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Statistical
Methods
Week 6
Sebaran Normal dan Sebaran Kontinu Lainnya /
The Normal Distribution and
Other Continuous Distributions
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-1 / 82
Chapter Goals
After completing this chapter, you should be
able to:
 Describe the characteristics of the normal distribution
 Translate normal distribution problems into standardized
normal distribution problems
 Find probabilities using a normal distribution table
 Evaluate the normality assumption
 Recognize when to apply the uniform and exponential
distributions
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-2 / 82
Chapter Goals
(continued)
After completing this chapter, you should be
able to:
 Define the concept of a sampling distribution
 Determine the mean and standard deviation
_ for the
sampling distribution of the sample mean, X
 Determine the mean and standard deviation for the
sampling distribution of the sample proportion, ps
 Describe the Central Limit Theorem and its importance
_
 Apply sampling distributions for both X and ps
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-3 / 82
Probability Distributions
Probability
Distributions
Wk. 5
Discrete
Probability
Distributions
Continuous
Probability
Distributions
Binomial
Normal
Poisson
Uniform
Hypergeometric
Statistical Methods © 2004 Prentice-Hall, Inc.
Wk. 6
Exponential
Week 6-4 / 82
Continuous Probability Distributions
 A continuous random variable is a variable that
can assume any value on a continuum (can
assume an uncountable number of values)




thickness of an item
time required to complete a task
temperature of a solution
height, in inches
 These can potentially take on any value,
depending only on the ability to measure
accurately.
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-5 / 82
The Normal Distribution
Probability
Distributions
Continuous
Probability
Distributions
Normal
Uniform
Exponential
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-6 / 82
The Normal Distribution
‘Bell Shaped’
 Symmetrical
 Mean, Median and Mode
are Equal
Location is determined by the
mean, μ

Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
+  to  
Statistical Methods © 2004 Prentice-Hall, Inc.
f(X)
σ
X
μ
Mean
= Median
= Mode
Week 6-7 / 82
Many Normal Distributions
By varying the parameters μ and σ, we obtain
different normal distributions
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-8 / 82
The Normal Distribution
Shape
f(X)
Changing μ shifts the
distribution left or right.
σ
μ
Statistical Methods © 2004 Prentice-Hall, Inc.
Changing σ increases
or decreases the
spread.
X
Week 6-9 / 82
The Normal Probability
Density Function
 The formula for the normal probability density
function is
1
(1/2)[(Xμ)/σ]2
f(X) 
e
2π
Where
e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
μ = the population mean
σ = the population standard deviation
X = any value of the continuous variable
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-10 / 82
The Standardized Normal

Any normal distribution (with any mean and
standard deviation combination) can be
transformed into the standardized normal
distribution (Z)

Need to transform X units into Z units
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-11 / 82
Translation to the Standardized
Normal Distribution

Translate from X to the standardized normal
(the “Z” distribution) by subtracting the mean
of X and dividing by its standard deviation:
X μ
Z
σ
Z always has mean = 0 and standard deviation = 1
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-12 / 82
The Standardized Normal
Probability Density Function
 The formula for the standardized normal
probability density function is
f(Z) 
Where
1 (1/2)Z 2
e
2π
e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
Z = any value of the standardized normal distribution
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-13 / 82
The Standardized
Normal Distribution



Also known as the “Z” distribution
Mean is 0
Standard Deviation is 1
f(Z)
1
0
Z
Values above the mean have positive Z-values,
values below the mean have negative Z-values
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-14 / 82
Example

If X is distributed normally with mean of 100
and standard deviation of 50, the Z value for
X = 200 is
X  μ 200  100
Z

