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Chap06 normal distributions & continous
- 1. Statistics for Managers
Using Microsoft® Excel
4th Edition
Chapter 6
The Normal Distribution and
Other Continuous Distributions
Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc.
Chap 6-1
- 2. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-2
- 3. 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
_
Statistics for Managers Using
Apply sampling distributions for both X and p
s
Microsoft Excel, 4e © 2004
Chap 6-3
Prentice-Hall, Inc.
- 5. 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
Statisticsaccurately. Using
for Managers
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-5
- 7. The Normal Distribution
‘Bell Shaped’
Symmetrical
Mean, Median and Mode
are Equal
f(X)
Location is determined by the
mean, μ
Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
Statistics for Managers Using
+ ∞ to − ∞
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
σ
μ
Mean
= Median
= Mode
Chap 6-7
X
- 8. Many Normal Distributions
By varying the parameters μ and σ, we obtain
different normal distributions
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-8
- 9. The Normal Distribution
Shape
f(X)
Changing μ shifts the
distribution left or right.
σ
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
μ
Changing σ increases
or decreases the
spread.
X
Chap 6-9
- 10. The Normal Probability
Density Function
The formula for the normal probability density
function is
f(X) =
Where
1
−(1/2)[(X −μ)/σ] 2
e
2πσ
e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
μ = the population mean
Statistics forσManagers Using
= the population standard deviation
Microsoft Excel, 4e © 2004 continuous variable
X = any value of the
Prentice-Hall, Inc.
Chap 6-10
- 11. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-11
- 12. 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 =
Statistics for Managers Using 0 and standard deviation = 1
Microsoft Excel, 4e © 2004
Chap 6-12
Prentice-Hall, Inc.
- 13. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-13
- 14. 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,
Statistics for Managers Using
values below the mean have negative Z-values
Microsoft Excel, 4e © 2004
Chap 6-14
Prentice-Hall, Inc.
- 15. 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 Managers Using
Statistics formean of 100.
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-15
- 16. 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
Statistics for Managers or in standardized units (Z)
original units (X) Using
Microsoft Excel, 4e © 2004
Chap 6-16
Prentice-Hall, Inc.
- 17. 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
Statistics for Managers Using
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b
X
Chap 6-17
- 18. 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
Statistics for Managers Using
P( −∞
Microsoft Excel, 4e © 2004 <
Prentice-Hall, Inc.
P(μ < X < ∞) = 0.5
0.5
μ
X
X < ∞ ) = 1.0
Chap 6-18
- 19. Empirical Rules
What can we say about the distribution of values
around the mean? There are some general rules:
f(X)
σ
σ
μ ± 1σ encloses about
68% of X’s
μ-1σ μ μ+1σ
Statistics for Managers Using
68.26%
Microsoft Excel, 4e © 2004
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X
Chap 6-19
- 20. 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%
Statistics for Managers Using
Microsoft Excel, 4e © 2004
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x
3σ
μ
x
99.72%
Chap 6-20
- 21. The Standardized Normal Table
The Standardized Normal table in the
textbook (Appendix table E.2) gives the
probability less than a desired value for Z
(i.e., from negative infinity to Z)
.9772
Example:
P(Z < 2.00) = .9772
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
0
2.00
Z
Chap 6-21
- 22. 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
0.00
0.01
0.02 …
0.0
0.1
.
.
.
2.0
P(Z Managers2.0
Statistics for < 2.00) = .9772
Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
.9772
The value within the
table gives the
probability from Z = − ∞
up to the desired Z
value
Chap 6-22
- 23. 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
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Chap 6-23
- 24. Finding Normal Probabilities
Suppose X is normal with mean 8.0 and
standard deviation 5.0
Find P(X < 8.6)
Statistics for Managers Using
Microsoft Excel, 4e © 2004
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X
8.0
8.6
Chap 6-24
- 25. Finding Normal Probabilities
(continued)
Suppose X is normal with mean 8.0 and
standard deviation 5.0. Find P(X < 8.6)
Z=
X − μ 8.6 − 8.0
=
= 0.12
σ
5.0
μ=8
σ = 10
8 8.6
μ=0
σ=1
X
Statistics for Managers Using
P(X < 8.6)
Microsoft Excel, 4e © 2004
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0 0.12
Z
P(Z < 0.12)
Chap 6-25
- 26. 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
0.3 .6179 .6217 .6255
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Z
0.00
0.12
Chap 6-26
- 27. Upper Tail Probabilities
Suppose X is normal with mean 8.0 and
standard deviation 5.0.
