.................................................................................................................................2
12.1
.......................................................................................................................2
12.1.1
.............................................................................................2
12.1.2
.........................................................................................................2
12.2 (Standardized Normal Distribution)
.........................................................4
....................................................................................................6
12.3 (Skewness)
........................................................................................................7
12.4
(Kurtosis)......................................................................................................................9
12.5
.....................................................................................................11
12.6 (Normal Approximation to the
Binomial).......................................12 12.7 (Normal
Approximation to
Poisson)............................................13
1
//(Normal Distribution)(Gauss Distribution) (error of
measurement)( ) (bell shape)()( )
12.1 x (Normal random variable) X ~ N ( , ) N ( x, , ) 2 ( 1 f (
x) = e 2 2 1 x
)2
1 = e 2
( x )2 2 2
, < x3 (platykurtosis) 2 0.5 n(1-p)> 5 ,. X (z): Z=
x E( X )
x
=
x np npq
, (correction of continuity) Pr(a x b) = Pr(a - 0.5 Y b + 0.5)
0.5 (halfunit correction of continuity) Sample: X , n = 15, p =
0.4, (a)Pr(2 X 4) (b)Pr(X 10) Solution: n = 15, p = 0.4 np = 15*0.4
= 6 > 5 npq = 15*0.4*0.6 = 3.6 2 = 3.6, = 1.9 (a) Pr(2X4) =
Pr(X4) Pr(X1) = 0.21732-0.0052=0.2121 Pr(2X4)=Pr(1.5Y4.5) = Pr(
= Pr(-2.37Z-0.79) = Pr(Z-0.79) Pr(Z-2.37) = 0.2148 0.0089 =
0.2059 0.2121 0.2059 = 0.0062 (b) Pr(X10) = 1 - Pr(X9) = 1 0.9662 =
0.0338 Pr(X 10) = Pr(Y 9.5) = Pr(
1.5 6 4.5 6 Z ) 1.9 1.9
= Pr(Z1.84)= 1- Pr(Z1.84) = 1 - 0.9671 = 0.0329 0.0338 0.0329 =
0.0009
y 6 9.5 6 ) 1.9 1.9
n = 30 , p = 0.4, Pr(X20) = ? np = 300.4 = 12 > 5 2 = npq =
300.40.6 = 7.2 = 2.68 Pr(X20) = 1 - Pr(X 19) = 1 - 0.9974 = 0.0026
Pr(X 20) = Pr(Y 19.5) = Pr(Z
= 1 - Pr(Z2.80) = 1 - 0.9974 = 0.002612
19.5 12 ) = Pr(Z2.80) 2.68
n
12.7 (Normal Approximation to Poisson) X , = S, 2 =S.
(parameter), S , , S/S = Pr(Xa) = Pr(Ya - 0.5) = Pr( Sample
y
(a 0.5)
)
5,400,000 /dm3, 1/10,000 dm3, 500 600 ? =S = 5,400,000
SD = Pr(500X600) = Pr(499.5Y600.5) = Pr(
1 = 540 10000 = 540 = 23.24
= Pr(Z2.60) Pr(Z-1.74) = 0.9953 0.0409 = 0.9544
499.5 540 600.5 540 Z ) = Pr(-1.74Z2.60) 23.24 23.24
13
