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Chapter 6
6-1
The Normal Distribution and
Other Continuous Distributions
Learning Objectives
Probability Distributions
Probability
Distributions
In this chapter, you learn:
The Normal Distribution
Ch. 5
The Standardized Normal Distribution
Evaluating the Normality Assumption
The Uniform Distribution
The Exponential Distribution
Discrete
Probability
Distributions
Continuous
Probability
Distributions
Binomial
Normal
Poisson
Uniform
Hypergeometric
Ch. 6
Exponential
1
2
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
Values from interval of numbers
Absence of gaps
These can potentially take on any value, depending only on the
ability to measure accurately.
Continuous Probability Distribution
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
σ
X
µ
The random variable has an infinite
theoretical range: + ∞ to − ∞
Interquartile Range equals 1.33 σ
Online normal distributions – uniform to
normal
Distribution of continuous random variable
Mean
= Median
= Mode
Most Important Continuous Probability Distribution
The normal distribution
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© 2006 Prentice Hall, Inc.
Chapter 6
6-2
The (Standard) Normal Probability
Density Function
Many Normal Distributions
There are an Infinite Number of Normal Distributions
The formula for the normal
probability density function is
f(X) =
Where
The formula for the standardized
normal probability density function
is
2
1
e −(1/2)[(X −µ)/σ]
2πσ
f(Z) =
1
2π
e − (1/2)Z
2
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
By varying the parameters µ and σ, we obtain different
normal distributions
X = any value of the continuous variable
Z = any value of the standardized normal distribution
Changing µ shifts the distribution left or right.
Changing σ increases or decreases the spread.
5
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
The Standardized Normal
Distribution
Example
When X is normally distributed with a mean µ and a standard deviation σ, Z =
(X - µ)/σ follows a standardized (normalized) normal distribution with a mean
0 and a standard deviation 1 (always).
Values above the mean have positive Z-values, values below the mean have
negative Z-values
Z=
µ
µZ = 0
Basic Business Statistics, 10/e
X
If X is distributed normally with mean of 100
and standard deviation of 50, the Z value for
X = 200 is
Z=
σ
σZ =1
X ~ N ( µ ,σ 2 )
Z ~ N ( 0,1)
X −µ
σ
f(Z)
6
X − µ 200 − 100
=
= 2.0
σ
50
This says that X = 200 is two standard
deviations (2 increments of 50 units) above
the mean of 100.
Z
7
8
© 2006 Prentice Hall, Inc.
Chapter 6
6-3
Finding Normal Probabilities
Comparing X and Z units
Probability is the
Probability is measured
area under the
curve! under the curve
f(X)
100
0
200
2.0
X
Z
by the area
P (a ≤ X ≤ b)
= P (a < X < b)
(µ = 100, σ = 50)
(µ = 0, σ = 1)
(Note that the
probability of any
individual value is zero)
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)
a
b
X
9
10
Probability as
Area Under the Curve
Empirical Rules
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
What can we say about the distribution of values
around the mean? There are some general rules:
f(X)
P(µ < X < ∞ ) = 0.5
σ
0.5
σ
0.5
µ
X
µ-1σ
P( −∞ < X < ∞) = 1.0
µ µ+1σ
68.26%
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µ ± 1σ encloses about
68% of X’s
X
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© 2006 Prentice Hall, Inc.
Chapter 6
6-4
The Standardized Normal Table
The Empirical Rule
(continued)
µ ± 2σ covers about 95% of X’s
µ ± 3σ covers about 99.7% of X’s
2σ
3σ
2σ
x
µ
3σ
x
µ
95.44%
The Cumulative 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)
=NORMSDIST(2)
0.9772
Example:
P(Z < 2.00) = 0.9772
99.73%
0
2.00
Z
13
14
General Procedure for
Finding Probabilities
The Standardized Normal Table
(continued)
Z
The column gives the value of
Z to the second decimal point
To find P(a < X < b) when X is distributed normally:
0.00
Translate X-values to Z-values
0.01
Draw the normal curve for the problem in terms of X
0.02 …
Use the Standardized Normal Table
The row shows
the value of Z
to the first
decimal point
0.0
Suppose X is normal with mean 8.0 and standard deviation 5.0
0.1
Find P(X < 8.6)
.
