Quizzes & Puzzles4 mins ago
Standard Deviation
8 Answers
What is Standard Deviation and how is it calculated?
Answers
Best Answer
No best answer has yet been selected by jellytot01. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.Standard Deviation, as relating to probability and statistics, refers to the distribution of the data. A theorem states that the data will, invariably, be close to the mean of all the available data. It can be determined by applying a formula that is the square root of the variance. Understand, this is a simplified description... if you'd like to see the mathmatical formula for determing it's value, check here:
(Posted in 2 parts):
Clanad, as always, has provided an excellent answer (and, in so doing,disproves the widely held belief that there is no intelligent life on the other side of the Atlantic! LOL).
Please allow me to try to fill in a few details:
Let's assume that 6 people take a test. Their marks are 8, 8, 9, 10, 12 and 13. We can work out the mean (average) very easily by adding all their marks together (=60) and dividing by the number of people. So the mean = (60 divided by 6) = 10.
Now, if someone asks me how the group did, I can say "Not bad, their mean score was 10". Now, while this conveys some information to the person asking the question it doesn't tell him anything about how the marks were spread out. With the marks I've given, everyone is reasonably close to the 'average' - but look at this set of marks:
0, 2, 7, 14, 17, 20.
If you add them up, you still get 60, so the mean is still 10 but the 'spread' is much greater. Standard Deviation is a way of measuring this spread. It looks at the 'deviation' of each mark from the mean (i.e. how far away from the mean it is) and then 'averages' these deviations but, to give greater weight to the figures which are well away from the mean, the figures are squared before 'averaging' and then 'unsquared' at the end of the calculation.
Clanad, as always, has provided an excellent answer (and, in so doing,disproves the widely held belief that there is no intelligent life on the other side of the Atlantic! LOL).
Please allow me to try to fill in a few details:
Let's assume that 6 people take a test. Their marks are 8, 8, 9, 10, 12 and 13. We can work out the mean (average) very easily by adding all their marks together (=60) and dividing by the number of people. So the mean = (60 divided by 6) = 10.
Now, if someone asks me how the group did, I can say "Not bad, their mean score was 10". Now, while this conveys some information to the person asking the question it doesn't tell him anything about how the marks were spread out. With the marks I've given, everyone is reasonably close to the 'average' - but look at this set of marks:
0, 2, 7, 14, 17, 20.
If you add them up, you still get 60, so the mean is still 10 but the 'spread' is much greater. Standard Deviation is a way of measuring this spread. It looks at the 'deviation' of each mark from the mean (i.e. how far away from the mean it is) and then 'averages' these deviations but, to give greater weight to the figures which are well away from the mean, the figures are squared before 'averaging' and then 'unsquared' at the end of the calculation.
(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).
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).