# Variance and Standard Deviation (1) by drr10525

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• pg 1
```									    Variance and
Standard Deviation (1)

50%

Tells us the spread of the middle 50% of the data
A 12 day survey of Fred and
Doris’s toilet habits revealed
some exciting results
Doris's
Fred visits    Deviation visits to the Deviation
Day     to the loo    from Mean       loo      from Mean
1           5           -3.1          28         15.4
2          11            2.9          17          4.4
3           2           -6.1           5         -7.6
4           6           -2.1           7         -5.6
5           0           -8.1           4         -8.6
6           4           -4.1          14          1.4
7          14            5.9          13          0.4
8           9            0.9          22          9.4
9          16            7.9          16          3.4
10          20           11.9           9         -3.6
11           3           -5.1           6         -6.6
12           7           -1.1          10         -2.6

Mean       8.1             Mean      12.6

Mean = sum of data / pieces of data
Mean = 8.1
1.5
1
0.5
0                                                                Fred
-0.5                                                                Doris
-1   0   2   4   6   8 10 12 14 16 18 20 22 24 26 28 30
-1.5
-2
-2.5

Mean = 12.6                            Doris's
Fred visits visits to the

Day    to the loo       loo
1          5            28
can find the deviation                  2
3
11
2
17
5
from the mean                           4
5
6
0
7
4
6          4            14
The deviations may be positive             7
8
14
9
13
22
e.g. Day 10, Fred : 20 – 8.1 = 11.9        9         16            16
10         20             9
11          3             6
or negative                               12          7            10
e.g. Day 10, Doris : 9 – 12.6 = -3.6
Mean      8.1          12.6
Deviations from Mean (1)
Doris's

Day
Fred visits
to the loo
Deviation visits to the Deviation
from Mean       loo      from Mean
To summarise the
1           5           -3.1          28         15.4     total deviation,
2          11            2.9          17          4.4
3           2           -6.1           5         -7.6     you can’t just add
4
5
6
0
-2.1
-8.1
7
4
-5.6
-8.6     – since they may
6
7
4
14
-4.1
5.9
14
13
1.4
0.4
cancel
8           9            0.9          22          9.4
9          16            7.9          16          3.4
10          20           11.9           9         -3.6
11           3           -5.1           6         -6.6
12           7           -1.1          10         -2.6

Mean      8.1             Mean      12.6

Either: ignore the signs and add
Or:    square the deviations (to make positive) then add
Deviations from Mean (2)
Doris's
Fred visits    Deviation   Deviation   visits to the Deviation     Deviation
Day   to the loo    from Mean    Squared          loo      from Mean     Squared
1         5           -3.1         9.5            28         15.4        237.7
2        11            2.9         8.5            17          4.4         19.5
3         2           -6.1        37.0             5         -7.6         57.5
4         6           -2.1         4.3             7         -5.6         31.2
5         0           -8.1        65.3             4         -8.6         73.7
6         4           -4.1        16.7            14          1.4          2.0
7        14            5.9        35.0            13          0.4          0.2
8         9            0.9         0.8            22          9.4         88.7
9        16            7.9        62.7            16          3.4         11.7
10        20           11.9       142.0             9         -3.6         12.8
11         3           -5.1        25.8             6         -6.6         43.3
12         7           -1.1         1.2            10         -2.6          6.7

Sum of deviation squared    408.9       Sum of deviation squared    584.9

Since two sets of data may have a different number of
items, you divide by the number of items in the data set
… this is known as the variance
Variance
Variance
= sum of (the deviations from mean)2
number of pieces of data

For Fred,
variance = 408.9 / 12 = 34.1

For Doris,
variance = 584.9 / 12 = 48.7
Standard Deviation
If the data were cm , then the variation would be in cm2

To keep it in the same units
– we square root the variance.

This is known as the Standard Deviation

Standard Deviation = variance
For Fred,
standard deviation = variance =  34.1 = 5.8
For Doris,
standard deviation = variance = 48.7 = 7.0
Activity

•   Page 26 of your Statistics 1
book and try …
• Exercise A
•   Calculating Variance and Standard Deviation
Notation (1)
 (sigma) denotes sum
If data has n values - x1, x2, x3, …. , xn

The sum of the data is written   xi

x (‘x bar’) denotes its mean
x = xi (sum of numbers divided by pieces of data)
n
Notation (2)
The sum of the squared deviations from the mean is written

(xi -     x)2

Hence, variance and standard deviation are :

Variance =  (xi - x)2
n
Standard Deviation = (xi - x)
2

n
Standard Deviation is denoted by s or sx
Variance is usually denoted by s2 or sx2
Notation (3)
We may as well get all the notation out the way
Sometimes calculators or computer programs use different symbols

mean x is sometimes denoted by 

Standard Deviation is denoted by s or sx

.. but sometimes by           n

HAVE A LOOK AT YOUR CALCULATOR
More on Standard Deviation
Standard Deviation =   (xi - x)2
n
Look at page 27-28 of your Statistics 1 book for the proof
(if I make a Slide of this, it’ll take me till midnight)

Standard Deviation can more conveniently be written

sx =      xi 2-   nx 2
or        xi 2   -   x 2

n                       n
… this makes manual calculations much simpler
Activity
Page 28 of your Statistics 1
book and try …
• Exercise C
• Calculating Standard Deviation