Download What does data from a normal distribution look like?

What does data from a
normal distribution look
like?
• The shape of histograms developed
from small samples drawn from a
normal population are somewhat
unpredictable. Yet, the bell pattern
is usually to some extent apparent.
• For large samples the
representation of the bell curve is
usually more visible.
Small Sample
n = 25
6
Large Sample
n = 200
35
30
5
25
4
20
3
15
2
10
1
5
0
0
-20
-6
8
22
36
50
64
78
92
106
0
2
4
6
8
10
12
14
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The Normal Probability
Density Function
1
f(y) 
2
e

y  m 2

2s
2
• The two parameters are
m (mean) and s2 (variance).
• The mean defines the location
and the variance determines
the dispersion.
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N(mu, sigma-squared)
or x ~ N(m, s2 )
• [1/{s(2)}] exp[ (xm)2/ 2s2]
• Where exp( ..) means e( .. ),
where e is the transcendental
constant.
• x ~ N(m, s2 ) means random
variable x is distributed as
Normal with the indicated
mean and variance separated
by comma in parentheses.
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Variance and Mean of
Normal Distributions
• The figure above illustrates several
normal distributions with identical
variances. The only difference in
the distributions is the central
location, the mean.
In the figure to the right there
are two distributions with
s=1
identical means, but
with different variances.
s=2
8-
Using the Normal
Probability Density
Function
• Using the normal probability
density function to determine
the probability of some interval
would be complicated.
• A special normal distribution,
called the standard normal,
can be used to determine
probabilities for any normal
random variable.
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The Standard Normal
Distribution N(0,1)
Sometimes called the
Z - distribution
8-
Standard Normal
Distribution: N(0,1)
• The standard normal
distribution provides a basis for
computing probabilities for all
normal distributions.
• It has a mean of zero and a
variance of one.
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Map from f(X) to f(Z)
• The techniques used to map or
translate any normal random
variable into a standard normal
random variable is called a
z-transform.
z  x s- m
• Because the z-transform gives
the standard normal unique
status among normals, the
standard normal is also referred
to as the z-distribution.
8-
Reverse Map from f(Z)
to f(X)
• The techniques used to map
standard normal random
variable to any normal random
variable is reverse ztransform.
zs =xm , zs +m =x
x is obtained by adding z
standard deviations to the
mean, x = zs +m
When z= -4, x is m-4s
z= 4, x is m+4s
z= 0, x is m.
8-
Standard Normal
Distribution Table
• A table in the appendix (p. 521)
contains probability calculations for
various points in the distribution.
• The table provides the probability
that a standard normal random
variable will be between 0 and a
specified value.
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
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Example 3
• To compute the probability that a
standard normal random variable
will be between 0 and 1, look up
the value 1.00 in the table.
• The table value of .3413 is the
area under the curve between 0
and 1, which is also the
probability that the random
variable will assume a value in
that interval.
.3413
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Example 4
Using the table, compute the
probability that a standard
normal random variable is
between 0 and 2.34.
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Example 4 - Solution
• First, draw a picture.
• The table is constructed to give the
probability between 0 and z.
• To determine the area under the
standard normal curve use the table
and find the area associated with
z = 2.34.
P(0 < z < 2.34) = .4904
8-
Example 5
Using the table, determine
the probability that a
standard normal random
variable is between -1.29
and 0.
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Example 5 - Solution
• First, draw a picture.
• Unfortunately, z = -1.29 is not
given in the table.
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Ex5 – Exploit symmetry
• Since the normal distribution is
symmetric,
P(-1.29 < z < 0) = P(0 < z < 1.29).
• P(0 < z < 1.29) = .4015.
• Thus, P(-1.29 < z < 0) = .4015.
8-
Area between two
positive z values (Ex6)
What is the probability that a
standard normal random
variable will be between 1.0
and 2.0?
Problem: The table only
contains probabilities from 0
to some value z.
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Ex6 First get two areas
• Determine the probability that z is
between 0 and 2.0.
P(0 < z < 2.0) = .4772
• Determine the probability that z is
between 0 and 1.0.
P(0 < z < 1.0) = .3413
8-
Ex6 – Subtract the
smaller # from Bigger #
• Subtract the unwanted probability.
P(0 < z < 2) - P(0 < z < 1) = P(1 < z < 2)
.4772 - .3413 = .1359
what we
got from
the table
the part
we don’t
want
leaving
the area
of interest
8-
Z – transformations
in computing
probabilities
8-
Z - transformation
• The z-transformation can
transform any normal random
variable into a standard normal
random variable.
• The z-transformation is
denoted by z and given by the
formula
x-m
z =
.
s
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How does the
transformation work?
• The numerator, x - m, centers
the z-distribution around zero.
• By subtracting the mean of the
random variable from each
data value, the mean of the
resulting z random variable will
be zero.
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Example of x before using
the z transformation
Find the probability that a normal
random variable with a mean of
10 and a standard deviation of
20 will lie between 10 and 40.
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Applying z transform to
each inequality term
Applying the z-transformation
to each term yields
(10-10
P( 10  X  40 ) = P(
20
(10-10)
(40-10)
 40 ) = P(
 z 
) = P( 0
20
20
(40-10)

