Chapter 2
The Normal Distribution
Lesson 2-1
Density Curve
Review
Graph the data
 Calculate a numerical summary of the data
 Describe the shape, center, spread and
outliers of the data

Histogram with Curve
Area Under the Curve
area = proportion (or percent) = probability
Density Curve
Is always on or above the horizontal
axis, and
 The area underneath is exactly 1.

Median and Mean of a
Density curve
Median is the equal-areas point, the point
that divides the area under the curve in half
 Mean is the balance point, at which the
curve would balance if made of solid
material.
 Median and mean are the same for a
symmetric density curve
 Mean of a skewed curve is pulled away from
the median in the direction of the long tail.

Mean and Median
Symmetric
Skewed to the Right
Notation
Actual Observation
Idealized Distribution
Mean  x 
Mean
Standard Deviation  s 
 
Standard Deviation  
Example – Page 83, #2.2
The density curve of a uniformed distribution. The curve
takes the constant value 1 over the interval from 0 to 1
and is zero outside the range of values. This means
that the data describe by this distribution takes values that
are uniformly spread between 0 and 1. Use the areas
under this density curve to answer the following questions
0
1
Example – Page 83, #2.2
A) Why is the total area under this curve equal to 1?
The area under the curve is a rectangle with
height 1 and width 1.
1
1
0
1
Example – Page 83, #2.2
B) What percent of the observation lie above 0.8?
(1  0.80)  0.20
.20
0
1
A  (0.20)(1)  0.20  20%
Example – Page 83, #2.2
C) What percent of the observation lies below 0.60?
.60
(.60  0)  .60  60%
0
1
Example – Page 83, #2.2
D) What percent of the observation lie between 0.25 and
0.75
.25
0
.75
1
(.75  .25)  .50  50%
Example – Page 83, #2.2
E) What is the mean μ of this distribution?
Mean = ½ or 0.50, the “balance point” of the density
curve.
0
1
Example, Page 84, #2.4
The following figure displays three density curves, each
with three points indicated. At which of these points on
each curve do the mean and median fall?
Example, Page 84, #2.4
Median B
Mean C
Median A
Mean A
Median B
Mean A
Example – Density Curve
I’m thinking about a density curve that consists of a straight
line segment from the point (0, 2/3) to the point (1,4/3) in
the x-y plane.
A). Sketch this density curve
4
3
1
2
3
1
3
1
4
1
2
3
4
1
Example – Density Curve
B). What percent of the observation lie below ½?
Area of the Rectangle
  
A bh  1 2  1
2
3
3
4
Area of the Triangle
1  1 

bh
3 1
A
 2
2
2
12
AT  1  1
5
 41.67%
3
12
12
3
1
2
3
1
3
1
4
1
2
3
4
1
Example – Density Curve
C). What percent of the observation lie below 1?
AT  1  100%
4
3
1
2
3
1
3
1
4
1
2
3
4
1
Example – Density Curve
C). What percent of the observation lie between ½ and 1?
Area of the Rectangle
 
A bh  1 1  1
2
2
Area of the Triangle
1 1
bh
A
 3 2 1
12
2
2
  
AT  1  1
7
 58.3%
2
12
12
4
3
1
2
3
1
3
1
4
1
2
3
4
1
Lesson 2-1
Normal Distribution
Normal Distribution




Symmetric, single peak
and bell shape
Tails fall quickly – so we
do no expect outliers
Mean and median lie
together at the peak in the
center of the curve
Standard deviation
determines the shape of
the curve
Standard Devotion
Inflection Point
Inflection Point
Why the Normal Distribution?

Are good descriptions for some distribution
of real data
SAT Score
 Characteristics of a biological population
 Tossing coins many times


Works well for other roughly symmetric
distributions
68 – 95 – 99.7 Rule
68 – 95 – 99.7 Rule
Example, Page 89, #2.6
The distribution of heights of adult American men is
approximately normal with mean 69 inches and
standard deviation 2.5 inches. Draw a normal curve
on which the mean and standard deviation are
correctly located.
  69
  2.5
Example, Page 89, #2.6
  69
  2.5
2.5
66.5
69
71.5
Example, Page 89, #2.8
Scores on the Wechsler Adult Intelligence Scale (WAIS,
a standard “IQ test”) for 20 to 34 age group are
approximately normally distributed with μ = 110 and
σ = 25. Use the 68 – 95 – 99.7 rule to answer these
questions.
Example, Page 89, #2.8
A) About what percent of people in this age group
have scores about 110?
about 68%
85 110 135
Example, Page 89, #2.8
B) About what percent have scores above 160?
160
135
110
about 2.35%
Example, Page 89, #2.8
C) In what range do the middle 95% of IQ scores lie?
160
135
110
85
60
between 60 to 160
or 110  50
Example, Page 91, #2.14
160
135
110
85
60
Wechsler Adult Intelligence Scale (WAIS) scores for
young adults are N(110, 25).
Example, Page 91, #2.14
A. If someone’s score were reported as the 16th percentile
about what score would that individual have.
160
135
110
85
60
approximately 85
Example, Page 91, #2.14
B. If someone’s score were reported as the 84th percentile
and 97.5th percentile about what score would that
individual have.
160
135
110
85
60
approximately 135 and
160
Lesson 2-2
The Standard Normal Distribution
How do I find the area under
the curve?
Use calculus (find the integral). Too
advance for the class.
 Use a series of tables to find areas for
every possible mean and standard
deviation. Infinitely many tables.
 Z-Scores

