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ECO 72 - INTRODUCTION TO ECONOMIC
STATISTICS
Topic 6
The Normal
Distribution
These slides are copyright © 2003 by Tavis Barr. This material may
be distributed only subject to the terms and conditions set forth in the
Open Publication License, v1.0 or later (the latest version is presently
available at http://www.opencontent.org/openpub/).
The Normal Distribution
●
Introduction to Continuous Random
Variables
●
The Normal Distribution
●
The Standard Normal Distribution
●
Using the Normal to approximate
the Binomial
Continuous Random
Variables
●
●
Random variables map
events (results of
experiments) to the number
line
Discrete random variables:
we divide events into
categories, and assign a
number to each category
–
Example: We ship out five
cars. What is the probability
that one is defective? Two?
Three?
Fuzzy vague
sample space:
Product qualities
1
2
3
4
5
Random Variable:
Number defective
6
Continuous Random
Variables
●
Discrete random variables:
we divide events into
categories, and assign a
number to each category
–
●
Example: We ship out five
cars. What is the probability
that one is defective? Two?
Three?
Sometimes we want to map
events straight to numbers
rather than categories.
–
Fuzzy vague
sample space:
Jobs
For example: Height, Sales,
Wages.
1
2
3
4
5
Random Variable:
Wages
6
Continuous Random
Variables
●
●
A continuous random
variable maps events to
ranges of numbers
Fuzzy vague
sample space:
Jobs
The way it takes on
probabilities is different.
–
–
If a person walks into the room
right now, what is the
probability that that person will
be exactly 6 feet tall, down to
the subatomic particle?
What is the probability that that
person will be between 5'11”
and 6'1” tall?
1
2
3
4
5
Random Variable:
Wages
6
Continuous Random
Variables
●
The way it takes on
probabilities is different.
–
If a person walks into the room
right now, what is the
probability that that person
will be exactly 6 feet tall, down
to the subatomic particle?
ZERO.
–
What is the probability that
that person will be between
5'11” and 6'1” tall? SMALL,
BUT NOT ZERO.
Fuzzy vague
sample space:
Jobs
1
2
3
4
5
Random Variable:
Wages
6
Describing Probabilities –
Discrete vs. Continuous
●
A discrete random variable takes on
specific values with positive
probabilities. So we can tabulate
the probabilities:
Passengers
Probability
0
1
2
3
4
.30
.35
.15
.10
.10
Describing Probabilities –
Discrete vs. Continuous
●
●
With a continuous random variable, no
outcome happens with a positive probability.
So we can't tabulate them this way:
Exact Height
Probability
5'2”
5'3”
5'4”
5'5”
5'6”
...
.00
.00
.00
.00
.00
...
We need a different way of describing
outcomes
Probability Density Function
f(x)
P(11<X<16)
P(5<X<9)
4
5
6
7
8
9
10
11
12
13
14
15
X
16
A probability density function (pdf) is a
function where the area under the curve
between any two numbers shows the
probability that the variable will take on a
value between the two numbers.
Probability Density Function
f(x)
Total area of 1
4
●
●
5
6
7
8
9
10
11
12
13
14
15
16
X
Probabilities are never negative; the pdf
function is never negative.
The total sample space has probability
one; the area under the whole real
number line equals one
Pdf's and Historgrams
A histogram is a
graph of a sample
where the heights of
each column reflects
the fraction of the
sample in that
interval on the
number line
Histogram
0.4
Fraction of Observations
●
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0-1.99
2-3.99
4-5.99
Number interval
6-7.99
Pdf's and Histograms
●
A histogram is a graph of
a sample where the
heights of each column
reflects the fraction of
the sample in that
interval on the number
line
A frequency polygon is a
line connecting the tops
of the columns, anchored
to zero at the edges.
0.4
Fraction of Observations
●
Histogram
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0-1.99
2-3.99
4-5.99
6-7.99
Number interval
Pdf's and Histograms
Pdf
Histogram / Frequency Polygon
f(x)
4
5
●
6
7
8
9
10
11
12
4
5
6
7
8
9
10
11
As the sample gets bigger and
includes more and more of the
population, the frequency polygon
begins to approximate the pdf.
12
Cumulative Distribution
Function
F(t)
P(X≤16)
P(X≤9)
4
5
6
7
8
9
10
11
12
13
14
15
16
t
The cumulative distribution function
, usually written F(t), shows the
probability that a random variable
will be less than some number X,
i.e., P(X≤t).
Cumulative Distribution
Function
F(t)
P(X≤16)
P(X≤9)
4
5
6
7
8
9
10
11
12
13
14
15
16
t
The cdf is never less than 0 or
greater than 1
●
The cdf never gets smaller as t gets
bigger
●
Cumulative Distribution
Function
●
●
This means that for any
two numbers a and b
P(a≤X≤b) = F(b) – F(a).