Now Find P(X > 8.6)
Statistics for Managers Using
Microsoft Excel, 4e © 2004
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X
8.0
8.6
Chap 6-27
- 28. 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
Z
Statistics for Managers Using
0
Microsoft Excel, 4e © 2004
0.12
Prentice-Hall, Inc.
1.0 - .5478
= .4522
Z
0
0.12
Chap 6-28
- 29. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
8 8.6
X
0 0.12
Z
P(8 < X < 8.6)
= P(0 < Z < 0.12)
Chap 6-29
- 30. Solution: Finding P(0 < Z < 0.12)
Standardized Normal Probability
Table (Portion)
Z
.00
.01
.02
0.0 .5000 .5040 .5080
0.1 .5398 .5438 .5478
P(8 < X < 8.6)
= P(0 < Z < 0.12)
= P(Z < 0.12) – P(Z ≤ 0)
= .5478 - .5000 = .0478
.0478
.5000
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
0.00
0.12
Z
Chap 6-30
- 31. 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)
Statistics for Managers Using
Microsoft Excel, 4e © 2004
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7.4
8.0
X
Chap 6-31
- 32. 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
Statistics for Managers Using
is the same as P(0 < Z < 0.12)
Microsoft Excel, 4e © 2004
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.4522
X
Z
7.4 8.0
-0.12 0
Chap 6-32
- 33. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
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Chap 6-33
- 34. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
?
?
8.0
0
X
Z
Chap 6-34
- 35. Find the Z value for
20% in the Lower Tail
1. Find the Z value for the known probability
Standardized Normal Probability
Table (Portion)
Z
-0.9
…
.03
.04
.05
20% area in the lower
tail is consistent with a
Z value of -0.84
… .1762 .1736 .1711
-0.8 … .2033 .2005 .1977
-0.7 … .2327 .2296 .2266
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
.2000
?
8.0
-0.84 0
Chap 6-35
X
Z
- 36. 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
Statistics for Managers Using 3.80
Microsoft Excel, 4e © 2004
Chap 6-36
Prentice-Hall, Inc.
- 37. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
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Chap 6-37
- 38. 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?
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Chap 6-38
Prentice-Hall, Inc.
- 39. 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 Using
Statistics for Managers slope?
Microsoft Excel, 4e © 2004
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Chap 6-39
- 40. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Chap 6-40
Prentice-Hall, Inc.
- 41. The Normal Probability Plot
(continued)
A normal probability plot for data
from a normal distribution will be
approximately linear:
X
90
60
30
-2 -1
Statistics for Managers Using 0
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
1
2
Z
Chap 6-41
- 44. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-44
- 45. 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
for Managers Using X
Statistics
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-45
- 46. Properties of the
Uniform Distribution
The mean of a uniform distribution is
a +b
μ=
2
The standard deviation is
σ=
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
(b - a)2
12
Chap 6-46
- 47. 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
Statistics for Managers Using6
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
X
σ=
a +b 2 +6
=
=4
2
2
(b - a)2
=
12
(6 - 2)2
= 1.1547
12
Chap 6-47
- 49. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
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Chap 6-49
- 50. 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
Statistics for Managers Using continuous variable where 0 < X < ∞
X = any value of the
Microsoft Excel, 4e © 2004
Chap 6-50
Prentice-Hall, Inc.
- 51. 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
Statistics for Managers Using
minutes
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-51
- 53. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
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Chap 6-53
- 54. Developing a
Sampling Distribution
Assume there is a population …
Population size N=4
B
C
D
Random variable, X,
is age of individuals
A
Values of X: 18, 20,
22, 24 (years)
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Chap 6-54
- 55. 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)
i
N
.2
.1
0
2
= 2.236
Statistics for Managers Using
Microsoft Excel, 4e © 2004
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18
20
22
24
A
B
C
D
Uniform Distribution
Chap 6-55
x
- 56. 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 Using
for Managerswith
replacement)
Statistics
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
22 20 21 22 23
24 21 22 23 24
Chap 6-56
- 57. 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
Statistics for Managers Using
24 21 22 23 24
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
_
P(X)
.3
.2
.1
0
18 19
20 21 22 23
24
(no longerChap 6-57
uniform)
_
X
- 58. Developing a
Sampling Distribution
(continued)
Summary Measures of this Sampling Distribution:
μX
∑X
=
σX =
N
i
18 + 19 + 21 + + 24
=
= 21
16
( X i − μ X )2
∑
N
(18 - 21)2 + (19 - 21)2 + + (24 - 21)2
=
= 1.58
Statistics for Managers Using
16
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Chap 6-58
- 59. 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
Statistics for Managers Using X
18
20
22
24
A
B
C
Microsoft Excel, 4e © 2004D
Prentice-Hall, Inc.
