Formula for Standard Deviation
We will learn how to calculate the standard deviation by using the two previous examples:
Sample A: 9, 8, 7, 6, 5, 4, 3, 2, 1
Sample B: 9, 5, 5, 5, 5, 5, 5, 5, 1
There is no point in compiling a frequency table for Sample A since each value appears exactly once.
We will therefore calculate the standard deviation from the sample given in the list.
First, we need to calculate the average: (1+2+3+4+5+6+7+8+9) / (9) = 5
Now we subtract the average from each of the values, and then calculate the square of the result:
The Value 
The Average 
The Difference Between the Value and the Average 
The Square of the Difference: 
(1) 
(2) 
(3) = (1) – (2) 
(4) = (3)2 
1 
5 
4 
16 
2 
5 
3 
9 
3 
5 
2 
4 
4 
5 
1 
1 
5 
5 
0 
0 
6 
5 
1 
1 
7 
5 
2 
4 
8 
5 
3 
9 
9 
5 
4 
16 
We can now calculate the average of the numbers in column 4 (the average of the square of the differences): (16+9+4+1+0+1+4+9+16) / (9) = 6.67
The standard deviation represents the square root of the number that we received.
In our case:
σ = The square root of 6.67 = 2.582
For Sample B, we will first compile a frequency table in order to calculate the average:
The Value 
The Frequency 
The Relative Frequency 
The Contribution to the Average 
1 
1 
11.11% 
0.1111 
5 
7 
77.78% 
3.8889 
9 
1 
11.11% 
1 
9 
100% 
5 
The average is 5.
We will now compile another frequency table to calculate the standard deviation:
The Value 
The Frequency 
The Relative Frequency 
The Average 
The Difference Between the Value and the Average 
The Square of the Difference 
The Contribution to the Dispersal 
(1) 
(2) 
(3) 
(4) 
(5) = (1) – (4) 
(6) = (5)2 
(7) = (3) x (6) 
1 
1 
11.11% 
5 
1 – 5 = 4 
(4)2 = 16 
11.11% x 16 = 1.78 
5 
7 
77.78% 
5 
5 5 = 0 
02 = 0 
77.78% x 0 = 0 
9 
1 
11.11% 
5 
9 – 5 = 4 
42 = 16 
11.11% x 16 = 1.78 
9 
100% 
3.56 
The standard deviation: = The square root of 3.56 = 1.886.
As we expected, this is below the standard deviation of Sample A. We have seen that this calculation has also verified our intuition that Sample A is more widely dispersed than Sample B.
The Significance of the Standard Deviation as a Measure of Dispersal in the Decisionmaking Process
StDev is the shorthand for Standard Deviation
Example:
In country A, a competition is held every year to choose the champion halfcourt basketball shooter. Every basketball team sends one representative to the competition.
Each contestant in the competition shoots 10 times from midcourt.
The person who will makes the most baskets receives a prize of $1 million, and his coach gets the same amount.
Spurs basketball team, the coach chooses his representative from among four outstanding players by using the following method: He has each player take five rounds of 10 halfcourt shots each. He chooses the player with the highest average in the five rounds.
The following table shows the results of the rounds of shots by the players:
Player 1  Player 2  Player 3  Player 4  
Results of round 1  1 basket  3 baskets  6 baskets  8 baskets 
Results of round 2  0 baskets  2 baskets  4 baskets  2 baskets 
Results of round 3  2 baskets  4 baskets  6 baskets  8 baskets 
Results of round 4  0 baskets  0 baskets  4 baskets  1 basket 
Results of round 5  2 baskets  6 baskets  5 baskets  6 baskets 
Total baskets  5 baskets  15 baskets  25 baskets  25 baskets 
Average per round  1 basket  3 baskets  5 baskets  5 baskets 
The coach chooses the player with the highest average of making baskets, but there are two such players: Player 3 and player 4.
The coach of the Spurs must choose between player 3 and player 4, and asks you to help him make a choice.
He also tells you two important facts:

The winner of last year’s competition made an average of 4 baskets per round.

The coach will win $1 million if his representative wins.
Even without calculating the StDev of the shots of the two players, it is easy to see that the dispersal of player 3 is smaller than the dispersal of player 4. The significance of the differences in the measures of dispersion in statistics is that the stability of player 3 is greater, and he will therefore put the coach’s prize in less jeopardy. Had player no. 3 demonstrated his current skill in last year’s competition; he had a good chance of winning, or at least sharing first place, since in no round did he score less than four baskets.
If, however, it is known that a player who always scores on 7 out of 10 shots will participate in the competition, then we will prefer to send player 4 to the competition.
If he has a hot hand on the day of the competition, then he will win; if he does not, he will lose. Player 3, however, in spite of his stability will never make more than six shots in any given round, and so he is therefore sure to lose.