082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

942 Chapter 29 Statistics and probability distributions
Example29.6 Find the standard deviation of −2,7.2,6.9, −10.4, 5.3.
Solution x 1
= −2,x 2
= 7.2,x 3
= 6.9,x 4
= −10.4,x 5
= 5.3,x=1.4
x 1
−x = −3.4
x 2
−x=5.8
x 3
−x=5.5
x 4
−x = −11.8
x 5
−x=3.9
variance = (−3.4)2 + (5.8) 2 + (5.5) 2 + (−11.8) 2 + (3.9) 2
√ 5
229.9
standard deviation = = 6.78
5
= 229.9
5
Calculatingx i
−x,i = 1,2,...,n,istediousforlargenandsoamoretractableformof
the standard deviation is sought. Firstly observe that ∑ x i
=nx and ∑ x 2 =nx 2 , since
x 2 isaconstant. Now
∑
(xi −x) 2 = ∑ x 2 i
− ∑ 2x i
x + ∑ x 2
= ∑ x 2 i
−2x ∑ x i
+nx 2
= ∑ x 2 i
−2x(nx)+nx 2
= ∑ x 2 i
−nx 2
Hence,
and
variance =
∑ x
2
i
−nx 2
n
√ ∑x
2
i
−nx 2
standard deviation =
n
Using these formulae itisnot necessary tocalculatex i
−x.
Example29.7 RepeatExample 29.6 usingthe newly derived formulae.
Solution ∑ xi 2 = (−2) 2 + (7.2) 2 + (6.9) 2 + (−10.4) 2 + (5.3) 2 = 239.7
√
239.7 −5(1.4)
2
standard deviation =
= 6.78
5