 2.0
σ
50

This says that X = 200 is two standard
deviations (2 increments of 50 units) above
the mean of 100.
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-15 / 82
Comparing X and Z units
100
0
200
2.0
X
Z
(μ = 100, σ = 50)
(μ = 0, σ = 1)
Note that the distribution is the same, only the
scale has changed. We can express the problem in
original units (X) or in standardized units (Z)
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-16 / 82
Finding Normal Probabilities
Probability is the
Probability is measured
area under the
curve! under the curve
f(X)
by the area
P (a ≤ X ≤ b)
= P (a < X < b)
(Note that the
probability of any
individual value is zero)
a
Statistical Methods © 2004 Prentice-Hall, Inc.
b
X
Week 6-17 / 82
Probability as
Area Under the Curve
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below
f(X) P(   X  μ)  0.5
0.5
P(μ  X   )  0.5
0.5
μ
X
P(  X   )  1.0
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-18 / 82
Empirical Rules
What can we say about the distribution of values
around the mean? There are some general rules:
f(X)
σ
μ-1σ
μ ± 1σ encloses about
68% of X’s
σ
μ
μ+1σ
X
68.26%
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-19 / 82
The Empirical Rule
(continued)

μ ± 2σ covers about 95% of X’s

μ ± 3σ covers about 99.7% of X’s
2σ
3σ
2σ
μ
95.44%
Statistical Methods © 2004 Prentice-Hall, Inc.
x
3σ
μ
x
99.72%
Week 6-20 / 82
The Standardized Normal Table
 The Standardized Normal table in the
textbook gives the probability less than a
desired value for Z (i.e., from negative infinity
to Z)
.9772
Example:
P(Z < 2.00) = .9772
0
Statistical Methods © 2004 Prentice-Hall, Inc.
2.00
Z
Week 6-21 / 82
The Standardized Normal Table
(continued)
The column gives the value of
Z to the second decimal point
Z
The row shows
the value of Z
to the first
decimal point
0.01
0.02 …
0.0
0.1
.
.
.
2.0
2.0
P(Z < 2.00) = .9772
Statistical Methods © 2004 Prentice-Hall, Inc.
0.00
.9772
The value within the
table gives the
probability from Z =  
up to the desired Z
value
Week 6-22 / 82
General Procedure for
Finding Probabilities
To find P(a < X < b) when X is
distributed normally:

Draw the normal curve for the problem in
terms of X
 Translate X-values to Z-values
 Use the Standardized Normal Table
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-23 / 82
Finding Normal Probabilities
 Suppose X is normal with mean 8.0 and
standard deviation 5.0
 Find P(X < 8.6)
X
8.0
8.6
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-24 / 82
Finding Normal Probabilities
(continued)

Suppose X is normal with mean 8.0 and
standard deviation 5.0. Find P(X < 8.6)
X  μ 8.6  8.0
Z

 0.12
σ
5.0
μ=8
σ = 10
8 8.6
P(X < 8.6)
Statistical Methods © 2004 Prentice-Hall, Inc.
μ=0
σ=1
X
0 0.12
Z
P(Z < 0.12)
Week 6-25 / 82
Solution: Finding P(Z < 0.12)
Standardized Normal Probability
Table (Portion)
Z
.00
.01
P(X < 8.6)
= P(Z < 0.12)
.02
.5478
0.0 .5000 .5040 .5080
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871
Z
0.3 .6179 .6217 .6255
0.00
0.12
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-26 / 82
Upper Tail Probabilities
 Suppose X is normal with mean 8.0 and
standard deviation 5.0.
 Now Find P(X > 8.6)
X
8.0
8.6
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-27 / 82
Upper Tail Probabilities
(continued)
 Now Find P(X > 8.6)…
P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)
= 1.0 - .5478 = .4522
.5478
1.000
1.0 - .5478
= .4522
Z
0
0.12
Statistical Methods © 2004 Prentice-Hall, Inc.
Z
0
0.12
Week 6-28 / 82
Probability Between
Two Values

Suppose X is normal with mean 8.0 and
standard deviation 5.0. Find P(8 < X < 8.6)
Calculate Z-values:
X μ 8 8
Z

