Unformatted text preview: owever, for the beginner, this quantity has
certain difficulty in interpretation, as the unit of
measurement (here, it is “squared thousand $”) is unusual.
To go back to our familiar unit ($), we take the square root
(to offset the previous step of “squaring”), and call it the
σ
standard deviation, which is denoted by
standard
which σ = Variance = 23.076 = 4.8037
[ σ = sigma, the Greek “s”. ]
sigma,
Formula: σ = Variance = ( xi − µ ) 2
∑
N →(1) Computer steps:
Computer For Company A data ON MODE COMP (1) MODE SD (4)
15.2 M + 9.2 DT 6.7 DT 12.4 DT 20.5 DT
15.2
DT SHIFT SSUM n (3)
SHIFT EXE 1
SHIFT SVAR 2 xσn (2) EXE 5 4.803748536 Interpretation:
Interpretation The deviation is
x1 − µ = 15.2 − 12.8 = 2.4 because Small 2.4<4.8037 Small 3.6<4.8037
x2 − µ = 9.2 − 12.8 = −3.6 x3 − µ = 6.7 − 12.8 = −6.1 Large 6.1>4.8037 x4 − µ = 12.4 − 12.8 = − 0.4 Small 0.4<4.8037
x5 − µ = 20.5 − 12.8 = 7.7 Large 7.7>4.8037 Thus, the standard deviation
is a “notlarge”, “notsmall”
deviation.
The standard deviation is a
deviation standard, used
to tell the largeness or
smallness of any other
individual deviation.
individual
It can be seen that the
variance is the average of
the squared deviations.
The s.d. is the square root of
the variance.
the Example 2:
Example
Find the standard deviation for each of the data sets for
Company A and Company B.
Company
A : 6.7, 9.2, 12.4, 15.2, 20.5
µ = 20
B : 7.3, 8.7, 10.7, 16.5, 21.8, 55.0
Solution: For Company A, σ A = 4.8037 , found above.
Solution
For Company B,
For
(8.7 − 20) 2 + (10.7 − 20) 2 + ......... + (55 − 20) 2
∴σ B =
6
1615.96
6
= 16.4112
= Comment:
Comment
Data set B is much more variable than Set A since the
standard deviation for data company B is larger than
for company A.
for Computer steps : For Company B data
Computer
ON MODE COMP (1) MODE SD (4)
8.7 DT 10.7 DT 7.3 DT 16.5 DT 21.8
SHIFT SSUM n (3) 1
SHIFT SVAR 2 SHIFT SVAR 2 x EXE (1) EXE xσn (2) EXE DT 55.0 DT 6
20
16.41117505 For data, x1 , x2 ,.............., xN , the standard deviation
is:
is:
σ = Variance = ( xi − µ ) 2
∑
N Often the data x1 , x2 ,.............., xN are “neat”
numbers such as integers, but µ involves
a lot of decimal places. The results of ( xi − µ ) 2
will also be ugly, i.e. involving many
decimal places, so that, at the end, the
trouble of rounding will appear. It would be
useful to derive another formula that does
not call upon µ until a very late stage. We
shall derive a practical formula for σ .
practical 2
( x1 − µ) 2 = x1 − 2 µx1 + µ2 2
( x N − µ) 2 = x N − 2 µx N + µ2 _____________________________
( xi − µ) 2 = ∑xi2 − 2 µ∑xi + Nµ2
∑ Now, ∑ x , so that
µ=
i N ∑ x = Nµ
i ∴ ∑ ( xi − µ ) = ∑ x − 2 µ ( Nµ ) + Nµ = ∑ x − Nµ
2 ∴σ = 2
i ( xi − µ ) 2
∑
N 2 = xi2
∑
N 2
i − µ 2 →(2) 2 Example 3:
Use formula (2) to compute
Use σ for set A. Solution 1:
Solution
15.2 xi
xi2 9.2 6.7 12.4 20.5 ∑x i 231.04 84.64 44.89 153.76 420.25 ∑ x 2
i N =5 ∑x i ∑x 2
i ∑x
µ= i = 64
= 934.58 σ= N 64
=
5
= 12.8 = ∑x 2
i N = 64 = 934.58 − µ2 934.58
−12.8 2
5 = 186.916 −163.84
= 4.8037 Calculator steps:
Calculator For Company A data ON MODE COMP (1) MODE SD (4)
15.2 DT 9.2 DT 6.7 DT 12.4 DT 20.5 DT
15.2
SHIFT
SHIFT
SHIFT
SHIFT SSUM n (3) EXE
SSUM ∑ x (2) EXE
SSUM ∑ x 2 (1) EXE SHIFT SVAR
SHIFT SVAR x
xσn (1) EXE
(2) EXE 5
64
934.58
12.8
4.803748536...
View
Full Document
 Fall '13
 Sets, Standard Deviation, Mean, Deviation, Errors and residuals in statistics

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