Outline
Normal Probability Distribution
 Standard Normal Probability Distribution
 Standardization (Z-score)
 Illustrations of Normal Distributions
 Applications Standard Normal
Distribution
 Central Limit Theorem

(c) 2007 IUPUI SPEA K300 (4392)
Continuous Probability Distribution

There are a nondenumerable number of
values; there are a indefinite number of
possible values that fall between any two
values
(c) 2007 IUPUI SPEA K300 (4392)
Normal Probability Distribution






A continuous, symmetric, bell-shaped
distribution of a random variable
Defined by a mean µ and variance σ2,
x~N(µ, σ2)
Mean=median=mode
Ranges from -∞ to +∞ (positive infinity)
No P(x) is zero; the normal curve never
touches the x axis.
1
P( x) 
e
 2
( x   ) 2
2 2
(c) 2007 IUPUI SPEA K300 (4392)
Standard Normal Probability
Distribution

Normal distribution with the mean 0 and
variance 1
 Why? Easy to get probability by standardizing
the scale
 P(x)(transforming)P(z)
z
x
P( z ) 

1
e
2
z2
2
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-1
0
.1
y
.2
.3
.4
P(0<=z<=2.34) = .4904
-4
-3
-2
-1
0
x
1
(c) 2007 IUPUI SPEA K300 (4392)
2
3
4
Example 6-3
0
.1
y
.2
.3
.4
P(-1.75<=z<=0) = .4599
-4
-3
-2
-1
0
x
1
2
(c) 2007 IUPUI SPEA K300 (4392)
3
4
Example 6-4
.1335 = .5-.3665
0
.1
y
.2
.3
.4
P(z>=1.11) = .1335
-4
-3
-2
-1
0
x
1
2
(c) 2007 IUPUI SPEA K300 (4392)
3
4
Example 6-5
.0268 = .5-.4732
0
.1
y
.2
.3
.4
P(z<=-1.93) = .0268
-4
-3
-2
-1
0
x
1
(c) 2007 IUPUI SPEA K300 (4392)
2
3
4
Example 6-6
.0160 = .4932-.4772
0
.1
y
.2
.3
.4
P(2<=z<=2.47) = .0160
-4
-3
-2
-1
0
x
1
2
(c) 2007 IUPUI SPEA K300 (4392)
3
4
Example 6-7
.1967 = .4934-.2967
0
.1
y
.2
.3
.4
P(-2.48<=z<=-.83) = .1967
-4
-3
-2
-1
0
x
1
(c) 2007 IUPUI SPEA K300 (4392)
2
3
4
Example 6-8
.8682 = .4535-.4147
0
.1
y
.2
.3
.4
P(-1.37<=z<=1.68) = .8682
-4
-3
-2
-1
0
x
1
(c) 2007 IUPUI SPEA K300 (4392)
2
3
4
Example 6-9
.9767 = .5+.4767
0
.1
y
.2
.3
.4
P(z<=1.99) = .9767
-4
-3
-2
-1
0
x
1
(c) 2007 IUPUI SPEA K300 (4392)
2
3
4
Example 6-10
.8770 = .5+.3770
0
.1
y
.2
.3
.4
P(z>=-1.16) = .8770
-4
-3
-2
-1
0
x
1
(c) 2007 IUPUI SPEA K300 (4392)
2
3
4
.10 Significance Level (α=.10)
.10 = .05 +.05
0
.1
y
.2
.3
.4
P(-1.645<=z<=1.645) = .10
-4
-3
-2
-1
0
x
1
(c) 2007 IUPUI SPEA K300 (4392)
2
3
4
.05 Significance Level (α=.05)
.05 = .025 +.025
0
.1
y
.2
.3
.4
P(-1.96<=z<=1.96 = .05
-4
-3
-2
-1
0
x
1
(c) 2007 IUPUI SPEA K300 (4392)
2
3
4
.01 Significance Level (α=.01)
.01 = .005 +.005
0
.1
y
.2
.3
.4
P(-2.58<=z<=2.58 = .01
-4
-3
-2
-1
0
x
1
(c) 2007 IUPUI SPEA K300 (4392)
2
3
4
PDF versus CDF

Probability Density Function (PDF)
b
p(a  x  b)   f ( x)dx  0
a
p(  x  )  



f ( x)dx  1
Cumulative Distribution Function (CDF)
F ( x)  P ( x  v)
(c) 2007 IUPUI SPEA K300 (4392)
Normal Distribution and Others