0
σ
5
X  μ 8.6  8
Z

 0.12
σ
5
8 8.6
X
0 0.12
Z
P(8 < X < 8.6)
= P(0 < Z < 0.12)
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-29 / 82
Solution: Finding P(0 < Z < 0.12)
Standardized Normal Probability
Table (Portion)
Z
.00
.01
.02
P(8 < X < 8.6)
= P(0 < Z < 0.12)
= P(Z < 0.12) – P(Z ≤ 0)
= .5478 - .5000 = .0478
0.0 .5000 .5040 .5080
.0478
.5000
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
Z
0.00
0.12
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-30 / 82
Probabilities in the Lower Tail
 Suppose X is normal with mean 8.0 and
standard deviation 5.0.
 Now Find P(7.4 < X < 8)
X
8.0
7.4
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-31 / 82
Probabilities in the Lower Tail
(continued)
Now Find P(7.4 < X < 8)…
P(7.4 < X < 8)
= P(-0.12 < Z < 0)
.0478
= P(Z < 0) – P(Z ≤ -0.12)
= .5000 - .4522 = .0478
The Normal distribution is
symmetric, so this probability
is the same as P(0 < Z < 0.12)
Statistical Methods © 2004 Prentice-Hall, Inc.
.4522
7.4 8.0
-0.12 0
X
Z
Week 6-32 / 82
Finding the X value for a
Known Probability
 Steps to find the X value for a known
probability:
1. Find the Z value for the known probability
2. Convert to X units using the formula:
X  μ  Zσ
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-33 / 82
Finding the X value for a
Known Probability
(continued)
Example:
 Suppose X is normal with mean 8.0 and
standard deviation 5.0.
 Now find the X value so that only 20% of all
values are below this X
.2000
?
?
Statistical Methods © 2004 Prentice-Hall, Inc.
8.0
0
X
Z
Week 6-34 / 82
Find the Z value for
20% in the Lower Tail
1. Find the Z value for the known probability
Standardized Normal Probability  20% area in the lower
Table (Portion)
tail is consistent with a
Z
-0.9
…
.03
.04
.05
… .1762 .1736 .1711
-0.8 … .2033 .2005 .1977
-0.7
Z value of -0.84
.2000
… .2327 .2296 .2266
?
8.0
-0.84 0
Statistical Methods © 2004 Prentice-Hall, Inc.
X
Z
Week 6-35 / 82
Finding the X value
2. Convert to X units using the formula:
X  μ  Zσ
 8.0  ( 0.84)5.0
 3.80
So 20% of the values from a distribution
with mean 8.0 and standard deviation
5.0 are less than 3.80
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-36 / 82
Assessing Normality
 Not all continuous random variables are
normally distributed
 It is important to evaluate how well the data set
is approximated by a normal distribution
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-37 / 82
Assessing Normality
(continued)
 Construct charts or graphs
 For small- or moderate-sized data sets, do stem-andleaf display and box-and-whisker plot look
symmetric?
 For large data sets, does the histogram or polygon
appear bell-shaped?
 Compute descriptive summary measures
 Do the mean, median and mode have similar values?
 Is the interquartile range approximately 1.33 σ?
 Is the range approximately 6 σ?
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-38 / 82
Assessing Normality
(continued)
 Observe the distribution of the data set
 Do approximately 2/3 of the observations lie within
mean  1 standard deviation?
 Do approximately 80% of the observations lie within
mean  1.28 standard deviations?
 Do approximately 95% of the observations lie within
mean  2 standard deviations?
 Evaluate normal probability plot
 Is the normal probability plot approximately linear
with positive slope?
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-39 / 82
The Normal Probability Plot
 Normal probability plot
 Arrange data into ordered array
 Find corresponding standardized normal quantile
values
 Plot the pairs of points with observed data values on
the vertical axis and the standardized normal quantile
values on the horizontal axis
 Evaluate the plot for evidence of linearity
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-40 / 82
The Normal Probability Plot
(continued)
A normal probability plot for data
from a normal distribution will be
approximately linear:
X
90
60
30
-2
Statistical Methods © 2004 Prentice-Hall, Inc.
-1
0
1
2
Z
Week 6-41 / 82
Normal Probability Plot
(continued)
Left-Skewed
Right-Skewed
X 90
X 90
60
60
30
30
-2 -1 0
1
2 Z
-2 -1 0
1
2 Z
Rectangular
Nonlinear plots
indicate a deviation
from normality
X 90
60
30
-2 -1 0
Statistical Methods © 2004 Prentice-Hall, Inc.
1
2 Z
Week 6-42 / 82
The Uniform Distribution
Probability
Distributions
Continuous
Probability
Distributions
Normal
Uniform
Exponential
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-43 / 82
The Uniform Distribution
 The uniform distribution is a
probability distribution that has equal
probabilities for all possible
outcomes of the random variable
 Also called a rectangular distribution
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-44 / 82
The Uniform Distribution
(continued)
The Continuous Uniform Distribution:
1
ba
if a  X  b
0
otherwise
f(X) =
where
f(X) = value of the density function at any X value
a = minimum value of X
b = maximum value of X
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-45 / 82
Properties of the
Uniform Distribution
 The mean of a uniform distribution is
ab
μ
2
 The standard deviation is
σ
Statistical Methods © 2004 Prentice-Hall, Inc.
(b - a)2
12
Week 6-46 / 82
Uniform Distribution Example
Example: Uniform probability distribution
over the range 2 ≤ X ≤ 6:
1
f(X) = 6 - 2 = .25 for 2 ≤ X ≤ 6
f(X)
μ
.25
2
Statistical Methods © 2004 Prentice-Hall, Inc.
6
X
σ
ab 26

