Normal Random Variable - Random Signals and Processes Primer with MATLAB - page 237

Solution The variance and the standard deviation are:

s 2 ¼ EfX 2

2

gðEfXgÞ ¼ 89 25 ¼ 64 ; s ¼ 8 :

From ( 4.30 ) we have:

1
2

5
8 2

PfX 10 g¼F X ð 10 Þ¼ 1 þ erf p ¼ 0 : 734 ; (4.35)

where

¼ 0

5
8 2

erf p : 468 : (4.36)

From ( 4.31 ), we have:

erf 7

1
2

3
8 2

p

8 2

p

Pf 2 X 8 g¼ erf

þ erf

1
2

3
8 2

7
8 2

¼ erf p p

1
2 0

¼ ½ : 2923 þ 0 : 6184 ¼ 0 : 4554 : (4.37)

4.2.2.2 Complementary Error Function

Complementary error function is denoted as erfc and is related to error function as:

erfc ðxÞ¼ 1 erf ðxÞ: (4.38)

From ( 4.23 ) and ( 4.24b ), it follows:

1 ð

x

1

2
p

2
p

2
p e u 2 d u e u 2 d u ¼ e u 2 d u: erfc ðxÞ¼ p p p (4.39)

x

0 0

From ( 4.30 ) and ( 4.38 ), we have:

x 1 m
p s

1
2

PfX x 1 g¼F X ðx 1 Þ¼ 1 þ 1 erfc

:

1
2 erfc

x 1 m
p s

¼ 1 (4.40)