A random variable x follows the uniform,
binomial, normal, and other distributions
 If x~N(µ,σ2), z=(x-µ)/σ~N(0,1)
 If x~N(µ,σ2), x2~Chi-squared(1)

Mean1 (df) and variance 2 (2*df)
If x1~N(µ,σ2) and x2~N(µ,σ2), x1/x2~F(n1, n2)
 If z~N(0,1) and x~Chi-squared(n) independent
of z, z/sqrt(x/n)~t(n), t distribution with degrees
of freedom n

(c) 2007 IUPUI SPEA K300 (4392)
Standardization



What is the probability that x is greater than 21.2161?
Integration from 21.2161 through the positive infinity?
2.0167=(21.2161-20)/.6030
Read probability from the standard normal probability
table.
(c) 2007 IUPUI SPEA K300 (4392)
Application 1

Transforming a normal distribution to the
standard normal distribution with the mean 0
and variance 1
 What is the z in the standard normal
distribution of a 10 of the random variable x in
a normal distribution with the mean 5 and
variance 4? How about 3 of x?
z
x

10  5

 2.5
2
z
x
(c) 2007 IUPUI SPEA K300 (4392)

35

 1
2
0
.1
.2y
.3
.4
Application 2
-3
0
3
Mu
x
7
10
0
.1
.2y
.3
.4
Y~N(5,2^2)
-3
-2
-1
Mu
x
1
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
2
3
0
.1 .2 .3 .4y .5 .6 .7 .8
Example 6-14
1
3.1
x
3.5
5
0
.1
.2y
.3
.4
Y~N(3.1,.5^2)
-3
-2
-1
0
x
.8
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
2
3
0
.1
.2y
.3
.4
Example 6-15
22
25
27
28
x
31
34
0
.1
.2y
.3
.4
Y~N(28,2^2)
-3
-.5
0
x
1.5
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
3
0
.05
y
.1
.15
Ex 6-16, p(x<15)=.0132=.5-.4868
15
25
x
0
.1
.2y
.3
.4
Y~N(25,4.5^2)
-2.22
0
x
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
2.22
0
.1
.2y
.3
.4
Ex 6-17, p(z>?)=.1=p(x>??)
-1.28
0
x
1.28
0
.01
.02
y
.03
.04
.05
Y~N(0,1^2)
200
x
226
Y~N(200,20^2)
(c) 2007 IUPUI SPEA K300 (4392)
Distribution of Sample Means

A sample distribution of sample means is a
distribution using the means computed from all
possible random samples of a specific size
taken from a population
 Sample error is the difference between the
sample measure and the corresponding
population measure due to the fact that the
sample is not a perfect representation of the
population
(c) 2007 IUPUI SPEA K300 (4392)
Properties of the Distribution of
Sample Means

The mean of the sample means will be the
same as the population mean.
 The standard deviation of the sample means
will be smaller than the standard deviation of
the population, and it will be equal to the
population standard deviation divided by the
square root of the sample size.
x 

n
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Central Limit Theorem

As the sample size n increases without limit,
the shape of the distribution of the sample
means taken with replacement from a
population with mean µ and standard deviation
σ will approach a normal distribution.
 This distribution will have a mean µ and a
standard deviation σ/sqrt(n)
zx 
x

n
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-21
x bar=25, σ=3, n=20
zx 
x
x

x
26.3  25 1.3


 1.94

3
.671
n
20
0
.1
.2y
.3
.4
p( z x  1.94)  .5  p(0  z x  1.94)  .5  .4738  .0262
-2.58 -1.96
0
x
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
1.94
2.58
Example 6-22
x bar=96, σ=16, n=36
zx 
x


n
90  96
6
100  96
4

 2.25 z x 

 1.50
16
16
.2.6667
2.6667
36
36
0
.1
.2y
.3
.4
p(2.25  z x  1.50)  .4878  .4332  .921
-2.58 -1.96
0
x
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
1.5 1.96
2.58
Example 6-23
x=218.4, σ=25
zx 
x


224  218.4
 .22
25
0
.1
.2y
.3
.4
p(  z x  .22)  .5  .0871  .5871
-2.58 -1.96
0.22
x
1.96
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
2.58
Example 6-23
x bar=218.4, σ=25, n=40
zx 
224  218.4
5.6

 1.42
25
3.9528
40
0
.1
.2y
.3
.4
p(  z x  1.42)  .5  .4222  .9222
-2.58 -1.96
0
x
1.42 1.96
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
2.58