We can prove this using
the laws of probability:
–
–
(X<a) and (a≤X≤b) are
mutually exclusive
P(X<a)P(a≤X≤b)
Same
P(X≤b)
a
X<a
b
a≤X≤b
X
X>b
So P[(X<a) or (a≤X≤b)]
= P(X<a) +P(a≤X≤b)
Mutually exclusive
Cumulative Distribution
Function
●
Want to prove:
P(a≤X≤b) = F(b) – F(a).
–
–
P[(X<a) or (a≤X≤b)] =
P(X<a) +P(a≤X≤b)
P(X<a)
F(a)
P(a≤X≤b)
Same
P(X≤b) F(b)
From def'n of cdf:
P(X<a) = F(a)
● P(X≤b) = F(b)
a
●
X<a
b
a≤X≤b
X
X>b
Mutually exclusive
Cumulative Distribution
Function
●
Want to prove:
P(a≤X≤b) = F(b) – F(a).
–
–
P[(X<a) or (a≤X≤b)] =
P(X<a) +P(a≤X≤b)
P(X<a)
F(a)
Same
P(X≤b) F(b)
From def'n of cdf:
P(X<a) = F(a)
● P(X≤b) = F(b)
So: F(a) +P(a≤X≤b) = F(b)
a
●
●
●
P(a≤X≤b)
Subtr. F(a) from both sides
X<a
b
a≤X≤b
X
X>b
Mutually exclusive
Cumulative Distribution
Function
F(x)
.8
↕P(5≤X≤9)
.4
P(x≤9)
↕P(X<5)
4
5
6
7
8
9
10
11
12
13
14
15
16
X
A graphical illustration: P(a≤X≤b) = F(b)–F(a)
●
Example: If F(9) = 0.8 and F(5) = 0.4,
then P(5≤X≤9) = 0.8 – 0.4 = 0.4
The Normal Distribution
●
Tends to be useful in describing
quantities that result from the
average or combined effect of a
whole lot of small, independent
events.
–
Test scores
–
Height
–
Product quality
Three Features of the
Normal
The normal distribution has a few important
properties:
50% less 50% more
than 
than 

f(x)
X
(1) Its density (pdf) is symmetric; the mean,
median, and mode are all the same
Three Features of the
Normal
The normal distribution has a few important
properties:
Never exactly zero
f(x)

X
(2) Probability is never zero, though it is
effectively zero not too far away from mean (5 or
6 Std Devs)
Three Features of the
Normal
The normal distribution has a few important
properties:


f(x)

X
(3) There are many Normal distributions; they
are uniquely specified by a mean and a
standard deviation.
The Standard Normal
=1 =1
­2
●
­1
=0
1
(z)
z
2
The standard Normal distribution is the
Normal distribution with mean zero and
standard deviation one
The Standard Normal
=1 =1
­2
●
●
­1
=0
1
(z)
z
2
Its cdf is denoted by (z) and its pdf by 
(z)
(z) is well-known and easily available; this
is not true of other normal distributions
The Standard Normal
=1 =1
­2
●
­1
=0
1
(z)
z
2
However, the standard normal can be used
to calculate the cdf of any normally
distributed variable, whether it is standard
normal or not
Standard Normal Example
●
Before using the standard normal to
approximate other distibutions, let's
practice using it alone
–
Suppose we play at a slot machine
whose winnings are normally
distributed with mean $0 and standard
deviation $1
–
What is the probability of winning
between 50 cents and $1.50?
Calculating P(a≤X≤b)
●
●
●
●
On the web site, there
is a table that tells us
P(X≤a)
From this table,
F(1.5) =.9332
Also, F(0.5) = .6915
So P(0.5≤z≤1.5) =
.9332 - .6915 = .2417
P(X<1.5)
­2
­1
Z
0
(z)
z
2
1
Probability
.00
.01
.02
0.0 .5000 .5040 .5080
...
0.5 .6915 .6950 .6985
...
1.0 .8413 .8438 .8461
...
1.5 .9332 .9345 .9357
...
.03
.5120
.7019
.8485
.9370
Standard Normal Table
(cont'd)
CALCULATING
P(-1<z<1):
● P(z<1) = .8413
P(z<1)
= .8413
­2
­1
Z
0
1
Probability
.00
.01
.02
0.0 .5000 .5040 .5080
...
0.5 .6915 .6950 .6985
...
1.0 .8413 .8438 .8461
...
1.5 .9332 .9345 .9357
...
(z)
z
2
.03
.5120
.7019
.8485
.9370
Standard Normal Table
(cont'd)
CALCULATING
P(-1<z<1):
● P(z<1) = .8413
● P(z≤-1) =
.1587
● P(-1≤z≤1)
= .8413 - .1587
= .6826
.1587.
