Example:

Calculate the variance and mean variance of the following set of hourly tank levels. Assume the tank is a 100 gal. tank.  Based on the mean and the mean variance, would you expect the tank to be able to accept a 40% (40 gal.) increase in level at any time?

1:00 - 40%       6:00 - 38%       11:00- 34%

2:00 - 38%       7:00 - 34%       12:00- 30%

3:00 - 28%       8:00 - 28%       1:00 - 40%

4:00 - 28%       9:00 - 40%       2:00 - 36%

5:00 - 40%       10:00- 38%     

Solution:

The mean is

[40(4) +38(3) +36+34(2) +30+28(3)]/14= 492/14 = 35.1

The mean variance is:

1/14 (¦40-35.1¦ + ¦38 - 35.1¦ + ¦28 - 35.1¦ + ...¦36 - 35.1¦) =

1/14 (57.8) = 4.12

From the tank mean of 35.1%, it can be seen that a 40% increase in level will statistically fit into the tank; 35.1 + 40 <100%.  But, the mean doesn't tell us if the level varies significantly over time.  Knowing the mean variance is 4.12% gives the additional information.  Understanding the mean variance also allows us to infer in which the level at any given time (most likely) will not be greater than 35.1 + 4.12 = 39.1%; and 39.1 + 40 is still less than 100%.  Thus, it is a good supposition that, in the near future, a 40% level increase will be accepted through the tank without any spillage.