Answer: EUnless we don't know the values or the mean of the two sets, we cannot find out the Standard Deviation.
(1) The standard deviation of the 10 scores for the free routine was also 0.7 points.
We don't know the score distribution or the mean. Thus we cannot really find out the SD of the combined set.
For example, {4,4,4} has SD = 0. {5,5,5} has SD = 0, Combined set C1 is {4,4,4,5,5,5}
and, {4,4,4} has SD = 0, {6,6,6} has SD = 0, Combined set C2 is {4,4,4,6,6,6}.
From the above example we can clearly see that the SD of both combined sets are different. (mean of C1 is 4.5, mean of C2 is 5, from a quick glance we can say that SD of C1 is 0.5 and that of C2 is 1)
(2) The range of scores for each of the compulsory routine and free routine was 3 points.
Lets take small sets of 3 variables each. S1 = {3,4,6} and S2 = {5,6,8} Combined C1 = {3,4,5,6,6,8}
And S11 = {3,4,6} and S12 = {7,8,10} Combined C2 = {3,4,6,7,8,10}.
We see SD of C1 is not equal to SD of C2, C2 must be greater than C1. We don't know the values but it can be deduced from the way C1 and C2 are distributed.
(1) and (2) combined.
SD of S1 = 0.7, SD of S2 = 0.7 and range of S1 and S2 is 3 points.
Let's take smaller sets of 3 variables. S1 = {2,2,5} S2 = {3,3,6} Mean of S1 is 3 and S2 is 4. Thus we can see SD of S1 = S2.
C1 = {2,2,3,3,5,6} which has some SD, let's say SD1.
Now we take another set S3 = {2,2,5} S4 = {7,7,10}. SD of S3 = S4
C2 = {2,2,5,7,7,10}
We see from above that S1 and S2 satisfy both conditions and give C1 as the combined set. And S3 and S4 also satisfy both conditions and give C2 as the combined set. We can clearly see SD of C2 > SD of C1.
Thus, unless we don't know the actual values or mean, we cannot find out the SD.