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Normal Distribution Curve
symetri axes
x=µ•
area = 1• • ••••
x=µ+σ x=µ+2σ x=µ+3
σ
x=µ-σx=µ-2σ
x=µ-3σ
Standard Normal Distribution Table
0 z
This area can be found by using a standard normal distribution tablel
This area can be thought as a probability appearing Z between 0 and z, written as P(Z|0<Z<z)
Example using a standard normal distribution table
0 1.35
Area = ?
P(Z|0<Z<1.35) = 0.4115
P(Z|Z>1.35) = 0.5000-0.4115 = 0.0885
1.3
0.05
.4115
0.4115
Example using a standard normal distribution table
0-1.24
0.98
Area =?
Area =
0.3925 +
0.3365 =
0.7290
On a group of 1000 students, the mean of their score is 70.0 and the standard deviation is 5.0. Assuming that the score are normally distributed. How many students have score between 73.6 dan 81.9?
Problem
Solution
µ = 70.0; σ = 5.0; X1 = 73.6;
X2 = 81.9;
We transform X into z by using the formulae:
0 0.72
Area = 0.4913 – 0.2642 = 0.2271
P(73.6<X<81.9) = P(Z|0.72<Z<2.38) = 0.2271
2.38
So, the number of students having score between 73.6 and 81.9 is 0.2271 x 1000 = 227 student
STANDARD NORMAL DISTRIBUTION N(0,1)
z=0 1• • •
2•3
•-1
•-2
•-3
0.4772
0.4987
0.0013
z0.0013z0.0228z0.1587z0.5000z0.8413
0.3413
Critical Value and Crtitical Region on N(0,1)
Significance level, usually
denoted by α
•
It is called critical value (nilai kritis)
(CV), denoted by zα
It is called critical region (daerah kritis), denoted by CR
CR = {z | z > zα}
Critical Values for t distribution
•tα ; Ʋ
t0.10 ; 12 =
1.356 t0.05 ; 12 =
1.782
t0.01 ; 24 =
2.492 t0.005 ; 28 =
2.763
α
Seen from the table
Critical Value for Chi-Square Distribution
•
α
Properties:Example
Seen from the table
11.070 48.278
α
•