.
.
2.0
.9772
2.0
P(Z < 2.00) = 0.9772
The value within the
table gives the
probability from Z = − ∞
up to the desired Z
value
X
8.0
8.6
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© 2006 Prentice Hall, Inc.
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Chapter 6
6-5
Finding Normal Probabilities
Solution: Finding P(Z < 0.12)
(continued)
Suppose X is normal with mean 8.0 and standard deviation 5.0. Find
P(X < 8.6)
=NORMDIST(8.6,8,10,TRUE)
=NORMSDIST(0.12)
Z=
Standardized Normal Probability
Table (Portion)
X − µ 8.6 − 8.0
=
= 0.12
σ
5.0
Z
.01
.02
.5478
0.0 .5000 .5040 .5080
µ=0
σ=1
µ=8
σ = 10
.00
P(X < 8.6)
= P(Z < 0.12)
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871
8 8.6
X
P(X < 8.6)
Z
Z
0 0.12
0.00
0.3 .6179 .6217 .6255
0.12
P(Z < 0.12)
17
More Examples of Normal
Distribution Using Excel
Upper Tail Probabilities
A set of final exam grades was found to be normally
distributed with a mean of 73 and a standard deviation of 8.
Suppose X is normal with mean 8.0 and
standard deviation 5.0.
Now Find P(X > 8.6)
What is the probability of getting a grade no higher than 91
on this exam? =NORMDIST(91,73,8,TRUE)
X ~ N ( 73,82 )
Mean
Standard Deviation
P ( X ≤ 91) = ?
18
σ =8
73
8
Probability for X <=
X Value
91
Z Value
2.25
P(X<=91)
0.9877756
X
µ = 73 91
0
X
8.0
Z
8.6
2.25
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© 2006 Prentice Hall, Inc.
Chapter 6
6-6
Probability Between
Two Values
Upper Tail Probabilities
(continued)
Suppose X is normal with mean 8.0 and standard deviation 5.0.
Now Find P(X > 8.6)…
=1-NORMDIST(8.6,8,10,TRUE) or =1-NORMSDSIT(0.12)
Suppose X is normal with mean 8.0 and standard deviation 5.0.
Find P(8 < X < 8.6)
=NORMDIST(8.6,8,5,TRUE)-0.5 or =NORMSDIST(0.12)-0.5
P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)
Calculate Z-values:
= 1.0 - 0.5478 = 0.4522
0.5478
1.000
1.0 - 0.5478
= 0.4522
8.0 8.6
X
.00
.01
.02
22
Probabilities in the Lower Tail
P(8 < X < 8.6)
= P(0 < Z < 0.12)
= P(Z < 0.12) – P(Z ≤ 0)
= 0.5478 - .5000 = 0.0478
0.0 .5000 .5040 .5080
Z
21
Solution: Finding P(0 < Z < 0.12)
Standardized Normal Probability
Table (Portion)
X
0 0.12
= P(0 < Z < 0.12)
0.12
0.12
8 8.6
P(8 < X < 8.6)
0
0
Z
X −µ 8 −8
=
=0
σ
5
X − µ 8.6 − 8
Z=
= 0.12
=
σ
5
Z
Z
Z=
Suppose X is normal with mean 8.0 and
standard deviation 5.0.
Now Find P(7.4 < X < 8)
0.0478
0.5000
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
X
Z
8.0
0.00
7.4
0.12
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Chapter 6
6-7
More Examples of Normal
Distribution Using Excel
Probabilities in the Lower Tail
(continued)
P(7.4 < X < 8)
= P(-0.12 < Z < 0)
X ~ N ( 73,82 )
0.0478
= P(Z < 0) – P(Z ≤ -0.12)
= 0.5000 - 0.4522 = 0.0478
7.4 8.0
-0.12 0
P ( 65 ≤ X ≤ 89 ) = ?
