(c) 2007 IUPUI SPEA K300 (4392)
Outline
Normal Probability DistributionStandard Normal Probability DistributionStandardization (Z-score)Illustrations of Normal DistributionsApplications Standard Normal
DistributionCentral 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.
2
2
2
)(
2
1)(
x
exP
(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)
2
2
2
1)(
z
ezP
x
z
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-1
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(0<=z<=2.34) = .4904
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-3
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(-1.75<=z<=0) = .4599
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-4
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(z>=1.11) = .1335
.1335 = .5-.3665
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-5
.0268 = .5-.4732
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(z<=-1.93) = .0268
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-6
.0160 = .4932-.4772
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(2<=z<=2.47) = .0160
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-7
.1967 = .4934-.2967
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(-2.48<=z<=-.83) = .1967
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-8
.8682 = .4535-.4147
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(-1.37<=z<=1.68) = .8682
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-9
.9767 = .5+.4767
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(z<=1.99) = .9767
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-10
.8770 = .5+.3770
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(z>=-1.16) = .8770
(c) 2007 IUPUI SPEA K300 (4392)
.10 Significance Level (α=.10)
.10 = .05 +.05
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(-1.645<=z<=1.645) = .10
(c) 2007 IUPUI SPEA K300 (4392)
.05 Significance Level (α=.05)
.05 = .025 +.025
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(-1.96<=z<=1.96 = .05
(c) 2007 IUPUI SPEA K300 (4392)
.01 Significance Level (α=.01)
.01 = .005 +.005
0.1
.2.3
.4y
-4 -3 -2 -1 0 1 2 3 4x
P(-2.58<=z<=2.58 = .01
(c) 2007 IUPUI SPEA K300 (4392)
PDF versus CDF
Probability Density Function (PDF)
Cumulative Distribution Function (CDF)
0)()( b
adxxfbxap
1)()(
dxxfxp
)()( vxPxF
(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?
5.22
510
xz 1
2
53
x
z
(c) 2007 IUPUI SPEA K300 (4392)
Application 2
0.1
.2.3
.4
y
-3 0 3 Mu 7 10x
Y~N(5,2^2)
0.1
.2.3
.4
y
-3 -2 -1 Mu 1 2 3x
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-14
0.1
.2.3
.4.5
.6.7
.8
y
1 3.1 3.5 5x
Y~N(3.1,.5^2)
0.1
.2.3
.4
y
-3 -2 -1 0 .8 2 3x
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-15
0.1
.2.3
.4
y
22 25 27 28 31 34x
Y~N(28,2^2)
0.1
.2
.3
.4
y
-3 -.5 0 1.5 3x
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
Ex 6-16, p(x<15)=.0132=.5-.4868
0.0
5.1
.15
y
15 25x
Y~N(25,4.5^2)
0.1
.2
.3
.4
y
-2.22 0 2.22x
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
Ex 6-17, p(z>?)=.1=p(x>??)0
.1
.2
.3
.4
y
-1.28 0 1.28x
Y~N(0,1^2)
0.0
1.0
2.0
3.0
4.0
5
y
200 226x
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.
nx
(c) 2007 IUPUI SPEA K300 (4392)
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)
n
xzx
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-21
0.1
.2.3
.4
y
-2.58 -1.96 0 1.94 2.58x
Y~N(0,1^2)
x bar=25, σ=3, n=20
94.1671.
3.1
20
3253.26
n
xxz
xx
0262.4738.5.)94.10(5.)94.1( xx zpzp
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-22
x bar=96, σ=16, n=36
25.26667.2.
6
36
169690
n
xzx
921.4332.4878.)50.125.2( xzp
0.1
.2.3
.4
y
-2.58 -1.96 0 1.5 1.96 2.58x
Y~N(0,1^2)
50.16667.2
4
36
1696100 xz
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-23
x=218.4, σ=25
22.25
4.218224
x
zx
5871.0871.5.)22.( xzp
0.1
.2.3
.4
y
-2.58 -1.96 0.22 1.96 2.58x
Y~N(0,1^2)
(c) 2007 IUPUI SPEA K300 (4392)
Example 6-23
x bar=218.4, σ=25, n=40
42.19528.3
6.5
40
254.218224 xz
9222.4222.5.)42.1( xzp
0.1
.2.3
.4
y
-2.58 -1.96 0 1.42 1.96 2.58x
Y~N(0,1^2)