­2
­1
Z
P(z<1)
= .8413
0
1
Probability
.00
.01
.02
0.0 .5000 .5040 .5080
...
0.5 .6915 .6950 .6985
...
1.0 .8413 .8438 .8461
...
1.5 .9332 .9345 .9357
...
(z)
z
2
.03
.5120
.7019
.8485
.9370
Standard Normal Table
Z
Probability
.00 .01
0.0 .0000 .0040
...
0.5 .1915 .1950
...
1.0 .3413 .3438
...
1.5 .4332 .4345
...
.02 .03
.0080 .0120
.1985 .2019
P(0<z<1.5)
=.4332
.3461 .3485
.4357 .4370
­2
­1
0
1
(z)
z
2
OUR TABLE IS NON-STANDARD
Most statistics textbooks have a table that tells
us not the cdf of a standard normal variable, but
the probability that the variable is between 0 and
the number we're interested in
Standard Normal Table
(cont'd)
Z
Probability
.00 .01
0.0 .0000 .0040
...
0.5 .1915 .1950
...
1.0 .3413 .3438
...
1.5 .4332 .4345
...
.02 .03
.0080 .0120
.1985 .2019
.3461 .3485
.4357 .4370
0.5 prob.
­2
­1
P(0<z<1.5)
= .4332
0
1
(z)
z
2
OUR TABLE IS NON-STANDARD
● Because 0 is the median of a standard normal,
the probability it is less than zero is 0.5
● So we can add 0.5 to the number in the table
to get the value of the cdf at that number
The Normal, In General
●
●
If X is normally distributed with
mean  and standard deviation ,
then (X - )/ is standard normally
distributed
This means that if we want to figure
out P(X<t) for some number t, all we
have to figure out is P(z< [t - ]/)
Normal Distribution –
Example 1
●
In the Czech Republic in 2002, the average
region had an unemployment rate of
9.94%, with a standard deviation of 4.15%.
Source: http://www.cazv.cz/2003/ZE%2012_03/4-Dufek.pdf
●
●
Assume that regional unemployment rates
are Normally distributed
What fraction of regions would you expect
to have an unemployment rate of 5 to 15
percent?
Normal Distribution –
Example 1
●
In the Czech Republic in 2002, the average
region had an unemployment rate of
9.94%, with a standard deviation of 4.15%.
Source: http://www.cazv.cz/2003/ZE%2012_03/4-Dufek.pdf
●
What fraction of regions would you expect
to have an unemployment rate of 5 to 15
percent?
–
Here, we want to know P(5<X<15)
= F(15) – F(5)
–
So one “t” is 5 and the other is 5
–
Want to know ([15-9.94]/4.15) - ([5-9.94]/4.15)
Normal Distribution –
Example 1
●
In the Czech Republic in 2002, the average
region had an unemployment rate of
9.94%, with a standard deviation of 4.15%.
Source: http://www.cazv.cz/2003/ZE%2012_03/4-Dufek.pdf
●
What fraction of regions would you expect
to have an unemployment rate of 5 to 15
percent?
–
Here, we want to know P(5<X<15)
= F(15) – F(5)
–
So one “t” is 5 and the other is 5
–
Want to know ([15-9.94]/4.15) - ([5-9.94]/4.15)
i.e., (1.22) - (-1.19)
Normal Distribution –
Example 1
●
In the Czech Republic in 2002, the average
region had an unemployment rate of
9.94%, with a standard deviation of 4.15%.
Source: http://www.cazv.cz/2003/ZE%2012_03/4-Dufek.pdf
●
What fraction of regions would you expect
to have an unemployment rate of 5 to 15
percent?
–
Want to know ([15-9.94]/4.15) - ([5-9.94]/4.15)
i.e., (1.22) - (-1.19)
–
From the chart: (1.22)= 0.8888 and (-1.19) = 0.1170
–
So, the answer is 0.8888 – 0.1170, i.e., 0.7718.
Normal Distribution –
Example 2
●
Average taxi time at Logan airport is
19.3 minutes, with a standard
deviation of 7.3 minutes.
Source: http://dspace.mit.edu/bitstream/1721.1/37322/1/TaxiOutModel.pdf
●
●
Suppose that taxi time is Normally
distributed.
What percentage of flights are on
the runway for at least 30 mins?
Normal Distribution –
Example 2
●
●
Average taxi time at Logan airport is 19.3
minutes, with a standard deviation of 7.3
minutes.
What percentage of flights are on the
runway for at least 30 mins?
–
Here, we want to know P(X > 30), i.e., our “t”
is 30.
–
So we want to look at P(z > [30-]/), i.e.,
P(z > [30 – 19.3]/7.3), i.e., P(z > 1.47).
