1. Standard Deviation A measure of variability
Commonly used to calculate other statistical measures
Shows the distance (difference) between the mean and the steepest part of a normal curve
Indicates dispersion, or spread, of scores in a distribution

2. Standard Deviation
The smaller the standard deviation, the more tightly clustered the scores
The larger the standard deviation, the more spread out the scores

3. Standard Deviation
On an average day in Los Angeles, the number of sudden deaths from cardiac causes is about 4.6. On the day of the Northridge earthquake, there were 25 such deaths, 24 of which occurred after 4:31 a.m., when the earthquake struck.

4. Standard Deviation Make your prediction:
Will the average male or the average female who died from sudden cardiac causes on the day of the earthquake be older?
A. the average female was older
B. the average male was older

5. Predict whether the ages of the males or the ages of the females who died on the day of the earthquake were more variable.
A. ages of males were more variable
B. ages of females were more variable

6. The Data Ages of females who died: 66 92 75 84 83 80
Ages of males who died:

7. Calculate the Mean 66 92
/6 = 80 84 83 80
480
1111
1111/18 = 61.7

8. Subtract the mean from each score
X - X = s
66 – 80 = -14
92 – 80 = 12
75 – 80 = -5
84 – 80 = 4
83 – 80 = 3
80 – 80 = 0
s s2
-142 = 196
122 = 144
- 52 = 25
42 = 16
32 = 9
02 = 0
399 (sum of squares)

9. Find the variance; find the square root of the variance
There it is! The standard deviation for the ages of the females who died that day is 8.93.
Now you try the same formula for the ages of the males.
s2/N (N-1)
399/5 = 79.8
sd = s2
79.8 = 8.93