Z-Scores
Assume that your variable is normally
distributed.
 Use your mean and standard deviation to
convert your data into z-scores such that
the new distribution has a mean of 0 and a
standard deviation of 1

Standard Normal Distribution
The standard normal distribution is the normal distribution
N(0,1) with mean 0 and standard deviation 1.
If a variable x has any normal distribution N(μ, σ) with
mean μ and standard deviation σ then the standardized
variable is
z
x

Standard Normal Distribution
Example – Page 95, #2.20
Three landmarks of baseball achievement are Ty Cobb’s
batting average of .420 in 1911, Ted Williams’s .406 in
1941, and George Brett’s 0.390 in 1980. These batting
averages cannot be compared directly because the
distribution of major league batting averages has
changed over the years. The distributions are quite
symmetric and (except for outliers such as Cobb, Williams
and Brett) reasonably normal. While the mean batting
average has been held roughly constant by rule changes
and the balance between hitting and pitching, the
standard deviation has dropped over time. Here are
the facts:
Example – Page 95, #2.20
Decade
1910s
1940s
Mean Std. Dev.
.266
.0371
.267
.0326
1970s
.261
.0317
Compute the standardized batting averages for Cobb,
Williams, and Brett to compare how far each stood above
his peers.
Example – Page 95, #2.20
z
x

Cobb:
.420  .266
z
 4.15
.0371
Williams
.406  .267
z
 4.26
.0326
Decade
1910s
1940s
1970s
Mean
.266
.267
.261
Std. Dev.
.0371
.0326
.0317
Brett
z
.390  .261
 4.07
.0317
Williams’s z-score is the
highest.
Example, Page 109, #2.28
Use Table A to find the proportion of observation from a
Standard normal distribution that falls in each of the
following regions. In each case sketch a standard normal
Curve and shade the area representing the region
a) z  2.25
b) z  2.25
c) z  1.77
d) 2.25  z  1.77
Example, Page 109, #2.28
a) z  2.25
-2.25
0
Example, Page 109, #2.28
Example, Page 109, #2.28
a) z  2.25
The area to the left
of -2.25 is 0.0122
-2.25
0
Example, Page 109, #2.28
b) z  2.25
Find the area to the left of -2.25
z  2.25  0.0122
1 0.0122  0.9878
-2.25
0
Area greater than -2.25 = 1 – area below -2.25
Example, Page 109, #2.28
c) z  1.77
A  1   z  1.77 
0 1.77
Example, Page 109, #2.28
Example, Page 109, #2.28
c) z  1.77
A  1   z  1.77 
 1  0.9616
 0.0384
0 1.77
Area greater than 1.77 = 1 – area below 1.77
Example, Page 109, #2.28
d) 2.25  z  1.77
A   z  1.77    z  2.25 
A  0.9616  0.0122
 0.9494
-2.25
0
1.77
Area between -2.25 and 1.77 = area below 1.77 – area below -2.25
TI – 83 (Area)
z  2.25
2nd Vars
TI – 83 (Area)
z  2.25
2.25  z  1.77
Example – Page 103, #2.22
Use Table A to find the value of z of a standard normal
Variable that satisfies each of the following conditions.
Use the value of z from Table A that comes closet to
satisfying the condition. In each case sketch a standard
curve with your value of z marked on the axis.
a) The point z with 25% of the observation falling below it.
b) The point z with 40% of the observation falling above it.
Example – Page 103, #2.22
a) The point z with 25% of the observation falling below it.
A  0.25
z?
Example – Page 103, #2.22
Example – Page 103, #2.22
a) The point z with 25% of the observation falling below it.
A  0.25
z  0.67
Example – Page 103, #2.22
b) The point z with 40% of the observation falling above it.
Find the area below 0.40
A  0.40
1  0.40  .60
z?
Example – Page 103, #2.22
Example – Page 103, #2.22
b) The point z with 40% of the observation falling above it.
Find the area below 0.40
A  0.40
1  0.40  .60
z  0.25
TI – 83 (Z – Value)
25% below
2nd Vars
TI – 83 (Z – Value)
40% Above
Example, Page 103, #2.24
Scores on the Wechsler Adult Intelligence Scale (a
standard IQ test) for the 20 to 34 age group are
approximately normally distributed with μ = 110 and
σ = 25
A. What percent of people age 20 to 34 have IQ
scores above 100?
B. What percent have scores above 150?
C. How high an IQ score is needed to be the highest 25%?
Example, Page 103, #2.24
μ = 110 and σ = 25
A. What percent of people age 20 to 34 have IQ
scores about 100?
z
x  μ 100  110

