The Normal Distribution and Other Continuous Distributions

Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 6-1
Chapter 6
The Normal Distribution and
Other Continuous Distributions
Statistics for Managers
Using Microsoft®
Excel
4th
Edition

Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc. Chap 6-2
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

Chapter Goals
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
_
_
(continued)

Probability Distributions
Continuous
Probability
Distributions
Binomial
Hypergeometric
Poisson
Probability
Distributions
Discrete
Probability
Distributions
Normal
Uniform
Exponential
Ch. 5 Ch. 6

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.

The Normal Distribution
Probability
Distributions
Normal
Uniform
Exponential
Continuous
Probability
Distributions

 ‘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 − ∞
Mean
= Median
= Mode
X
f(X)
μ
σ

By varying the parameters μ and σ, we obtain
different normal distributions
Many Normal Distributions

Shape
X
f(X)
μ
σ
Changing μ shifts the
distribution left or right.
Changing σ increases
or decreases the
spread.

The Normal Probability
Density Function
 The formula for the normal probability density
function is
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
2
μ)/σ](1/2)[(X
e
2π
1
f(X) −−
σ
=

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

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:
σ
μX
Z
−
=
Z always has mean = 0 and standard deviation = 1

The Standardized Normal
Probability Density Function
 The formula for the standardized normal
probability density function is
Where e = the mathematical constant approximated by 2.71828
π = the mathematical constant approximated by 3.14159
Z = any value of the standardized normal distribution
2
(1/2)Z
e
2π
1
f(Z) −
=

The Standardized
Normal Distribution
 Also known as the “Z” distribution
 Mean is 0
 Standard Deviation is 1
Z
f(Z)
0
1
Values above the mean have positive Z-values,
values below the mean have negative Z-values

Example
 If X is distributed normally with mean of 100
and standard deviation of 50, the Z value for
X = 200 is
 This says that X = 200 is two standard
deviations (2 increments of 50 units) above
the mean of 100.
2.0
50
100200
σ
μX
Z =
−
=
−
=

Comparing X and Z units
Z
100
2.00
200 X
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)
(μ = 100, σ = 50)
(μ = 0, σ = 1)

Finding Normal Probabilities
Probability is the
area under the
curve!
a b X
f(X)
P a X b( )≤
Probability is measured by the area
under the curve
≤
P a X b( )<<=
(Note that the probability
of any individual value is
zero)

f(X)
Xμ
Probability as
Area Under the Curve
0.50.5
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below
1.0)XP( =∞<<−∞
0.5)XP(μ =∞<<0.5μ)XP( =<<−∞

Empirical Rules
μ ± 1σ encloses about
68% of X’s
f(X)
X
μ μ+1σμ-1σ
What can we say about the distribution of values
around the mean? There are some general rules:
σσ
68.26%

The Empirical Rule
 μ ± 2σ covers about 95% of X’s
 μ ± 3σ covers about 99.7% of X’s
xμ
2σ 2σ
xμ
3σ 3σ
95.44% 99.72%
(continued)

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)
Z0 2.00
.9772
Example:
P(Z < 2.00) = .9772

The Standardized Normal Table
The value within the
table gives the
probability from Z = − ∞
up to the desired Z
value
.9772
2.0P(Z < 2.00) = .9772
The row shows
the value of Z
to the first
decimal point
The column gives the value of
Z to the second decimal point
2.0
.
.
.
(continued)
Z 0.00 0.01 0.02 …
0.0
0.1

General Procedure for
Finding Probabilities
 Draw the normal curve for the problem in
terms of X
 Translate X-values to Z-values
 Use the Standardized Normal Table
To find P(a < X < b) when X is
distributed normally:

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

standard deviation 5.0. Find P(X < 8.6)
Z0.120X8.68
μ = 8
σ = 10
μ = 0
σ = 1
(continued)
0.12
5.0
8.08.6
σ
μX
Z =
−
=
−
=
P(X < 8.6) P(Z < 0.12)

Z
0.12
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
Solution: Finding P(Z < 0.12)
.5478.02
0.1 .5478
Standardized Normal Probability
Table (Portion)
0.00
= P(Z < 0.12)
P(X < 8.6)

Upper Tail Probabilities
standard deviation 5.0.
 Now Find P(X > 8.6)
X
8.6
8.0

 Now Find P(X > 8.6)…
(continued)
Z
0.12
0
Z
0.12
.5478
0
1.000 1.0 - .5478
= .4522
P(X > 8.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)
= 1.0 - .5478 = .4522
Upper Tail Probabilities

Probability Between
Two Values
standard deviation 5.0. Find P(8 < X < 8.6)
P(8 < X < 8.6)
= P(0 < Z < 0.12)
Z0.120
X8.68
0
5
88
σ
μX
Z =
−
=
−
=
0.12
5
88.6
σ
μX
Z =
−
=
−
=
Calculate Z-values:

