Chap06 normal distributions & continous

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

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

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 p
s
Chap 6-3
Prentice-Hall, Inc.


Probability Distributions
Probability
Distributions
Ch. 5

Discrete
Probability
Distributions

Continuous
Probability
Distributions

Binomial

Normal

Poisson

Ch. 6

Uniform

Statistics for Hypergeometric
Managers Using
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Exponential
Chap 6-4

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


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Chap 6-5

The Normal Distribution
Probability
Distributions
Continuous
Probability
Distributions
Normal
Uniform
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Exponential
Chap 6-6

‘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:
+ ∞ to − ∞
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σ
μ
Mean
= Median
= Mode
Chap 6-7

X

Many Normal Distributions

By varying the parameters μ and σ, we obtain
different normal distributions
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Chap 6-8

Shape
f(X)

Changing μ shifts the
distribution left or right.

σ

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μ

Changing σ increases
or decreases the
spread.
X
Chap 6-9

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

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

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Chap 6-11

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
Chap 6-12
Prentice-Hall, Inc.

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

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Chap 6-13

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
Chap 6-14
Prentice-Hall, Inc.

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.


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Chap 6-15

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
Chap 6-16
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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
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b

X
Chap 6-17

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( −∞
Microsoft Excel, 4e © 2004 <
Prentice-Hall, Inc.

P(μ < X < ∞) = 0.5

0.5

μ

X

X < ∞ ) = 1.0
Chap 6-18

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
68.26%
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X
Chap 6-19

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%
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x

3σ
μ

x

99.72%

Chap 6-20

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
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0

2.00

Z
Chap 6-21

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
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.9772

The value within the
table gives the
probability from Z = − ∞
up to the desired Z
value
Chap 6-22

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





Suppose X is normal with mean 8.0 and
standard deviation 5.0
Find P(X < 8.6)

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X

8.0
8.6

Chap 6-24

(continued)


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

P(X < 8.6)
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0 0.12

Z

P(Z < 0.12)
Chap 6-25

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
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Z

0.00
0.12

Chap 6-26

Upper Tail Probabilities




standard deviation 5.0.
Now Find P(X > 8.6)

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X

8.0
8.6

Chap 6-27

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

0
0.12
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1.0 - .5478
= .4522

Z

0
0.12

Chap 6-28

Probability Between
Two Values


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
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8 8.6

X

0 0.12

Z

P(8 < X < 8.6)
= P(0 < Z < 0.12)
Chap 6-29

Solution: Finding P(0 < Z < 0.12)
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
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0.00
0.12

Z

Chap 6-30

Probabilities in the Lower Tail




Now Find P(7.4 < X < 8)

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7.4

8.0

X

Chap 6-31

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)
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.4522

X
Z

7.4 8.0
-0.12 0

Chap 6-32

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
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Chap 6-33

Finding the X value for a
Known Probability
(continued)

Example:
 Suppose X is normal with mean 8.0 and
 Now find the X value so that only 20% of all
values are below this X
.2000
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?
?

8.0
0

X
Z

Chap 6-34

Find the Z value for
20% in the Lower Tail
1. Find the Z value for the known 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
Prentice-Hall, Inc.

.2000

?
8.0
-0.84 0

Chap 6-35

X
Z

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
Chap 6-36
Prentice-Hall, Inc.

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

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Chap 6-37

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?
Chap 6-38
Prentice-Hall, Inc.


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?


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Chap 6-39

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
Chap 6-40
Prentice-Hall, Inc.


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
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1

2

Z
Chap 6-41

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
X 90
60

30
Microsoft Excel, -1 0© 1 2 Z
4e 2004
-2
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Nonlinear plots indicate
a deviation from
normality
Chap 6-42

The Uniform Distribution
Probability
Distributions
Continuous
Probability
Distributions
Normal
Uniform
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Exponential
Chap 6-43



The uniform distribution is a
probability distribution that has equal
probabilities for all possible
outcomes of the random variable



Also called a rectangular distribution

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Chap 6-44

(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
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Statistics
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Chap 6-45

Properties of the
Uniform Distribution


The mean of a uniform distribution is
a +b
μ=
2



The standard deviation is
σ=

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(b - a)2
12
Chap 6-46

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
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

The Exponential Distribution
Probability
Distributions
Continuous
Probability
Distributions
Normal
Uniform
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Exponential
Chap 6-48



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

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Chap 6-49





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
Chap 6-50
Prentice-Hall, Inc.

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


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Chap 6-51

Sampling Distributions
Sampling
Distributions

Sampling
Distributions
of the
Mean

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Sampling
Distributions
of the
Proportion
Chap 6-52



A sampling distribution is a
distribution of all of the possible
values of a statistic for a given size
sample selected from a population

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Chap 6-53

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

Developing a
(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

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18

20

22

24

A

B

C

D

Uniform Distribution
Chap 6-55

x

Developing a
(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
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22 20 21 22 23
24 21 22 23 24
Chap 6-56

Developing a
(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
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_

P(X)
.3
.2
.1
0

18 19

20 21 22 23

24

(no longerChap 6-57
uniform)

_

X

Developing a
(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
16
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Chap 6-58

Comparing the Population with its
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

of the Mean
Sampling
Distributions

Sampling
Distributions
of the
Mean

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Sampling
Distributions
of the
Proportion
Chap 6-60

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
Chap 6-61
Prentice-Hall, Inc.


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)
Chap 6-62
Prentice-Hall, Inc.

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
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Chap 6-63

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
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Chap 6-64

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

Sampling Distribution Properties
(continued)


For sampling with replacement:
As n increases,

Larger
sample size

σ x decreases
Smaller
sample size

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x

μ
Chap 6-66

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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σ
σx =
n
Chap 6-67

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

If the Population is not Normal
(continued)

Sampling distribution
properties:

Population Distribution

Central Tendency

μx = μ
Variation

σ
σx =
n

x

μ
(becomes normal as n increases)
Larger
sample
size

Smaller
sample size

(Sampling with

replacement)
Prentice-Hall, Inc.

μx

Chap 6-69

x

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

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?

Prentice-Hall, Inc.

Chap 6-71

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 =
n
36
Chap 6-72
Prentice-Hall, Inc.

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

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

of the Proportion
Sampling
Distributions

Sampling
Distributions
of the
Mean

Prentice-Hall, Inc.

Sampling
Distributions
of the
Proportion
Chap 6-74

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
Chap 6-75
Prentice-Hall, Inc.

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


.2

.4

.6

p(1 − p)
=
n

8

1

Microsoft Excel, 4e ©(where p = population proportion)
2004
Chap 6-76
Prentice-Hall, Inc.

ps

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:

ps − p
p(1 − p)
n



Prentice-Hall, Inc.

σ ps =

p(1 − p) N − n
n
N −1
Chap 6-77

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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Chap 6-78

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)
Chap 6-79
Prentice-Hall, Inc.

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


.4251
Standardize

.40
Statistics for Managers .45 ps
Using
Prentice-Hall, Inc.

0

1.44

Z

Chap 6-80

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

Prentice-Hall, Inc.

Chap 6-81

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

Prentice-Hall, Inc.

Chap 6-82

Chap06 normal distributions & continous

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