0
18 19
20 21 22 23
24
Chap 6-59
_
X
- 60. Sampling Distributions
of the Mean
Sampling
Distributions
Sampling
Distributions
of the
Mean
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Sampling
Distributions
of the
Proportion
Chap 6-60
- 61. 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
Statistics forsample sizeUsing
the Managers increases
Microsoft Excel, 4e © 2004
Chap 6-61
Prentice-Hall, Inc.
- 62. 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
Statistics (This assumes that sampling is with replacement or
for Managers Using
sampling is without replacement from an infinite population)
Microsoft Excel, 4e © 2004
Chap 6-62
Prentice-Hall, Inc.
- 63. 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 = Using
Statistics for Managers sample size
Microsoft Excel, 4e © 2004
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Chap 6-63
- 64. 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
( X − μ)
Then
Z=
σ N −n
Statistics for Managers Using n N − 1
Microsoft Excel, 4e © 2004
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Chap 6-64
- 65. Sampling Distribution Properties
Normal Population
Distribution
μx = μ
(i.e.
x is unbiased )
μ
x
μx
x
Normal Sampling
Distribution
(has the same mean)
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Chap 6-65
- 67. 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:
μ x = μUsing
and
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Microsoft Excel, 4e © 2004
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σ
σx =
n
Chap 6-67
- 68. Central Limit Theorem
As the
sample
size gets
large
enough…
n↑
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the sampling
distribution
becomes
almost normal
regardless of
shape of
population
Chap 6-68
x
- 69. If the Population is not Normal
(continued)
Sampling distribution
properties:
Population Distribution
Central Tendency
μx = μ
Variation
σ
σx =
n
x
μ
Sampling Distribution
(becomes normal as n increases)
Larger
sample
size
Smaller
sample size
(Sampling with
Statistics for Managers Using
replacement)
Microsoft Excel, 4e © 2004
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μx
Chap 6-69
x
- 70. 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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Chap 6-70
- 71. 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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Chap 6-71
- 72. 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
… with mean μx = 8
x
is
σ
3
=
= 0.5
…and standard deviation σ x =
Statistics for Managers Using
n
36
Microsoft Excel, 4e © 2004
Chap 6-72
Prentice-Hall, Inc.
- 73. Example
(continued)
Solution (continued):
μX -μ
7.8 - 8
8.2 - 8
P(7.8 < μ X < 8.2) = P
<
<
3
σ
3
36
n
36
= P(-0.5 < Z < 0.5) = 0.3830
Population
Distribution
???
?
??
?
??
?
?
Sampling
Distribution
?
Sample
Statistics for Managers Using
7.8
8.2
μ=8
X © 2004
μX = 8
Microsoft Excel, 4e
Prentice-Hall, Inc.
Standard Normal
Distribution
.1915
+.1915
Standardize
x
-0.5
μz = 0
0.5
Chap 6-73
Z
- 74. Sampling Distributions
of the Proportion
Sampling
Distributions
Sampling
Distributions
of the
Mean
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Sampling
Distributions
of the
Proportion
Chap 6-74
- 75. Population Proportions, p
p = the proportion of the population having
some characteristic
ps =
Sample proportion ( ps ) provides an estimate
of p:
X
number of items in the sample having the characteristic of interest
=
n
sample size
0 ≤ ps ≤ 1
ps has a binomial distribution
(assuming sampling with replacement from a finite
Statistics for Managers Usingan infinite population) population or
without replacement from
Microsoft Excel, 4e © 2004
Chap 6-75
Prentice-Hall, Inc.
- 76. Sampling Distribution of p
Approximated by a
normal distribution if:
.3
.2
.1
0
np ≥ 5
and
n(1− p) ≥ 5
where
μps = p
P( ps)
and
0
σps
Sampling Distribution
.2
.4
.6
p(1 − p)
=
n
8
1
Statistics for Managers Using
Microsoft Excel, 4e ©(where p = population proportion)
2004
Chap 6-76
Prentice-Hall, Inc.
ps
- 77. 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
Statistics for Managers Using
factor:
ps − p
p(1 − p)
n
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
σ ps =
p(1 − p) N − n
n
N −1
Chap 6-77
- 78. 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) ?
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-78
- 79. 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:
Statistics for Managers Using
= P(0 ≤ Z ≤ 1.44)
Microsoft Excel, 4e © 2004
Chap 6-79
Prentice-Hall, Inc.
- 80. 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
Statistics for Managers .45 ps
Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
0
1.44
Z
Chap 6-80
- 81. 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 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
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc.
Chap 6-82