Probability for a Range
From X Value
65
To X Value
89
Z Value for 65
-1
Z Value for 89
2
P(X<=65)
0.1587
P(X<=89)
0.9772
P(65<=X<=89)
0.8186
0.4522
The Normal distribution is
symmetric, so this probability
is the same as P(0 < Z < 0.12)
(continued)
What percentage of students scored between 65 and 89
inclusively? Mean=73 and Std=8
=NORMDIST(89,73,8,TRUE)-NORMDIST(65,73,8,TRUE)
Now Find P(7.4 < X < 8)…
=0.5-NORMDIST(7.4,8,5,TRUE) or =0.5-NORMSDIST(-0.12)
X
Z
X
65
µ = 73
89
-1 0
Z
2
25
Finding the X value for a
Known Probability
26
Finding the X value for a
Known Probability
(continued)
Given a probability value 0.975, to find the corresponding z value = 1.96?
1). Draw a Normal Curve
2). Identify the area on the normal curve
3). Use the Normal Table to find the answer
=NORMSINV(0.975)
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
=NORMINV(0.20,8,5)
=NORMSINV(-0.84)
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:
=NORMINV(p*,µ,σ) or =µ+NORMSINV(p*)*σ
0.2000
X = µ + Zσ
?
?
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Basic Business Statistics, 10/e
8.0
0
X
Z
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© 2006 Prentice Hall, Inc.
Chapter 6
6-8
Find the Z value for
20% in the Lower Tail
Finding the X value
1. Find the Z value for the known probability
2. Convert to X units using the formula:
Standardized Normal Probability 20% area in the lower
Table (Portion)
tail is consistent with a
Z
-0.9
…
.03
.04
.05
Z value of -0.84
= 8.0 + ( −0.84)5.0
… .1762 .1736 .1711
-0.8 … .2033 .2005 .1977
-0.7
X = µ + Zσ
= 3.80
0.2000
… .2327 .2296 .2266
X
Z
?
8.0
-0.84 0
So 20% of the values from a distribution
with mean 8.0 and standard deviation
5.0 are less than 3.80
29
More Examples of Normal
Distribution Using Excel
More Examples of Normal Distribution
Using Excel
(continued)
Only 5% of the students taking the test scored higher than what
grade? Mean=73 and STD=8
=NORMINV(0.95,73,8)
X ~ N ( 73,82 )
30
Online normal distribution calculator
The middle 50% of the students scored between what two
scores?
=NORMINV(0.25,73,8) and =NORMINV(0.75,73,8)
X ~ N ( 73,82 )
P ( X > ? ) = .05
P ( a ≤ X ≤ b ) = .50
Find X and Z Given Cum. Pctage.
Cumulative Percentage
25.00%
Z Value
-0.67449
X Value
67.60408
Find X and Z Given Cum. Pctage.
Cumulative Percentage
95.00%
Z Value
1.644853
X Value
86.15882
.25
.25
X
µ = 73 ? =86.16
0
Z
1.645
X
Find X and Z Given Cum. Pctage.
Cumulative Percentage
75.00%
Z Value
0.67449
X Value
78.39592
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Basic Business Statistics, 10/e
(continued)
67.6 µ = 73 78.4
-0.67 0
Z
0.67
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© 2006 Prentice Hall, Inc.
Chapter 6
6-9
Evaluating Normality
Assessing Normality
(continued)
Observe the distribution of the data set
Not all continuous random variables are normally
distributed
It is important to evaluate how well the data set is
approximated by a normal distribution
Construct charts or graphs
Do approximately 2/3 of the observations lie within
mean ±1 standard deviation?
Do approximately 4/5 or 80% of the observations lie
within mean ±1.28 standard deviations?
Do approximately 19/20 or 95% of the observations
lie within mean ±2 standard deviations?
For small- or moderate-sized data sets, do stem-and-leaf
display and box-and-whisker plot look symmetric?
For large data sets, does the histogram or polygon appear bellshaped?
Evaluate normal probability plot
Compute descriptive summary measures
Is the normal probability plot approximately linear
with positive slope?
Do the mean, median and mode have similar values?
Is the interquartile range approximately 1.33 σ?
Is the range approximately 6 σ?
33
The Normal Probability Plot
34
The Normal Probability Plot
(continued)
A normal probability plot for data
from a normal distribution will be
approximately linear:
Normal probability plot
Arrange data into ordered array
Find corresponding standardized normal quantile
values
X
Plot the pairs of points with observed data values on
the vertical axis and the standardized normal quantile
values on the horizontal axis
90
60
30
Evaluate the plot for evidence of linearity
-2
-1
0
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1
2
Z
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© 2006 Prentice Hall, Inc.