4
2
2
(b - a)2

12
(6 - 2)2
 1.1547
12
Week 6-47 / 82
The Exponential Distribution
Probability
Distributions
Continuous
Probability
Distributions
Normal
Uniform
Exponential
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-48 / 82
The Exponential Distribution
 Used to model the length of time between two
occurrences of an event (the time between
arrivals)
 Examples:
 Time between trucks arriving at an unloading dock
 Time between transactions at an ATM Machine
 Time between phone calls to the main operator
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-49 / 82
The Exponential Distribution
 Defined by a single parameter, its mean λ
(lambda)
 The probability that an arrival time is less than
some specified time X is
P(arrival time  X)  1 e
where
 λX
e = mathematical constant approximated by 2.71828
λ = the population mean number of arrivals per unit
X = any value of the continuous variable where 0 < X <
Statistical Methods © 2004 Prentice-Hall, Inc.

Week 6-50 / 82
Exponential Distribution
Example
Example: Customers arrive at the service counter at
the rate of 15 per hour. What is the probability that the
arrival time between consecutive customers is less
than three minutes?

The mean number of arrivals per hour is 15, so λ = 15

Three minutes is .05 hours

P(arrival time < .05) = 1 – e-λX = 1 – e-(15)(.05) = .5276

So there is a 52.76% probability that the arrival time
between successive customers is less than three
minutes
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-51 / 82
Sampling Distributions
Sampling
Distributions
Sampling
Distributions
of the
Mean
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Sampling
Distributions
of the
Proportion
Week 6-52 / 82
Sampling Distributions
 A sampling distribution is a
distribution of all of the possible
values of a statistic for a given size
sample selected from a population
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-53 / 82
Developing a
Sampling Distribution
 Assume there is a population …
 Population size N=4
A
B
C
D
 Random variable, X,
is age of individuals
 Values of X: 18, 20,
22, 24 (years)
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-54 / 82
Developing a
Sampling Distribution
(continued)
Summary Measures for the Population Distribution:
X

μ
P(x)
i
N
.3
18  20  22  24

 21
4
σ
 (X  μ)
0
2
i
N
.2
.1
 2.236
18
20
22
24
A
B
C
D
x
Uniform Distribution
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-55 / 82
Developing a
Sampling Distribution
(continued)
Now consider all possible samples of size n=2
1st
Obs
2nd Observation
18
20
22
24
18 18,18 18,20 18,22 18,24
16 Sample
Means
20 20,18 20,20 20,22 20,24
1st 2nd Observation
Obs 18 20 22 24
22 22,18 22,20 22,22 22,24
18 18 19 20 21
24 24,18 24,20 24,22 24,24
20 19 20 21 22
16 possible samples
(sampling with
replacement)
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22 20 21 22 23
24 21 22 23 24
Week 6-56 / 82
Developing a
Sampling Distribution
(continued)
Sampling Distribution of All Sample Means
Sample Means
Distribution
16 Sample Means
1st 2nd Observation
Obs 18 20 22 24
18 18 19 20 21
20 19 20 21 22
22 20 21 22 23
24 21 22 23 24
_
P(X)
.3
.2
.1
0
18 19
20 21 22 23
24
_
X
(no longer uniform)
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-57 / 82
Developing a
Sampling Distribution
(continued)
Summary Measures of this Sampling Distribution:
μX
X


