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The deviation of the first reading x I is
d
i
and that of second reading x2 is d2 and so on
lhe deviation fiom mean can be expressed
1'he algebraic sum
as
d,
:
x, - x, d, :az - x
etc -
of deviation is zero.
Deviations : It is def,ned of an infinite number of data is defined as the square root
of the sum of individual deviations squared divided by the number of readings.
3" Average
Thus standard deviation is expressed as
o=
Practically the number of reading is finite, when the number of reading is small (n < 30) the
denominator is (n - 1) and the standard deviation is represented by S.
di +d?r+d]+......+d:
S-
The standard deviation is also called as mean square deviation.
5" Variance : The square ofstandard deviation is called variance.
v = (o)'
=
[,8.l=
n
\r
)
"
Forsmallnumberof rea drng
t 9.6
n
Id'
V=5"=n_l
Problems on Statistical Analysis
:
For the given readings 1.34, 1.38 , 1.56, 1.47, 1.42, 1.43, 1.54, 1.48, 1.49,
1.
Calculate.
L."Arithmetic Mean
3. Standard deviation
Sol
rd'
:
1.50.
2. Average deviation
4. Variance
Given: Readings ; 1.34, 1.38, 1.56, L47, L43,1.54,1"48,1"49, 1.50
n:
1.
10
[
10 readings]
Arithmetic Mean
:
n
Y"
- i=, =a[x, + x2 + x, +,..... + xn]
n n'
,(-J'^t
x
I
1.34 + 1.38 + 1.56 + 1.47 + 1.42 + I "43 + I.54 + 1.48 + 1.49 + 1.50
10
ii. Averagc deviation
dr = Xr
-x
:
= I .34 -1.461 = -0"121
d, -- \, - x = 1.38 - I 461= -0.081
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3
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