Social Research Methods: Qualitative and Quantitative Approaches

FIGURE 4 The Standard Deviation

STEPS IN COMPUTING THE STANDARD DEVIATION

1. Compute the mean.


2. Subtract the mean from each score.


3. Square the resulting difference for each score.


4. Total up the squared differences to get the sum of squares.


5. Divide the sum of squares by the number of cases to get the variance.


6. Take the square root of the variance, which is the standard deviation.

EXAMPLE OF COMPUTING THE STANDARD DEVIATION

[8 respondents, variable = years of schooling]

Score Score – Mean Squared (Score – Mean)

15 15 – 12.5 = 2.5 6.25
12 12 – 12.5 =0.5 .25
12 12 – 12.5 =0.5 .25
10 10 – 12.5 =2.5 6.25
16 16 – 12.5 = 3.5 12.25
18 18 – 12.5 = 5.5 30.25
8 8 – 12.5 = 4.5 20.25
9 9 – 12.5 = –3.5 12.25
Mean 15 + 12 + 12 + 10 + 16 + 18 + 8 + 9 =100, 100/8 =12.5
Sum of squares 6.25 + .25 + .25 + 6.25 + 12.25 + 30.25 + 20.25 + 12.25 = 88
Variance =Sum of squares/Number of cases =88/8 = 11
Standard deviation Square root of variance = 11 =3.317 years.
Here is the standard deviation in the form of a formula with symbols.

Symbols:
X=SCORE of case Σ=Sigma (Greek letter) for sum, add together
X ̄ ̄=MEAN N=Number of cases
Formula:a
Standard deviation 



S(X –X) ̄ ̄^2

N–1

aThere is a slight difference in the formula depending on whether one is using data for the
population or a sample to estimate the population parameter.

ANALYSIS OF QUANTITATIVE DATA

a standard deviation of .50, whereas the mean

grade-point average at Queens College is 3.24 with

a standard deviation of .40. The employer suspects

that grades at Queens College are inflated. Suzette

from Kings College has a grade-point average of

3.62; Jorge from Queens College has a grade-point

average of 3.64. Both students took the same

courses. The employer wants to adjust the grades

for the grading practices of the two colleges (i.e.,

create standardized scores). She calculates z-scores

by subtracting each student’s score from the mean

and then divides by the standard deviation. For

example, Suzette’s z-score is 3.62  2.62 1.00/.50

2, whereas Jorge’s z-score is 3.64  3.24. 

.40/.40 1. Thus, the employer learns that Suzette

is two standard deviations above the mean in her