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Chapter 9 Normal Distribution
9.1 Continuous distribution
 9.2 The normal distribution
 9.3 A check for normality
 9.4 Application of the normal
distribution
 9.5 Normal approximation to
Binomial

9.1 Continuous Distribution

For a discrete distribution, for example
Binomial distribution with n=5, and p=0.4,
the probability distribution is
x
f(x)
0
0.07776
1
0.2592
2
3
0.3456 0.2304
4
0.0768
5
0.01024
A probability histogram
0.3
0.2
P(x)
0.1
0.0
0
1
2
3
x
4
5
How to describe the distribution of
a continuous random variable?

For continuous random variable, we also represent
probabilities by areas—not by areas of rectangles, but by
areas under continuous curves.

For continuous random variables, the place of histograms
will be taken by continuous curves.

Imagine a histogram with narrower and narrower classes.
Then we can get a curve by joining the top of the
rectangles. This continuous curve is called a probability
density (or probability distribution).
Continuous distributions

For any x, P(X=x)=0. (For a
continuous distribution, the area
under a point is 0.)

Can’t use P(X=x) to describe the
probability distribution of X

Instead, consider P(a≤X≤b)

P(a≤X≤b) is the
area between a
and b
0.20
The area under the
curve is 1
0.15

0.00
0.05
0.10
A curve f(x):
f(x) ≥ 0
y

0.25
Density function
0
2
4
6
x
8
10
0.00
0.05
0.10
y
0.15
0.20
0.25
P(2≤X≤4)= P(2≤X<4)=
P(2<X<4)
0
2
4
6
x
8
10
9.2 The normal distribution


A normal curve: Bell shaped
Density is given by
2

1
(x  ) 
f ( x) 
exp  

2
2
 2



μand σ2 are two parameters: mean and
standard variance of a normal population
(σ is the standard deviation)
0.06
0.04
0.02
0.00
fx
0.08
0.10
0.12
The normal—Bell shaped
curve: μ=100, σ2=10
90
95
100
x
105
110
0.2
0.1
0.0
fx1
0.3
0.4
Normal curves:
(μ=0, σ2=1) and (μ=5, σ 2=1)
-2
0
2
4
x
6
8
Normal curves:
0.2
0.1
0.0
y
0.3
0.4
(μ=0, σ2=1) and (μ=0, σ2=2)
-3
-2
-1
0
x
1
2
3
Normal curves:
0.0
0.2
0.4
fx1
0.6
0.8
1.0
(μ=0, σ2=1) and (μ=2, σ2=0.25)
-2
0
2
4
6
8
0.2
0.1
0.0
y
0.3
0.4
The standard normal curve:
μ=0, and σ2=1
-3
-2
-1
0
x
1
2
3
How to calculate the probability of
a normal random variable?

Each normal random variable, X, has a density
function, say f(x) (it is a normal curve).

Probability P(a<X<b) is the area between a and
b, under the normal curve f(x)

Table I in the back of the book gives areas for a
standard normal curve with =0 and =1.

Probabilities for any normal curve (any  and )
can be rewritten in terms of a standard normal
curve.
Table I: Normal-curve Areas
Table I on page 494-495
 We need it for tests
 Areas under standard normal curve
 Areas between 0 and z (z>0)
 How to get an area between a and b?
when a<b, and a, b positive
area[0,b]–area[0,a]

Get the probability from standard
normal table
z denotes a standard normal random
variable
 Standard normal curve is symmetric
about the origin 0
 Draw a graph

Table I: P(0<Z<z)
z
0.0
0.1
0.2
0.3
0.4
0.5
…
1.0
1.1
.00
.0000
.0398
.0793
.1179
.1554
.1915
…
.3413
.3643
.01
.0040
.0438
.0832
.1217
.1591
.1950
…
.3438
.3665
.02
.0080
.0478
.0871
.1255
.1628
.1985
…
.3461
.3686
.03
.0120
.0517
.0910
.1293
.1664
.2019
…
.3485
.3708
.04
.0160
.0557
.0948
.1331
.1700
.2054
…
.3508
.3729
.05
.0199
.0596
.0987
.1368
.1736
.2088
…
.3531
.3749
.06
.0239
.0636
.1026
.1404
.1772
.2123
…
.3554
.3770
Examples
Adobe Acrobat 7.0
Document
Example 9.1
P(0<Z<1)
= 0.3413
 Example 9.2
P(1<Z<2)
=P(0<Z<2)–P(0<Z<1)
=0.4772–0.3413
=0.1359

Examples
Adobe Acrobat 7.0
Document

Example 9.3
P(Z≥1)
=0.5–P(0<Z<1)
=0.5–0.3413
=0.1587
Examples
Adobe Acrobat 7.0
Document