Z
0.12
Solution: Finding P(0 < Z < 0.12)
.0478
0.00
= P(0 < Z < 0.12)
P(8 < X < 8.6)
= P(Z < 0.12) – P(Z ≤ 0)
= .5478 - .5000 = .0478
.5000
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.02
0.1 .5478
Table (Portion)

 Now Find P(7.4 < X < 8)
X
7.4
8.0
Probabilities in the Lower Tail

Probabilities in the Lower Tail
Now Find P(7.4 < X < 8)…
X7.4 8.0
P(7.4 < X < 8)
= P(-0.12 < Z < 0)
= P(Z < 0) – P(Z ≤ -0.12)
= .5000 - .4522 = .0478
(continued)
.0478
.4522
Z-0.12 0
The Normal distribution is
symmetric, so this probability
is the same as P(0 < Z < 0.12)

 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:
Finding the X value for a
Known Probability
ZσμX +=

Finding the X value for a
Known Probability
Example:
 Now find the X value so that only 20% of all
values are below this X
X? 8.0
.2000
Z? 0
(continued)

Find the Z value for
20% in the Lower Tail
 20% area in the lower
tail is consistent with a
Z value of -0.84Z .03
-0.9 .1762 .1736
.2033
-0.7 .2327 .2296
.04
-0.8 .2005
Table (Portion)
.05
.1711
.1977
.2266
…
…
…
…
X? 8.0
.2000
Z-0.84 0
1. Find the Z value for the known probability

2. Convert to X units using the formula:
Finding the X value
80.3
0.5)84.0(0.8
ZσμX
=
−+=
+=
So 20% of the values from a distribution
with mean 8.0 and standard deviation
5.0 are less than 3.80

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

Assessing Normality
 Construct charts or graphs
 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 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?
(continued)

Assessing Normality
 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?
(continued)
±
±
±

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

A normal probability plot for data
from a normal distribution will be
approximately linear:
30
60
90
-2 -1 0 1 2 Z
X
The Normal Probability Plot
(continued)

Normal Probability Plot
Left-Skewed Right-Skewed
Rectangular
30
60
90
-2 -1 0 1 2 Z
X
(continued)
30
60
90
-2 -1 0 1 2 Z
X
30
60
90
-2 -1 0 1 2 Z
X Nonlinear plots indicate
a deviation from
normality

The Uniform Distribution
Continuous
Probability
Distributions
Probability
Distributions
Normal
Uniform
Exponential

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

The Continuous Uniform Distribution:
otherwise0
bXaif
ab
1
≤≤
−
where
f(X) = value of the density function at any X value
a = minimum value of X
b = maximum value of X
(continued)
f(X) =

Properties of the
Uniform Distribution
 The mean of a uniform distribution is
 The standard deviation is
2
ba
μ
+
=
12
a)-(b
σ
2
=

Uniform Distribution Example
Example: Uniform probability distribution
over the range 2 ≤ X ≤ 6:
2 6
.25
f(X) = = .25 for 2 ≤ X ≤ 66 - 2
1
X
f(X)
4
2
62
2
ba
μ =
+
=
+
=
1547.1
12
2)-(6
12
a)-(b
σ
22
===

The Exponential Distribution
Continuous
Probability
Distributions
Probability
Distributions
Normal
Uniform
Exponential

 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

Xλ
e1X)timeP(arrival −
−=<
 Defined by a single parameter, its mean λ
(lambda)
 The probability that an arrival time is less than
some specified time X is
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 < ∞

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

Sampling Distributions
Sampling
Distributions
Sampling
Distributions
of the
Mean
Sampling
Distributions
of the
Proportion

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

Developing a
Sampling Distribution
 Assume there is a population …
 Population size N=4
 Random variable, X,
is age of individuals
 Values of X: 18, 20,
22, 24 (years)
A B C D

.3
.2
.1
0
18 20 22 24
A B C D
Uniform Distribution
P(x)
x
(continued)
Summary Measures for the Population Distribution:
Developing a
21
4
24222018
N
X
μ i
=
+++
=
=
∑
2.236
N
μ)(X
σ
2
i
=
−
=
∑

1st
2nd
Observation
Obs 18 20 22 24
18 18,18 18,20 18,22 18,24
20 20,18 20,20 20,22 20,24
22 22,18 22,20 22,22 22,24
24 24,18 24,20 24,22 24,24
16 possible samples
(sampling with
replacement)
Now consider all possible samples of size n=2
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
(continued)
Developing a
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
Sampling Distribution of All Sample Means
18 19 20 21 22 23 24
0
.1
.2
.3
P(X)
X
Sample Means
Distribution
16 Sample Means
_
Developing a
(continued)
(no longer uniform)
_

Summary Measures of this Sampling Distribution:
Developing a
(continued)
21
16
24211918
N
X
μ i
X
=
++++
==
∑ 
1.58
16
21)-(2421)-(1921)-(18
N
)μX(
σ
222
2
X
i
X
=
+++
=
−
=
∑


