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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
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
)( 

3.1 - 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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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 

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 


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Standard Deviation of Outdoor Paint Fading 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  