Example 9.4
P(Z ≥ -1)
=0.3413+0.50
=0.8413
Examples
Adobe Acrobat 7.0
Document

Example 9.5
P(-2<Z<1)
=0.4772+0.3413
=0.8185
Examples
Adobe Acrobat 7.0
Document

Example 9.6
P(Z ≤ 1.87)
=0.5+P(0<Z ≤ 1.87)
=0.5+0.4693
=0.9693
Examples
Adobe Acrobat 7.0
Document

Example 9.7
P(Z<-1.87)
= P(Z>1.87)
= 0.5–0.4693
= 0.0307
From non-standard normal to
standard normal

X is a normal random variable with
mean μ, and standard deviation σ

Set Z=(X–μ)/σ
Z=standard unit or z-score of X
Then Z has a standard normal
distribution and
Example 9.8
X is a normal random variable
with μ=120, and σ=15
Find the probability P(X≤135)
Solution:

x
x  120
Let z 


15
120  120
z is normal  z 
0
15
15
z  1
15
x   135  120
P( x  135)  P(

)  P( z  1)  0.5  0.3413  0.8413

15
XZ
x z-score of x
Example 9.8 (continued)

P(X≤150)
x=150  z-score z=(150-120)/15=2
P(X≤150)=P(Z≤2)
= 0.5+0.4772= 0.9772
9.3



Checking Normality
Most of the statistical tools we will use in
this class assume normal distributions.
In order to know if these are the right
tools for a particular job, we need to be
able to assess if the data appear to have
come from a normal population.
A normal plot gives a good visual check
for normality.
Simulation: 100 observations,
normal with mean=5, st dev=1

5
4
3
2
x
6
7
8

x<-rnorm(100, mean=5, sd=1)
qqnorm(x)
-2
-1
0
Quantiles of Standard Normal
1
2
Estimating a woman’s risk of having a preganancy
associated with Down’s syndrome using her age
and serum alpha-fetoprotein level
H.S.Cuckle, N.J.Wald, S.O.Thompson
The plot below shows results on alpha-fetoprotein (AFP) levels in
maternal blood for normal and Down’s syndrome fetuses.
Normal Plot
The way these normal plots work is
Straight means that the data appear
normal
 Parallel means that the groups have
similar variances.

Normal plot
In order to plot the data and check
for normality, we compare
•our observed data to
•what we would expect from a sample
of normal data.
To begin with, imagine taking n=5 random values from a
standard normal population (=0, =1)
Let Z(1) Z(2) Z(3) Z(4) Z(5) be the ordered values. Suppose we
do this over and over.
Sample
1
2
3
…
Forever
Mean
Z(1)
-1.7
-0.9
-2.3
…
___
-1.163
E(Z(1))
Z(2)
-0.2
0.2
-1.5
…
___
-0.495
E(Z(2))
Z(3)
0.8
0.5
-0.6
…
___
0
E(Z(3))
Z(4)
1.3
0.9
0.4
…
___
0.495
E(Z(4))
Z(5)
1.9
2.0
1.3
…
___
1.163
E(Z(5))
On average

the smallest of n=5 standard normal values is 1.163
standard deviations below average

the second smallest of n=5 standard normal values is 0.495
standard deviations below average

the middle of n=5 standard normal values is at the average,
0 standard deviations from average
The table of “rankits” from the
Statistics in Biology table gives
these expected values.
For larger n, space is saved by just
giving the positive values. The
negative values are a mirror image
of the positive values, since a
standard normal distribution is
symmetric about its mean of zero.
Check for normality
If X is normal, how do ordered values of X, X(i) ,
relate to expected ordered Z values, E( Z(i) ) ?
Z
X 

X   Z
For normal with mean  and standard deviation
, the expected values of the data, X(i), will be a
linear rescaling of standard normal expected
values
E(X(i)) ≈  +  E( Z(i) )
The observed data X(i) will be approximately a
linearly related to E( Z(i) ).
X(i) ≈  +  E( Z(i) )

If we plot the ordered X values
versus E( Z(i) ), we should see
roughly a straight line with
•intercept 
•slope 
Example
Example: Lifetimes of springs
under 900 N/mm2 stress
i
1
2
3
4
5
6
7
8
9
10
E( Z(i) )
-1.539
-1.001
-0.656
-0.376
-0.123
0.123
0.376
0.656
1.001
1.539
X(i)
153
162
189
216
216
216
225
225
243
306
Lifetime of Springs at Stress 900
350
Lifetime
300
250
900 stress
200
150
100
-2.000
-1.000
0.000
1.000
2.000
E(Z)
The plot is fairly linear indicating that the data are
pretty similar to what we would expect from
normal data.
To compare results from different treatments, we can
put more than one normal plot on the same graph.
350
Lifetime
300
250
950 stress
900 stress
200
150
100
-2.000
-1.000
0.000
1.000
2.000
E(Z)
The intercept for the 900 stress level is above the intercept for
the 950 stress group, indicating that the mean lifetime of the 900
stress group is greater than the mean of the 950 stress group.
The slopes are similar, indicating that the variances or standard
deviations are similar.

