(2nd part):
So, let's look again at the first set of marks, these were:
8, 8, 9, 10, 12, 13
Let's now see how far away each of these marks is from the mean (10):
-2, -2, -1, 0, +2, +3
So that we give more weight to the figures furthest from the mean, we square everything in the last list. This gives:
4, 4, 1, 0, 4, 9.
Now we find the average (mean) of these figures by adding them up and dividing by 6:
22 divided by 6 = 3.66666.......
The last job is to 'unsquare' the information by finding the square root of 3.66666..... This is 1.91 (to 3 significant figures).
So the first set of data has a mean of 10 with a standard deviation of 1.91
Now let's return to the second set of marks:
0, 2, 7, 14, 17, 20
If we list how far away each number is from the mean (10), we get:
-10, -8, -3, +4, +7, +10
Now, so that we give greater weight to the marks which are furthest away from the mean, we square all the numbers in the last list:
100, 64, 9, 16, 49, 100.
The next task is to find the average (mean) of the last list of figures by adding them together and dividing by 6. So 338 divided by 6 = 56.33333....
The last stage is to 'unsquare' by taking the square root of 56.33333.... This is 7.51 (to 3 significant figures)
So, the second set of data still has a mean of 10 but has a much greater standard deviation of 7.51.
Hoping this helps,
Chris (a former maths teacher).