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Normal and Standard Normal
Distributions
June 29, 2004
Histogram
Percent of total that fall in the 10pound interval.
Data are divided into 10pound groups (called
“bins”).
With only one woman <100 lbs,
this 25
bin represents <1% of the
total 120-women sampled.
115-125
125-135
20
135-145
P
e 15
r
c
e
n 10
t
105-115
95-105
145-155
155-165
5
85-95
0
80
90
100
110
120
POUNDS
130
140
150
160
What’s the shape of the
distribution?
25
20
P
e 15
r
c
e
n 10
t
5
0
80
90
100
110
120
POUNDS
130
140
150
160
~ Normal Distribution
25
20
P
e
r
c
e
n
t
15
10
5
0
80
90
100
110
120
POUNDS
130
140
150
160
The Normal Distribution

Equivalently the shape is described as:
“Gaussian” or “Bell Curve”
 Every normal curve is defined by 2
parameters:
– 1. mean - the curve’s center
– 2. standard deviation - how fat the curve is
(spread)

X ~ N (, 2)
Examples:








height
weight
age
bone density
IQ (mean=100; SD=15)
SAT scores
blood pressure
ANYTHING YOU
AVERAGE OVER A
LARGE ENOUGH #
A Skinny Normal Distribution
More Spread Out...
Wider Still...
The Normal Distribution:
as mathematical function
f ( x) 
Note constants:
=3.14159
e=2.71828
1
 2
1 x 2
 (
)
e 2 
Integrates to 1



1
2
1 x 2
 (
)
 e 2  dx
1
Expected Value

E(X)=


x
1
 2
1 x 2
 (
)
e 2 
dx
=
Variance

Var(X)=

( x

2
1
 2
Standard Deviation(X)=
1 x 2
 (
)
 e 2  dx) 

2
= 2
normal curve with =3 and =1
**The beauty of the normal curve:
No matter what  and  are, the area between - and
+ is about 68%; the area between -2 and +2 is
about 95%; and the area between -3 and +3 is
about 99.7%. Almost all values fall within 3 standard
deviations.
68-95-99.7 Rule
68% of
the data
95% of the data
99.7% of the data
How good is rule for real data?
Check the example data:
The mean of the weight of the women = 127.8
The standard deviation (SD) = 15.5
68% of 120 = .68x120 = ~ 82 runners
In fact, 79 runners fall within 1-SD (15.5 lbs) of the mean.
112.3
127.8
143.3
25
20
P
e
r
c
e
n
t
15
10
5
0
80
90
100
110
120
POUNDS
130
140
150
160
95% of 120 = .95 x 120 = ~ 114 runners
In fact, 115 runners fall within 2-SD’s of the mean.
96.8
127.8
158.8
25
20
P
e
r
c
e
n
t
15
10
5
0
80
90
100
110
120
POUNDS
130
140
150
160
99.7% of 120 = .997 x 120 = 119.6 runners
In fact, all 120 runners fall within 3-SD’s of the mean.
81.3
127.8
174.3
25
20
P
e
r
c
e
n
t
15
10
5
0
80
90
100
110
120
POUNDS
130
140
150
160
Example

Suppose SAT scores roughly follows a normal
distribution in the U.S. population of collegebound students (with range restricted to 200-800),
and the average math SAT is 500 with a standard
deviation of 50, then:
– 68% of students will have scores between 450 and 550
– 95% will be between 400 and 600
– 99.7% will be between 350 and 650
Example
BUT…
 What if you wanted to know the math SAT
score corresponding to the 90th percentile
(=90% of students are lower)?
P(X≤Q) = .90 

Q
1 x 500 2
)
50
dx
 (
1
e 2
(50) 2
200

Solve for Q?….Yikes!
 .90
The Standard Normal
Distribution
“Universal Currency”

Standard normal curve: =0 and =1

Z ~ N (0, 1)
f ( z) 
1
2
1
 z2
e 2
The Standard Normal Distribution (Z)
All normal distributions can be converted into
the standard normal curve by subtracting the
mean and dividing by the standard deviation:
Z
X 

Somebody calculated all the integrals for the standard
normal and put them in a table! So we never have to
integrate!
Even better, computers now do all the integration.
Example

For example: What’s the probability of getting a math SAT
score of 575 or less, =500 and =50?
575  500
Z
 1.5
50
i.e., A score
of 575 is 1.5 standard deviations above the mean
575
 P( X  575) 
1
 (50)
200
2
1.5
1 x 500 2
 (
)
 e 2 50 dx 



1
2
1
 Z2
 e 2 dz
Yikes!
But to look up Z= 1.5 in standard normal chart (or enter
into SAS) no problem! = .9332
Use SAS to get area
You can also use also use SAS:
data _null_;
theArea=probnorm(1.5);
put theArea;
run;
0.9331927987
This function gives the
area to the left of X
standard deviations in a
standard normal curve.
In-Class Exercise
a.
b.
If birth weights in a population are
normally distributed with a mean of 109
oz and a standard deviation of 13 oz,
What is the chance of obtaining a birth
weight of 141 oz or heavier when
sampling birth records at random?
What is the chance of obtaining a birth
weight of 120 or lighter?
Answer
a.
What is the chance of obtaining a birth
weight of 141 oz or heavier when
sampling birth records at random?
141  109
Z
 2.46
13
From the chart  Z of 2.46 corresponds to a right tail (greater than)
area of: P(Z≥2.46) = 1-(.9931)= .0069 or .69 %
Answer
b. What is the chance of obtaining a birth
weight of 120 or lighter?
120  109
Z
 .85
13
From the chart  Z of .85 corresponds to a left tail area of:
P(Z≤.85) = .8023= 80.23%
Reading for this week

Walker: 1.3-1.6 (p. 10-22), Chapters 2 and 3
(p. 23-54)