# Mean vs Standard Deviation Exercise

#### StatisticsCenas

##### New Member
Hey everyone.

I would like to know your opinion on the following exercise:

We have 2 coders teams (A and B).
We want to choose the team who writes less errors. (this is the only criterion)

Here are the information from both teams:

• Mean A: 2.1
• Standard Deviation A: 1.14
• Mean B: 1.84
• Standard Deviation B: 1.58

Apparently the best option is the team A. Do you agree?

Thanks.

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#### rogojel

##### TS Contributor
Well, if this were a real life problem, I would say that there was a measurement issue at least with team B. Their distribution is almost V shaped and there is no mechanism, I believe, that would make a team either to program with no errors or with many (4) errors. Team A has a much more plausible distribution - but based on this data I would not know which team to pick because I could not trust the data.

#### Dragan

##### Super Moderator
Well, in my opinion, I would select Team A because the statistics associated with Team A yield a much smaller Coefficient of Variation (CV) i.e. Team A has a CV=54.29% and Team B has a CV of 85.87%. Specifically, what this implies is that Team A provides more efficient CV because it's CV has more precision because it has much less variance around the estimate of the mean.

#### rogojel

##### TS Contributor
Well, in my opinion, I would select Team A because the statistics associated with Team A yield a much smaller Coefficient of Variation (CV) i.e. Team A has a CV=54.29% and Team B has a CV of 85.87%. Specifically, what this implies is that Team A provides more efficient CV because it's CV has more precision because it has much less variance around the estimate of the mean.
So, you would ignore the distribution shape and decide on summary measures only?

#### hlsmith

##### Omega Contributor
Could you treat these data as Poisson and make up a range of samples sizes.

Next calculate the respective 95% CI for those values and plot them.

Wait, now I want to do this or try to simulate them.