02 PSBE3_PPT.Ch01_2_Examining Distribution.ppt

The Practice of Statistics for
Business and Economics
Third Edition
David S. Moore
George P. McCabe
Layth C. Alwan
Bruce A. Craig
William M. Duckworth
© 2011 W.H. Freeman and Company

Examining Distributions
The Normal Distributions
PSBE Chapter 1.3
© 2011 W.H. Freeman and Company

Objectives
Density Curves and Normal Distributions
 Density curves
 The Mean and Median of a density curve
 Normal Distributions
 The 68-95-99.7 rule
 The standard Normal Distribution
 Normal Distribution calculations
 Finding a value when given a proportion
 Assessing the Normality of data

Density Curves
A density curve is a mathematical model of a distribution.
The total area under the curve, by definition, is equal to 1, or 100%.
The area under the curve for a range of values is the proportion of
all observations for that range.
Histogram of a sample
with the smoothed
density curve describing
theoretically the
population.
What are the differences?

Density curves come in any
imaginable shape.
Some are well known
mathematically and others
aren’t.

Median & Mean of a Density Curve
• The Median of a density curve is the equal-areas point, the point
that divides the area under the curve in half.
• The Mean of a density curve is the balance point, at which the
curve would balance if made of solid material.
The Median & Mean are the same for a Symmetric Density Curve.
The Mean of a skewed curve is pulled in the direction of the long tail.

Normal Distributions
e = 2.71828… The base of the natural logarithm
π = pi = 3.14159…
Normal—or Gaussian—distributions are a family of
symmetrical, bell-shaped density curves defined by a Mean m
(mu) & a Standard Deviation s (sigma) : N (m , s).
2
2
1
2
1
)
(





 

 s
m

x
e
x
f
x
x

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
A family of Density Curves
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30
Here means are different
(m = 10, 15, and 20), whereas
standard deviations are the same
(s = 3).
Here means are the same (m = 15),
whereas standard deviations are
different (s = 2, 4, and 6).

µ = 64.5 s = 2.5
N(µ, s) = N(64.5, 2.5)
Inflection point
The “68-95-99.7” Rule
 About 68% of all
observations are within
1-s of the mean (m).
 About 95% of all
2-s of the mean m.
 Almost all (99.7%)
3-s of the m.

Because all Normal distributions share the same
properties, we can standardize our data to transform
any Normal curve N(m,s) into the Standard Normal
curve N(0,1).
The Standard Normal Distribution
For each x, we calculate a new value, z (called a z-score).
N(0,1)
=>
z
x
N(64.5, 2.5)
Standardized height (no units)
Why we standardize??

z 
(x m)
s
A z-score measures the number of Standard Deviations that a
data value x is from the mean m.
Standardizing: calculating z-scores
When x is larger than the Mean, z is positive. (x  m)
When x is smaller than the Mean, z is negative. (x  m)
1
, 






s
s
s
m
s
m
s
m z
x
for
• When x is 1-standard deviation
larger than the m , then z = 1.
2
2
2
,
2 






s
s
s
m
s
m
s
m z
x
for
• When x is 2-standard deviations
larger than the m, then z = 2.

µ = 64.5" , s = 2.5" , x (height) = 67"
We calculate z, the STANDARDIZED value of x:
mean
from
dev.
stand.
1
1
5
.
2
5
.
2
5
.
2
)
5
.
64
67
(
,
)
(






 z
x
z
s
m
Because of the (68-95-99.7_Rule), we can conclude that the percent of women shorter
than 67” should be, approximately:
0.68 + half of (1  0.68) = 0.84 or 84%.
Area= ???
Area = ???
N( µ , s) =
N (64.5, 2.5)
m = 64.5” x = 67”
z = 0 z = 1
Example: Women Heights
Women Heights follow N (64.5", 2.5")
distribution.
Q. What percent of women are
shorter than 67" tall?
2-sided

Using Table A
(…)
Table-A gives the area under the standard Normal curve to the left of any z
value.
0.0082 is the
area under
N(0,1) left of
z = 2.40
0.0080 is the area
under N(0,1) left
of z = 2.41
0.0069 is the area
under N(0,1) left
of z = 2.46

Area ≈ 0.84
Area ≈ 0.16
N(µ, s) =
N(64.5”, 2.5”)
m = 64.5” x = 67”
z = 1
Conclusion:
• Shorter: 84.13% are shorter than 67”.
• Taller: by subtraction (1  0.8413 =
15.87%) of women are taller than 67”.
For z = 1.00, the area
under the standard
Normal curve to the left
of z = (0.8413).
Percent of women shorter than 67"