i
N
σX 

18  19  21    24

 21
16
2
(
X

μ
)
i

X
N
(18 - 21)2  (19 - 21)2    (24 - 21)2
 1.58
16
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-58 / 82
Comparing the Population with its
Sampling Distribution
Population
N=4
μ  21
σ  2.236
Sample Means Distribution
n=2
μX  21
σ X  1.58
_
P(X)
.3
P(X)
.3
.2
.2
.1
.1
0
18
20
22
24
A
B
C
D
Statistical Methods © 2004 Prentice-Hall, Inc.
X
0
18 19
20 21 22 23
24
_
X
Week 6-59 / 82
Sampling Distributions
of the Mean
Sampling
Distributions
Sampling
Distributions
of the
Mean
Statistical Methods © 2004 Prentice-Hall, Inc.
Sampling
Distributions
of the
Proportion
Week 6-60 / 82
Standard Error of the Mean
 Different samples of the same size from the same
population will yield different sample means
 A measure of the variability in the mean from sample to
sample is given by the Standard Error of the Mean:
σ
σX 
n
 Note that the standard error of the mean decreases as
the sample size increases
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-61 / 82
If the Population is Normal
 If a population is normal with mean μ and
standard deviation σ, the sampling distribution
of X is also normally distributed with
μX  μ
and
σ
σX 
n
(This assumes that sampling is with replacement or
sampling is without replacement from an infinite population)
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-62 / 82
Z-value for Sampling Distribution
of the Mean
 Z-value for the sampling distribution of X :
Z
where:
( X  μX )
σX
( X  μ)

σ
n
X = sample mean
μ = population mean
σ = population standard deviation
n = sample size
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-63 / 82
Finite Population Correction
 Apply the Finite Population Correction if:
 the sample is large relative to the population
(n is greater than 5% of N)
and…
 Sampling is without replacement
Then
Statistical Methods © 2004 Prentice-Hall, Inc.
( X  μ)
Z
σ Nn
n N 1
Week 6-64 / 82
Sampling Distribution Properties
μx  μ

(i.e.
x is unbiased )
Statistical Methods © 2004 Prentice-Hall, Inc.
Normal Population
Distribution
μ
x
μx
x
Normal Sampling
Distribution
(has the same mean)
Week 6-65 / 82
Sampling Distribution Properties
(continued)
 For sampling with replacement:
As n increases,
Larger
sample size
σ x decreases
Smaller
sample size
μ
Statistical Methods © 2004 Prentice-Hall, Inc.
x
Week 6-66 / 82
If the Population is not Normal
 We can apply the Central Limit Theorem:
 Even if the population is not normal,
 …sample means from the population will be
approximately normal as long as the sample size is
large enough.
Properties of the sampling distribution:
μx  μ
Statistical Methods © 2004 Prentice-Hall, Inc.
and
σ
σx 
n
Week 6-67 / 82
Central Limit Theorem
As the
sample
size gets
large
enough…
n↑
the sampling
distribution
becomes
almost normal
regardless of
shape of
population
x
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-68 / 82
If the Population is not Normal
(continued)
Population Distribution
Sampling distribution
properties:
Central Tendency
μx  μ
σ
σx 
n
Variation
μ
Sampling Distribution
(becomes normal as n increases)
Larger
sample
size
Smaller
sample size
(Sampling with
replacement)
Statistical Methods © 2004 Prentice-Hall, Inc.
x
μx
x
Week 6-69 / 82
How Large is Large Enough?
 For most distributions, n > 30 will give a
sampling distribution that is nearly normal
 For fairly symmetric distributions, n > 15
 For normal population distributions, the
sampling distribution of the mean is always
normally distributed
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Example
 Suppose a population has mean μ = 8 and
standard deviation σ = 3. Suppose a random
sample of size n = 36 is selected.
 What is the probability that the sample mean is
between 7.8 and 8.2?
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Week 6-71 / 82
Example
(continued)
Solution:
 Even if the population is not normally
distributed, the central limit theorem can be
used (n > 30)
 … so the sampling distribution of
approximately normal
x
is
 … with mean μx = 8
σ
3
 …and standard deviation σ x  n  36  0.5
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-72 / 82
Example
(continued)
Solution (continued):


