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

29.8 Standard deviation of a random variable 947
Calculate
(a) the expected value
(b) the standard deviation
Solution (a) µ = ∑ x i
p i
= 1(0.12) +2(0.15) +3(0.23) +4(0.3) +5(0.2) = 3.31
(b) σ 2 = ∑ p i
(x i
− µ) 2
=0.12(1 −3.31) 2 +0.15(2 −3.31) 2 +0.23(3 −3.31) 2
+0.3(4 −3.31) 2 +0.2(5 −3.31) 2
=1.6339
standard deviation = σ = √ 1.6339 = 1.278
29.8.2 Standarddeviationofacontinuousrandomvariable
Wesimplystatetheformulaforthestandarddeviationofacontinuousrandomvariable.
It is analogous to the formula for the standard deviation of a discrete variable. Letxbe
a continuousrandom variablewith p.d.f. f (x),a xb. Then
√ ∫ b
σ =
a
(x − µ) 2 f(x)dx
Example29.11 Arandomvariable,x, has p.d.f. f (x)given by
f(x)=1
0x1
Calculate the standarddeviation ofx.
Solution Theexpected value, µ,is found:
∫ 1
[ x
2
µ = xf(x)dx= = 1 2 2
0
] 1
0
The variance can now befound:
∫ 1
(
variance = σ 2 = x − 1 2
1dx
2)
Hence
=
σ =
√
1
12 = 0.29
0
∫ 1
0
x 2 −x+ 1 4 dx
The standard deviation ofxis0.29.
[ x
3
=
3 − x2
2 4] + x 1
= 1
0
3 − 1 2 + 1 4 = 1
12