I would be very careful with what conclusions you try to draw from your data, considering you only have 5 pieces of data. Another thing to note is that you're acting under the assumption of normality for this and, having had some experience with measurement system analysis, I can tell you that I've run into quite a few interesting distributions which even after 100+ readings failed a normality test (both using KS and AD tests).
I went ahead and did a confidence interval for your data, as pointless as it is, to double check JGK and I found the same thing he did.
t-value for 95% confidence using a conservative (n-1) degrees of freedom is 2.7765 rounded to 2.78
where avg. is the average value you found and std. dev. is the standard deviation of the sample (using bessel's formula due to small sample size)
avg +- std.dev*t-value/sqrt(n)
average = 25/5 = 5
standard deviation = sqrt((5.2-5)^2 + (7.6-5)^2 + (3.1-5)^2 + (2.7-5)^2 + (6.4-5)^2)/sqrt(n-1) = sqrt(.04 + 6.76 + 3.61 + 5.29 + 1.96)/sqrt(n) = sqrt(17.66)/sqrt(4) = 2.10
sqrt(5) = 2.24
{5-2.61, 5+2.61} = {2.39, 7.61}
To JGK --
I haven't seen a Dixon Q test before so I looked it up. It seems to be a very weak tool in determining outliers, and here's why I think so:
Say you have 10 data points
1, 2, 94, 95, 96, 97, 98, 99, 100, 101
Using this test, you would not tag 1 or 2 as an outlier, as the magnitude of the distance between the two of them is 1 and the range is 100. Granted not all situations will be like this, but I feel it's worth mentioning nonetheless.
As far as your interpretation on the confidence interval for the mean, I believe you might be mistaken as to what it actually means. You wrote that 95% of the data should fall within the range, and this is not quite correct. It means that we are 95% confident that the true mean value of whatever it is this guy is measuring lies between 2.39 and 7.61. If you wanted to determine the range that 95% of the data would fall into, it would merely be avg-+t*statistic*st.dev. The values you get for that are something like {-.84, 10.84}.