μ
μ
 7.8 - 8
8.2 - 8 
X
P(7.8  μ X  8.2)  P



3
σ
3


36
n
36


 P(-0.5  Z  0.5)  0.3830
Population
Distribution
???
?
??
?
?
?
?
?
μ8
Sampling
Distribution
Standard Normal
Distribution
Sample
.1915
+.1915
Standardize
?
X
Statistical Methods © 2004 Prentice-Hall, Inc.
7.8
μX  8
8.2
x
-0.5
μz  0
0.5
Z
Week 6-73 / 82
Sampling Distributions
of the Proportion
Sampling
Distributions
Sampling
Distributions
of the
Mean
Statistical Methods © 2004 Prentice-Hall, Inc.
Sampling
Distributions
of the
Proportion
Week 6-74 / 82
Population Proportions, p
p = the proportion of the population having
some characteristic
 Sample proportion ( ps ) provides an estimate
of p:
ps 
X
number of items in the sample having the characteri stic of interest

n
sample size
 0 ≤ ps ≤ 1
 ps has a binomial distribution
(assuming sampling with replacement from a finite population or
without replacement from an infinite population)
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Week 6-75 / 82
Sampling Distribution of p
 Approximated by a
normal distribution if:

and
0
n(1  p)  5
μps  p
Sampling Distribution
.3
.2
.1
0
np  5
where
P( ps)
and
σps
.2
.4
.6
8
1
ps
p(1  p)

n
(where p = population proportion)
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-76 / 82
Z-Value for Proportions
Standardize ps to a Z value with the formula:
ps  p
Z

σ ps
 If sampling is without replacement
and n is greater than 5% of the
population size, then σ p must use
the finite population correction
factor:
Statistical Methods © 2004 Prentice-Hall, Inc.
ps  p
p(1  p)
n
σ ps 
p(1  p) N  n
n
N 1
Week 6-77 / 82
Example
 If the true proportion of voters who support
Proposition A is p = .4, what is the probability
that a sample of size 200 yields a sample
proportion between .40 and .45?
 i.e.: if p = .4 and n = 200, what is
P(.40 ≤ ps ≤ .45) ?
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Example
(continued)

Find σ ps:
if p = .4 and n = 200, what is
P(.40 ≤ ps ≤ .45) ?
σ ps
p(1  p)
.4(1  .4)


 .03464
n
200
Convert to
.45  .40 
 .40  .40
P(.40  p s  .45)  P
Z

standard
.03464 
 .03464
normal:
 P(0  Z  1.44)
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Week 6-79 / 82
Example
(continued)

if p = .4 and n = 200, what is
P(.40 ≤ ps ≤ .45) ?
Use standard normal table:
P(0 ≤ Z ≤ 1.44) = .4251
Standardized
Normal Distribution
Sampling Distribution
.4251
Standardize
.40
Statistical Methods © 2004 Prentice-Hall, Inc.
.45
ps
0
1.44
Z
Week 6-80 / 82
Chapter Summary
 Presented key continuous distributions
 normal, uniform, exponential
 Found probabilities using formulas and tables
 Recognized when to apply different distributions
 Applied distributions to decision problems
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Week 6-81 / 82
Chapter Summary
(continued)
 Introduced sampling distributions
 Described the sampling distribution of the mean
 For normal populations
 Using the Central Limit Theorem
 Described the sampling distribution of a
proportion
 Calculated probabilities using sampling
distributions
Statistical Methods © 2004 Prentice-Hall, Inc.
Week 6-82 / 82