To understand the reasoning more clearly, let's replace the given values with easier values:

The Department of Environmental Protection measured the volume of water in 4 similarly sized reservoirs in State X and found that the standard deviation of their volumes at the start of the year was x cubic gallons, where x>0. Was the standard deviation of those 4 volume measurements lower at the end of the year?

(1) At the end of the year, the average volume of the water in the 4 reservoirs had decreased by 20%.

(2) The percent decrease in the volume of the water in each reservoir during the year was the same.

SD reflects how far the data points DEVIATE from the mean.
A LOW SD implies that the data points are NOT VERY SPREAD OUT.
A HIGH SD implies that the data points are VERY SPREAD OUT.

Statement 1: At the end of the year, the average volume of the water in the 4 reservoirs had decreased by 20%.
Let the sum of the 4 volumes at the start of the year = 100 cubic gallons.
Then:
Average at the start of the year = 100/4 = 25.
Average at the end of the year = 25 - (0.2)25 = 20.
Sum at the end of the year = 4*20 = 80.

Case 1: Volumes at the start of the year = {1, 1, 49, 49}, volumes at the end of the year = {20, 20, 20, 20}
Here, the volumes at the end of the year are LESS spread out than are the volumes at the beginning of the year, so the SD DECREASES.

Case 2: Volumes at the start of the year = [24, 24, 26, 26}, volumes at the end of the year = {0, 0, 0, 80}
Here, the volumes at the end of the year are MORE spread out than are the volumes at the beginning of the year, so the SD INCREASES.
INSUFFICIENT.

Statement 2: The percent decrease in the volume of the water in each reservoir during the year was the same.
Case 3: Volumes at the start of the year = {10, 20, 30, 40}, percent decrease for each volume = 50%.
After each volume decreases by 50%, the resulting volumes at the end of the year = {5, 10, 15, 20}.
Since the volumes at the end of the year are LESS spread out than are the volumes at the beginning of the year, the SD DECREASES.

Case 4: Volumes at the start of the year = {100, 100, 200, 200}, percent decrease for each volume = 20%.
After each volume decreases by 20%, the resulting volumes at the end of the year = {80, 80, 160, 160}.
Since the volumes at the end of the year are LESS spread out than are the volumes at the beginning of the year, the SD DECREASES.

Cases 3 and 4 illustrate that, if each data point decreases by the same percentage, then the resulting data points will be LESS SPREAD OUT that the original data points, with the result that the SD DECREASES.
SUFFICIENT.

The correct answer is B.

No work is needed to evaluate statement 2 if we know the following:
If each data point decreases by x%, then the SD decreases by x%.
If each data point increases by x%, then the SD increases by x%.
Since statement 2 indicates that each volume decreases by the same percentage, the SD must also decrease by that percentage.

Statement 1:

First statement is not sufficient because the increase of 20% amount of water in each 10 reservoir can cause the 20% increase in the average. therefore keeping the standard deviation constant

However if the average has increased by increase in just one reservoir then standard deviation can increase as well

Also if the reservoirs with lower amount of water are filled in order to increase the average by 20% then standard deviation will decrease

Therefore no conclusion can be derived from statement 1

Statement 2:

If the water in each reservoir is decreased by a constant percentage will also decrease the deviation among the quantities of water in reservoirs thereby decreasing the standard deviation therefore SUFFICIENT


Right Answer Choice: B