Summaries of ten cartons
For the ten milk-carton volumes in litres, 1.035, 1.019, 1.068, 1.084, 1.077, 0.998, 1.041, 1.037, 1.051 and 1.102, compute the sample mean, the median, the quartiles, the bias-corrected sample standard deviation and the standard error of the mean.
The sum of the ten volumes is 10.512, so \overline x=1.0512. Sorted, the values are
0.998,\ 1.019,\ 1.035,\ 1.037,\ 1.041,\ 1.051,\ 1.068,\ 1.077,\ 1.084,\ 1.102.
With n=10 even, the median is the average of the 5th and 6th values, (1.041+1.051)/2=1.046. The lower quartile is the median of the five values below the median, 1.035, and the upper quartile is the median of the five above it, 1.077.
The squared deviations from 1.0512 sum to 0.009039, so
s^2=\frac{0.009039}{9}=0.001004,\qquad s=0.0317.
The standard error of the mean is s/\sqrt{10}=0.0100. The standard deviation describes the spread of single cartons; the standard error describes the uncertainty in \overline x as an estimate of \mu.