 0.40
σ
25
65.5%
100 110
-0.40 0
Z
Example, Page 103, #2.24
μ = 110 andσ = 25
B. What percent have scores above 150?
x  μ 150  110
z

 1.6
σ
25
5.5%
110
0
150
1.60
Z
Example, Page 103, #2.24
μ = 110 andσ = 25
C. How high an IQ score is needed to be the highest 25%?
xμ
z
σ
x  110
0.68 
25
A = 0.25
x  127
110
?
Example, Page 110, #2.30
The annual rate of return on stock indexes (which
combine many individual stocks) is approximately
normal. Since 1945, the Standard & Poor’s 500 Index
has had a mean yearly return of 12%, with a standard
deviation of 16.5%. Take this normal distribution to be
the distribution of yearly returns over a long period.
A. In what range do the middle 95% of all yearly returns lie?
B. The market is down for the year if the return on the index is
less than zero. In what proportion of years is the market
down?
Example, Page 110, #2.30
μ = 12% and σ = 16.5%
A. In what range do the middle 95% of all yearly returns
lie?
12%  2(16.5%) 
-21% to 45%
-21%
12%
45%
Example, Page 110, #2.30
μ = 12% and σ = 16.5%
B. The market is down for the year if the return on the
index is less than zero. In what proportion of years is
the market down?
x  μ 0  12
z

 0.727
σ
16.5
0.2335
0% 12%
-0.73 0
Z
Example, Page 110, #2.30
μ = 12% and σ = 16.5%
C. In what proportion of years does the index gain 25%
or more?
z
x  μ 25  12

 0.7878
σ
16.5
0.2154
12 25
0 0.79
Z
Lesson 2-2
Accessing Normality
Assessing Normality

Construct a frequency histogram or a
stemplot or boxplot


Check to see if graph is approximately
bell-shape and symmetric about the mean
Construct a normal probability plot

Use TI to see if plotted points lie close to a
straight line
Example, Page 108, #2.26
Repeated careful measurements of the same physical
quantity often have a distribution that is close to normal.
Here are Henry Cavendish’s 29 measurements of the
density of the earth, made in 1798. (The data gives the
density of the earth as multiple of the density of water.)
5.50 5.61 4.88 5.07 5.26 5.55 5.36 5.29 5.58 5.65
5.57 5.53 5.62 5.29 5.44 5.34 5.79 5.10 5.27 5.39
5.42 5.47 5.63 5.34 5.46 5.30 5.75 5.68 5.85
Example, Page 108, #2.26
5.50 5.61 4.88 5.07 5.26 5.55 5.36 5.29 5.58 5.65
5.57 5.53 5.62 5.29 5.44 5.34 5.79 5.10 5.27 5.39
5.42 5.47 5.63 5.34 5.46 5.30 5.75 5.68 5.85
A. Construct a stemplot to show that the data are
reasonably symmetric
Example, Page 108, #2.26
48 8
A. Construct a stemplot
to show that the data are
reasonably symmetric
4.88
48|8
49
50
51
52
53
7
0
6 7 9 9
0 4 4 6 9
54
55
56
57
58
2
0
1
5
5
4 6 7
3 5 7 8
2 3 5 8
9
Example, Page 108, #2.26
48 8
B. Now check how closely
they follow the 68-95-99.7 rule.
Find x and s, then count the
number of observations that
fall between x  s, between
x  2s, between x  3s .
Compare the percents of the
29 observations in each of
these intervals with the
68-95-99.7 rule.
4.88
48|8
49
50
51
52
53
7
0
6 7 9 9
0 4 4 6 9
54
55
56
57
58
2
0
1
5
5
4 6 7
3 5 7 8
2 3 5 8
9
Example, Page 108, #2.26
48 8
1
0%
2
11
11
4
6.9% 37.9% 37.9% 13.8%
5.89
x
5.67
5.45
5.23
5.01
x  2s x  s
0
0%
x  s x  2s
4.88
48|8
49
50
51
52
53
7
0
6 7 9 9
0 4 4 6 9
54
55
56
57
58
2
0
1
5
5
4 6 7
3 5 7 8
2 3 5 8
9
Example, Page 108, #2.26
C. Use your calculator to construct a normal probability
plot for Cavendish’s density of the earth data, write a
brief statement about the normality of the data. Does
the normal probability reinforce your findings in (a).
Example, Page 108, #2.26
The linearity of the normal probability indicates an
approximately normal distribution.