Chapter 6
6-10
Normal Probability Plot
The Uniform Distribution
(continued)
Left-Skewed
Right-Skewed
X 90
X 90
60
60
30
The uniform distribution is a
probability distribution that has equal
probabilities for all possible
outcomes of the random variable
30
-2 -1 0
1 2 Z
-2 -1 0
1
2 Z
Rectangular
Nonlinear plots
indicate a deviation
from normality
X 90
60
Also called a rectangular distribution
30
-2 -1 0
1
2 Z
37
38
Properties of the
Uniform Distribution
The Uniform Distribution
(continued)
The Continuous Uniform Distribution:
1
b−a
if a ≤ X ≤ b
0
otherwise
The mean of a uniform distribution is
µ=
f(X) =
a+b
2
The standard deviation is
where
f(X) = value of the density function at any X value
a = minimum value of X
b = maximum value of X
σ=
(b - a) 2
12
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© 2006 Prentice Hall, Inc.
Chapter 6
6-11
Uniform Distribution Example
Uniform Distribution Example
(continued)
Example: Uniform probability distribution
over the range 2 ≤ X ≤ 6:
Example: Using the uniform probability
distribution to find P(3 ≤ X ≤ 5):
1
f(X) = 6 - 2 = 0.25 for 2 ≤ X ≤ 6
P(3 ≤ X ≤ 5) = (Base)(Height) = (2)(0.25) = 0.5
f(X)
f(X)
a+b 2+6
µ=
=
=4
2
2
0.25
2
6
X
σ=
(b - a) 2
=
12
0.25
(6 - 2) 2
= 1 . 1547
12
2
3
4
5
6
X
41
The Exponential Distribution
42
The Exponential Distribution
Often used to model the length of time
between two occurrences of an event (the
time between arrivals)
Defined by a single parameter, its mean λ
(lambda)
The probability that an arrival time is less than
some specified time X is
Examples:
P(arrival time < X) = 1 − e − λX
Time between trucks arriving at an unloading dock
Time between transactions at an ATM Machine
Time between phone calls to the main operator
where
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 <
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∞
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© 2006 Prentice Hall, Inc.
Chapter 6
6-12
Exponential Distribution
Example
Normal Approximation to the
Binomial Distribution
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 0.05 hours
P(arrival time < .05) = 1 – e-λX = 1 – e-(15)(0.05) = 0.5276
So there is a 52.76% probability that the arrival time
between successive customers is less than three
minutes
The binomial distribution is a discrete
distribution, but the normal is continuous
To use the normal to approximate the binomial,
accuracy is improved if you use a correction for
continuity adjustment
Example:
X is discrete in a binomial distribution, so P(X = 4)
can be approximated with a continuous normal
distribution by finding
P(3.5 < X < 4.5)
45
Normal Approximation to the
Binomial Distribution
(continued)
The closer p is to 0.5, the better the normal
approximation to the binomial
The larger the sample size n, the better the
normal approximation to the binomial
General rule:
46
Normal Approximation to the
Binomial Distribution
(continued)
The mean and standard deviation of the
binomial distribution are
µ = np
σ = np(1 − p)
The normal distribution can be used to approximate
the binomial distribution if
Transform binomial to normal using the formula:
np ≥ 5
Z=
and
X −µ
X − np
=
σ
np(1 − p)
n(1 – p) ≥ 5
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© 2006 Prentice Hall, Inc.
Chapter 6
6-13
Using the Normal Approximation
to the Binomial Distribution
Chapter Summary
If n = 1000 and p = 0.2, what is P(X ≤ 180)?
Approximate P(X ≤ 180) using a continuity correction
adjustment:
P(X ≤ 180.5)
Transform to standardized normal:
Presented key continuous distributions
normal, uniform, exponential
Found probabilities using formulas and tables
Recognized when to apply different distributions
X − np
180.5 − (1000)(0.2 )
Z=
=
= −1.54
np(1 − p)
(1000)(0.2 )(1− 0.2)
Applied distributions to decision problems
So P(Z ≤ -1.54) = 0.0618
180.5
-1.54
Basic Business Statistics, 10/e
200
0
X
Z
49
50
© 2006 Prentice Hall, Inc.