) = P( 0  z  1.5).
20
8-
Map x~N(m, s2) on left
to the z~N(0,1) on right
P(10 < x < 40) = P(0 < z < 1.5)
m = 10
m + 1.5s = 40
mean m for z = 0
m + 1.5s = 1.5
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Geometry before z
transformation
• Find the probability that a
normal random variable with a
mean of 10 and standard
deviation of 20 will be greater
than 30.
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Algebra of z transform.
• Applying the z-transformation yields
30-1
P( X > 30 ) = P( z >
20
30-10
P( X > 30 ) = P( z >
)
20
= P(z > 1).
• Then, write as: Part1 Minus Part2
P(z > 1) = P(0 < z < ) - P(0 < z < 1)
= .5 - .3413 = .1587
8-
Map from x above to z
below (Ex.8)
• Note that the X
value of 30
transformed
into the z-value
of 1.
• In other words,
30 is one
standard
deviation away
from the mean.
8-
Example 9, x~ N(12,1)
A beer distributor
believes that the
amount of beer in a
12 ounce can of
beer has a normal
distribution with a
mean of 12 ounces
and a standard
deviation of 1
ounce. If a 12
ounce beer can is
randomly selected,
find the following
probabilities.
8-
Prob that beer makers are
cheating. Example 9-A
Find the probability that the 12
ounce can of beer will actually
contain less than 11 ounces of beer.
Let X = amount of beer in the 12 ounce
can. X has a normal distribution with
m = 12 and s = 1. x~ N(12,1)
P(X < 11) = P(X s- m < 11 - 12)
1
= P(z < -1) = .5 - P(-1 < z < 0)
= .5 - .3413 = .1587
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Prob that they are
cheating, x~ N(12,1)
.3413
.1587
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Prob that beer makers are
too generous: N(12,1)
Find the probability that the 12 ounce
can of beer will actually contain more
than 12.5 ounces of beer.
Let X = amount of beer in the 12 ounce
can. X has a normal distribution with
m = 12 and s = 1.
P(X > 12.5) = P(X s- m > 12.5 - 12)
1
= P(z > .5) = .5 - P(0 < z < .5)
= .5 - .1915 = .3085
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Prob that beer makers
are too generous
.1915
.3085
8-
Prob of slightly different
kind of cheating N(12,1)
Find the probability that the 12 ounce
can of beer will actually contain
between 10.5 and 11.5 ounces of beer.
Let X = amount of beer in the 12 ounce
can. X has a normal distribution with
m = 12 and s = 1.
P(10.5 < X < 11.5)
X-m
10.5 -<12
= P(
P(-1.5
z <<-.5)s < 11.5 - 12)
1
1
= P(-1.5 < z < 0) - P(-.5 < z < 0)
= .4332 - .1915 = .2417
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Slightly different
cheating prob from z
.1915
.2417
8-