Normal Distribution –
Example 2
●
●
Average taxi time at Logan airport is 19.3
minutes, with a standard deviation of 7.3
minutes.
What percentage of flights are on the
runway for at least 30 mins?
–
Here, we want to know P(X > 30), i.e., t = 30.
–
So we want to look at P(z > [30-]/), i.e.,
P(z > [30 – 19.3]/7.3), i.e., P(z > 1.47).
–
From Table, P(z < 1.47) = 0.9292
Normal Distribution –
Example 2
●
●
Average taxi time at Logan airport is 19.3
minutes, with a standard deviation of 7.3
minutes.
What percentage of flights are on the
runway for at least 30 mins?
–
Here, we want to know P(X > 30), i.e., t = 30.
–
So we want to look at P(z > [30-]/), i.e.,
P(z > [30 – 19.3]/7.3), i.e., P(z > 1.47).
–
From Table, P(z < 1.47) = 0.9292
–
So, P(z>1.47) = 1-P(z<1.47) = 1-.9292 =.0708
Normal Approx. to Binomial
●
●
●
●
Remember, when k is large enough, k! is
impossible to calculate
Most calculators go up to about 50!, my
computer goes up to 170! but is inaccurate
This makes Binomial probabilities
impossible to calculate for many practical
problems
It turns out many Binomial probabilities
can be approximated by the Normal
distribution.
Normal Approximation to
Binomial
●
●
●
This only works when both np and
n(1-p) are at least five
If so, then we just plug in the mean
and the standard deviation of the
Binomial.
Remember, if X is Binomial with n
and p:
 np 1−p
–
Its mean is np
–
Its standard deviation is
Approx. to Binomial –
Example
●
●
Suppose probability of an employee
stealing from a company is 0.2.
If there are 100 employees, what is
the probability that at least 25
steal?
Approx. to Binomial –
Example
●
●
Suppose probability of employee
stealing from company is 0.2.
If there are 100 employees, what is
the probability that at least 25
steal?
–
This is a Binomial with n = 100 and p =
0.2
Approx. to Binomial –
Example
●
●
Suppose probability of employee
stealing from company is 0.2.
If there are 100 employees, what is
the probability that at least 25
steal?
–
This is a Binomial with n = 100 and p =
np
1−p=
16=4


0.2
–
So we can approximate by a normal
with mean
np = 100(.2) = 20 and std. dev.
Approx. to Binomial –
Example
●
●
Suppose probability of employee stealing
from company is 0.2.
If there are 100 employees, what is the
probability that at least 25 steal?
–
This is a Binomial with n = 100 and p = 0.2
–
So we can approximate by a normal with mean
np = 100(.2) = 20 and std. dev. np 1−p= 16=4
–
So we want to know the probability that a
Normal variable with mean 20 and std. dev. 4
is at least 25
Approx. to Binomial –
Example
●
●
Suppose probability of employee stealing
from company is 0.2.
If there are 100 employees, what is the
probability that at least 25 steal?
–
This is a Binomial with n = 100 and p = 0.2
–
So we want to know the probability that a
Normal variable with mean 20 and std. dev. 4
is at least 25
–
We have to round up to 25.5
Approx. to Binomial –
Example
●
●
Suppose probability of employee stealing
from company is 0.2.
If there are 100 employees, what is the
probability that at least 25 steal?
–
So we want to know the probability that a
Normal variable with mean 20 and std. dev. 4
is at least 25.5
–
This is the same as 1-([25.5-20]/4) = 1-
(1.375)
= 1-0.9154 = .0846
Approximation to Binomial Another Example
●
●
The probability of someone being a
vegetarian is 6 percent.
In a group of 300 people, what is the
probability that there are less than 10
vegetarians?
Approximation to Binomial Another Example
●
●
The probability of someone being a
vegetarian is 6 percent.
In a group of 300 people, what is the
probability that there are less than 10
vegetarians?
–
Approximate using a normal with mean np
and standard deviation  np 1−p
Approximation to Binomial Another Example
●
●
The probability of someone being a
vegetarian is 6 percent.
In a group of 300 people, what is the
probability that there are less than 10
vegetarians?
–
Approximate using a normal with mean np and
standard deviation  np 1−p
–
n = 300, p = .06 so np = 18 and  np1−p=4.11
–
So we want to know Prob. that N(18,4.11) < 10
(- 0.5)
Approximation to Binomial Another Example
●
●
The probability of someone being a
vegetarian is 6 percent.
In a group of 300 people, what is the
probability that there are less than 10
vegetarians?
–
Approximate using a normal with mean np and standard
deviation  np 1−p
–
n = 300, p = .06 so np = 18 and
–
So we want to know Prob. that N(18,4.11) < 9.5
–
This is the same as ([10.5-18]/4.11)= (-2.06) = .0197
 np1−p=4.11