Tips on using Table A
Because the Normal
distribution is symmetrical,
there are two ways that you
can calculate the area under
the standard Normal curve
to the right of a z value.
area right of z = 1  area left of z
Area = 0.9901
Area = 0.0099
z = -2.33
area right of z = area left of z

Tips on using Table A
To calculate the area between two “z-values”:
1. To get the area under N(0,1) to the left for each z-value from
Table A, then
2. Subtract the smaller area from the larger area.
Area between z1 & z2 = Area left of z1 – Area left of z2
A common mistake made
by students is to subtract
both z-values, but the
Normal curve is not
uniform.
 The area under N(0,1) for a single value of z is zero
(Try calculating the area to the left of z minus that same area!)

The National Athletic Association (NCAA) requires Division-I athletes to score
at least 820 on the combined Math & verbal SAT exam, to compete in their
first college year.
SAT scores of 2013 were approximately normal with (m = 1026) & (s = 209).
What proportion of all students would be NCAA qualifiers (SAT ≥ 820)?
820
1026
209
820 1026
209
206
0 99
209
x
( x )
z
( )
z
z .
z
m
s
m
s








  
Table A: area under
N(0,1) to the left of
-.99 is 0.1611
or approx. 16%.
Note: The actual data may contain students
who scored exactly 820 on the SAT.
area right of 820 = total area - area left of 820
= 1 - 0.1611
≈ 84%
Standardized Normal
-0.99 = 0.8389
SAT: Standard Assessment Test

The NCAA defines a “partial qualifier” eligible to practice and receive an athletic
scholarship, but not to compete, as a combined SAT score of at least 720.
What proportion of all students who take the SAT would be partial qualifiers?
That is, what proportion have scores between 720 and 820?
720
1026
209
720 1026
209
306
1 46
209
x
x
z
z
z
z
m
s
m
s








  
( )
( )
.
Table A: area under
N(0,1) to the left of
-.99 is 0.0721
or approx. 7%.
About 9% of all students who take the SAT have scores
between 720 and 820.
area between = area left of 820 - area left of 720
720 and 820 = 0.1611 - 0.0721
≈ 9%
1- 0.9279 = 0.0721

N(0,1)
z 
(x m)
s
The cool thing about working with
normally distributed data is that
we can manipulate it and then find
answers to questions that involve
comparing seemingly non-
comparable distributions.
We do this by “standardizing” the
data. All this involves is changing
the scale so that the mean now = 0
and the standard deviation = 1. If
you do this to different distributions
it makes them comparable.

Finding a value when given a proportion
Inverse normal calculations: We may also want to find the observed
range of values that correspond to a given proportion under the curve.
For that, we use Table A backward:
 We first find the desired
area/proportion in
the body of the
table.
 We then read the
corresponding z-value
from the left column
and top row. For an area to the left of 1.25 % (0.0125),
the z-value is -2.24

Inverse Normal Calculations
 SAT Verbal test scores follow approximately the N (505, 110)
distribution. How high must a student score to place in the
top 10% of all students taking the SAT?
a. z = 1.28 is the
standardized value with area
0.9 to its left and 0.1 to its
right.
b. Un-standardize
505
1.28
110
x -
=
Solving for x gives a score
of at least 646 on the SAT.

One way to assess if a distribution is indeed approximately normal
is to plot the data on a Normal Quantile Plot.
The data points are ranked and the percentile ranks are converted
to z-scores with Table A. The z-scores are then used for the x-axis
against which the data are plotted on the y-axis of the normal
quantile plot.
 If the distribution is indeed normal the plot will show a straight line,
indicating a good match between the data and a normal
distribution.
 Systematic deviations from a straight line indicate a non-normal
distribution. Outliers appear as points that are far away from the
overall pattern of the plot.
Assessing the Normality of data

Normal quantile
plot of the IQ of
fifth graders. This
distribution is
roughly Normal.
Normal quantile plot
of the time to start a
business. This
distribution is skewed
to the right.

This distribution is roughly
Normal except for one low
outlier.
This distribution is
skewed, not normal.

02 PSBE3_PPT.Ch01_2_Examining Distribution.ppt

02 PSBE3_PPT.Ch01_2_Examining Distribution.ppt

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