Comparing the Population with its
18 19 20 21 22 23 24
0
.1
.2
.3
P(X)
X18 20 22 24
A B C D
0
.1
.2
.3
Population
N = 4
P(X)
X _
1.58σ21μ XX
==2.236σ21μ ==
Sample Means Distribution
n = 2
_

of the Mean
Sampling
Distributions
Sampling
Distributions
of the
Mean
Sampling
Distributions
of the
Proportion

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:
 Note that the standard error of the mean decreases as
the sample size increases
n
σ
σX
=

If the Population is Normal
 If a population is normal with mean μ and
standard deviation σ, the sampling distribution
of is also normally distributed with
and
(This assumes that sampling is with replacement or
sampling is without replacement from an infinite population)
X
μμX
=
n
σ
σX
=

Z-value for Sampling Distribution
of the Mean
 Z-value for the sampling distribution of :
where: = sample mean
= population mean
= population standard deviation
n = sample size
X
μ
σ
n
σ
μ)X(
σ
)μX(
Z
X
X −
=
−
=
X

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
1N
nN
n
σ
μ)X(
Z
−
−
−
=

Normal Population
Distribution
Normal Sampling
Distribution
(has the same mean)
Sampling Distribution Properties

(i.e. is unbiased )x
x
x
μμx =
μ
xμ

Sampling Distribution Properties
 For sampling with replacement:
As n increases,
decreases
Larger
sample size
Smaller
sample size
x
(continued)
xσ
μ

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:
andμμx =
n
σ
σx =

n↑
Central Limit Theorem
As the
sample
size gets
large
enough…
the sampling
distribution
becomes
almost normal
regardless of
shape of
population
x

Population Distribution
(becomes normal as n increases)
Central Tendency
Variation
(Sampling with
replacement)
x
x
Larger
sample
size
Smaller
sample size
If the Population is not Normal
(continued)
Sampling distribution
properties:
μμx =
n
σ
σx =
xμ
μ

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

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?

Example
Solution:
 Even if the population is not normally
distributed, the central limit theorem can be
used (n > 30)
 … so the sampling distribution of is
approximately normal
 … with mean = 8
 …and standard deviation
(continued)
x
xμ
0.5
36
3
n
σ
σx ===

Example
Solution (continued):
(continued)
0.38300.5)ZP(-0.5
36
3
8-8.2
n
σ
μ-μ
36
3
8-7.8
P8.2)μP(7.8 X
X
=<<=










<<=<<
Z7.8 8.2 -0.5 0.5
Sampling
Distribution
Standard Normal
Distribution .1915
+.1915
Population
Distribution
?
?
?
?
????
?
???
Sample Standardize
8μ = 8μX
= 0μz =xX

of the Proportion
Sampling
Distributions
Sampling
Distributions
of the
Mean
Sampling
Distributions
of the
Proportion

Population Proportions, p
p = the proportion of the population having
some characteristic
 Sample proportion ( ps ) provides an estimate
of p:
 0 ≤ ps ≤ 1
 ps has a binomial distribution
(assuming sampling with replacement from a finite population or
without replacement from an infinite population)
sizesample
interestofsticcharacterithehavingsampletheinitemsofnumber
n
X
ps ==

Sampling Distribution of p
 Approximated by a
normal distribution if:

where
and
(where p = population proportion)
P(ps)
.3
.2
.1
0
0 . 2 .4 .6 8 1 ps
pμ sp =
n
p)p(1
σ sp
−
=
5p)n(1
5np
and
≥−
≥

Z-Value for Proportions
 If sampling is without replacement
and n is greater than 5% of the
population size, then must use
the finite population correction
factor:
1N
nN
n
p)p(1
σ sp
−
−−
=
n
p)p(1
pp
σ
pp
Z s
p
s
s
−
−
=
−
=
Standardize ps to a Z value with the formula:
pσ

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) ?

Example
 if p = .4 and n = 200, what is
P(.40 ≤ ps ≤ .45) ?
(continued)
.03464
200
.4).4(1
n
p)p(1
σ sp =
−
=
−
=
1.44)ZP(0
.03464
.40.45
Z
.03464
.40.40
P.45)pP(.40 s
≤≤=





 −
≤≤
−
=≤≤
Find :
Convert to
standard
normal:
spσ

Example
Z
.45 1.44
.4251
Standardize
Standardized
Normal Distribution
 if p = .4 and n = 200, what is
P(.40 ≤ ps ≤ .45) ?
(continued)
Use standard normal table: P(0 ≤ Z ≤ 1.44) = .4251
.40 0
ps

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

Chapter Summary
 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
(continued)

The Normal Distribution and Other Continuous Distributions

More Related Content

What's hot (20)

Similar to The Normal Distribution and Other Continuous Distributions (20)

More from Yesica Adicondro (20)

Recently uploaded (20)

The Normal Distribution and Other Continuous Distributions