vdoc

244 Part IV: Quality Assurance
Continued
2 6 X 6 10
6
P( 20 . X 100
. )= P P 10 .
4 4 4
= Z 10 .
= P( 10 . Z 0)P( 0 Z 10
. )
( )
( )
= 2P
0Z
1. 0 = 2( 0. 3413)=
0. 6826.
c. Again, transforming X into Z and using Figure 18.40, we get
0 6 X 6 4
6
P( 0
X 4. 0)=
P P 150 Z
4 4 4
= . 050
.
( )
= P 05 . Z
150
.
( )
= P( 0Z 1. 50)P( 0 Z 0.
50)= 0. 4332 0. 1915 = 0. 2417.
–5 5 6 8 10 14 15 –3 –2 –1 0 0.5 1 2 3
Figure 18.37 Converting normal N(6,4) to standard normal N(0,1).
Part IV.A.5
–3 0 0.5 2.0 3
Figure 18.38 Shaded area showing P(0.5 Z 2.0).
–3 –1 0 1
3
Figure 18.39 Shaded area showing P(–1.0 Z 1.0).
–3 –1.5 –0.5 0
3
Figure 18.40 Shaded area showing P(–1.50 Z –0.50).