An experimental brand of outdoor paint was tested on 6 cans to determine how long it lasts before fading, measured in months. The standard deviation was calculated by first finding the mean of all values (35 months). Then the difference between each value and the mean was determined and squared. These squared differences were summed and divided by the number of values minus one, resulting in a standard deviation of 1.17 months.

1. 3.1 - 1
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20

2. 3.1 - 2
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 


3. 3.1 - 3
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Calculate the mean

4. 3.1 - 4
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Calculate the mean

5. 3.1 - 5
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Create a column of each
value minus the mean
µ = 35

6. 3.1 - 6
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Create a column of each
value minus the mean
µ = 35
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20

7. 3.1 - 7
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Create a column of each
value minus the mean
µ = 35
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35

8. 3.1 - 8
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Create a column of each
value minus the mean
µ = 35
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
10-3535
35
35
35
35
35

9. 3.1 - 9
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Create a column of each
value minus the mean
µ = 35
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25

10. 3.1 - 10
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Create a column of each
value minus the mean
µ = 35
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25

11. 3.1 - 11
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Create a column of each
value minus the mean
µ = 35
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15

12. 3.1 - 12
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Create a column of each
value minus the mean
µ = 35
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5

13. 3.1 - 13
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Create a column of each
value minus the mean
µ = 35
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5

14. 3.1 - 14
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Create a column of each
value minus the mean
µ = 35
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15

15. 3.1 - 15
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Square each deviation
from the mean
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15

16. 3.1 - 16
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Square each deviation
from the mean
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15
625

17. 3.1 - 17
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Square each deviation
from the mean
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15
625
625

18. 3.1 - 18
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Square each deviation
from the mean
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15
625
625
225

19. 3.1 - 19
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Square each deviation
from the mean
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15
625
625
225
25

20. 3.1 - 20
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Square each deviation
from the mean
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15
625
625
225
25
25

21. 3.1 - 21
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Square each deviation
from the mean
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15
625
625
225
25
25
225

22. 3.1 - 22
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Sum the squared
deviations from the
mean
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15
625
625
225
25
25
225

23. 3.1 - 23
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
N
x 

2
)( 

Sum the squared
deviations from the
mean
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15
625
625
225
25
25
225
1750

24. 3.1 - 24
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15
625
625
225
25
25
225
1750
6
1750
)( 2




N
x 


25. 3.1 - 25
Example 13: Outdoor Paint
An experimental brand of outdoor paint is tested to
see how long it will last before fading (in months). Six
cans of the brand constitutes a small population.
Find the standard deviation.
10, 60, 50, 30, 40, 20
Months, x µ x - µ (x - µ)2
10
60
50
30
40
20
35
35
35
35
35
35
-25
25
15
-5
5
-15
625
625
225
25
25
225
1750
1.17
6
1750
)( 2





N
x 
