Transcript Measures of Variance

2.4
http://www.youtube.com/watch?v=Rn_Oh
PKBjB0
Why we need to learn something so we
never sound like this.
 The
simplest measure of variance is the
range.
 The range of a data set is the difference
between the maximum and minimum data
entries in the set.

Range = (maximum data entry)-(Minimum data entry)
Midrange = (maximum data entry)-(Minimum data entry)/2

Two corporations each hired 10 graduates. The starting
salaries for each are shown. Find the range of the starting
salaries for Corporation A and B.
Starting Salaries for Corporation A (1000’s of Dollars)
Salary
41
38
39
45
47
41
44
41
37
42
29
52
58
Starting Salaries for Corporation B (1000’s of Dollars)
Salary
40
23
41
50
49
32
41
Ordering the data helps to find the least and greatest salaries.
37 38 39 41 41 41 42 44 45 47
Range = (Maximum salary) – (Minimum salary)
= 47 – 37
=10
Find Corporation B
 As
a measure of variation, the range has the
advantage of being easy to compute. Its
disadvantage, however, is that it uses only
two entries from the data set. Two measures
of variation that use all the entries in a data
set are the variance and the standard
deviation. However, before you learn about
these measures of variation, you need to
know what is meant by the deviation of an
entry in a data set.
 The
deviation of an entry x in a population
data set is the difference between the entry
and the mean μ of the data set.
 Deviation of x = x – μ
 Find the deviation in Corporation A
Starting Salaries for Corporation A (1000’s of Dollars)
Salary
41
38
39
45
47
41
44
41
37
42
29
52
58
Starting Salaries for Corporation B (1000’s of Dollars)
Salary
40
23
41
50
49
32
41
The mean starting salary is μ = 415/10 = 41.5. To find out how much
the salary deviates from the mean, subtract 41.5 from the salary. For
instance, the deviation of 41 (or $41,000) is
41 – 41.5 = -0.5 or (-$500)
Find the deviation of salary $49,000
 Notice
that the sum
of the deviations is
zero. Because this is
true for any set, it
doesn’t make sense
to find the average of
the deviations. To
overcome this
problem, you can
square each
deviation. In a
population data set,
the mean of the
square of the
deviations is called
the population
variance.
Salary
Deviation
(1000s of dollars)
(1000s of dollars)
x
x-μ
41
-0.5
38
-3.5
39
-2.5
45
3.5
47
5.5
41
-0.5
44
2.5
41
-0.5
37
-4.5
42
-0.5
Ʃx = 415
Ʃ(x-μ) = 0



Variance is often used as a measure of risk in finance.
It is a statistical concept whose purpose is to measure
the dispersion (the spread of the #’s) of observation
around the mean or the average value.
Risk is the possibility that actual returns might differ,
or vary, from expected returns. In fact, actual
returns will likely differ from expected returns. It is
important for decision-makers to estimate the
magnitude and likelihood of the difference between
actual and estimated returns. After all, there is a big
difference if your predictions result in an error of
only $100 versus an error of $1 million.
By using the concepts of variance and standard
deviation, investors can judge not only how wrong
their estimates might be, but also estimate the
likelihood, or probability, of favorable or unfavorable
outcomes. With the tools of expected return and
standard deviation, financial decision-makers are
better able to evaluate alternative investments based
on risk-return tradeoffs, and their own risk
preferences.
The preceding computational procedure reveals
several things. First, the square root of the
variance gives the standard deviation; and vice
versa, squaring the standard deviation gives the
variance.
 Second, the variance is actually the average of
the square of the distance that each value is
from the mean. Therefore, if the values are near
the mean, the variance will be small. In
contrast, if the values are far from the mean,
the variance will be large.
 Remember the sum of the distances will always
be zero!!!!

 The
variance is the average of the squares of
the distance each value is from the mean.
Population variance =
 σ2
is the lowercase Greek letter sigma
X=individual value
μ= population value
N= population size
 The
population standard deviation of a
population data set of N entries is the square
root of the population variance.
Population standard deviation =
http://www.youtube.com/watch?v=jYd38gALa04
Salary
Deviation
Squares
(x – μ)2
(1000s of
dollars)
(1000s of
dollars)
x
x-μ
41
-0.5
0.25
38
-3.5
12.25
39
-2.5
6.25
45
3.5
12.5

47
5.5
30.25

41
-0.5
0.25

44
2.5
6.25
41
-0.5
0.25
37
-4.5
20.25
42
-0.5
0.25
Ʃx = 415
Ʃ(x-μ) = 0
SSx=88.5


Find the population
variance and standard
deviation of the starting
salaries for Corporation A
given in Example 1. μ = 41.5
The table at the left
summarizes the steps used
to find SSx = 88.5
N= 10
Population Variance = 8.9
Population Standard
deviation = 2.98 or $2,999.

When interpreting the standard deviation,
remember that it is a measure of the typical
amount an entry deviates from the mean. The
more the entries are spread out, the greater the
standard deviation.
 The
sample variance formula and sample
standard deviation formula of a sample data
set of n entries are listed below.
 Sample variance =
 Sample
Standard Deviation =


Find the sample variance and standard deviation for the amount
of European auto sales for a sample of 6 years shown, The data
are in millions of dollars.
11.2, 11.9, 12.0, 12.8, 13.4, 14.3