SEV1, SEV2, SEV3+SEV4). again depending on your definitions of the severities.

Usually, you have the options to group tickets under one "parent ticket" if it is exactly the same issue, so it will be counted as one incident.