These plots were done in Excel. In Excel you can
either enter values from the table of E(Z) values or
generate approximations to these tables values.

One way to generate approximate E(Z) values is to
generate evenly spaced percentiles of a standard
normal, Z, distribution.

The ordered X values correspond roughly to
particular percentiles of a normal distribution.

For example if we had n=5 values, the 3rd ordered
values would be roughly the median or 50th
percentile.

A common method is to use percentiles
corresponding to 100  i  0.5 .
n
For n=5 this would give us



i
i  0.5
n
1
2
3
4
5
0.1
0.3
0.5
0.7
0.9
the 50th percentile
For E(Z) we would use corresponding percentiles
of a standard normal Z distribution.
Percentiles expressed as fractions are called
quantiles. The 0.5 quantile is the 50th
percentile.
Normal plots from this perspective are
sometimes called Q-Q plots, since we are
plotting standard normal quantiles versus the
associated quantiles of the observed data.
For n = 10 values for the spring data, the
corresponding normal percentiles would be
i  0.5
i
Z quantile
n
1
2
3
4
5
6
7
8
9
10
0.05
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
-1.64
-1.04
-0.67
-0.39
-0.13
0.13
0.39
0.67
1.04
1.64




For assessing whether a plotted line is fairly
parallel, either the E(Z) values or the normal
quantiles work fine.
If you are doing the plot by hand it’s easiest
to use the E(Z) table.
If you are doing these in Excel it’s easiest to
use the normal quantiles.
The function NORMINV(p,0,1) finds the Z
values corresponding to a given quantile.
This is the inverse of the function that finds
the cumulative probability for a given Z value.
Z
 NORMDIST  probability = NORMDIST(1.645, 0, 1, TRUE)  0.95
Probability  NORMINV  probability = NORMINV(0.95, 0, 1)  1.645
(The TRUE in NORMDIST says to return the cumulative probability rather
than density curve height.)
Excel File of Lifetime of Springs Data
n
10
i
1
2
3
4
5
6
7
8
9
10
(i-0.5)/n
0.05
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
Normal
Quantile
-1.645
-1.036
-0.674
-0.385
-0.126
0.126
0.385
0.674
1.036
1.645
Ordered
E(Z) 900 stress
-1.539
153
-1.001
162
-0.656
189
-0.376
216
-0.123
216
0.123
216
0.376
225
0.656
225
1.001
243
1.539
306
Ordered
950 stress
117
135
135
162
162
171
189
189
198
225
For data that are not normal
Many types of data tend to follow a normal
distribution, but many data sets aren’t particularly
normal.
If the data aren’t fairly normal we have several
options

Transform the data, meaning change the scale.

A log or ln scale is most common.
• Weights of fish
• Concentrations
• Bilirubin levels in blood
• pH is a log scale
• RNA expression levels in a microarray experiment


A reciprocal (1/Y) change of times to rates
Other powers
• Square root for Poisson variables
Non-normal data continued

Use a different distribution other
than a normal distribution

Weibull distribution for lifetimes
• Motors at General Electric
• Patients in a clinical trial
Weibull Distributions
(Time to Failure – Non-binomial & Nonnormal)
Infant Mortality: Fail
immediately or last a long
time
Early Failure: These do
not fail immediately, but
many do fail early
Old-age Wearout: Very
few of these fail until they
were out
Non-normal data continued

Use a nonparametric methods which
doesn’t assume any distribution

Finding a distribution that models the
data well rather than nonparametric
• Allows us to develop a more complete
model
• Allows us to generalize to other situations
• Gives us more precise information for the
same amount of effort




The methods in this class largely apply to normal
data or data that we can transform to normal.
The EPA fish example is a good example of
transforming data with a log transformation.
Geometric means and harmonic means arise
when we are working with transformed data.
For example fish weights are usually analyzed in
the log scale. Having a mean in the log scale we
want to put this value back into the original
scale, for example grams.
The back-transformed mean from the log scale is the
geometric mean.
The back-transformed mean from a reciprocal scale (rates),
is the harmonic mean.
Back-transformed differences between geometric means
correspond to ratios in the original scale.
Suppose ln(X) = Y ~ N(, 2).
This means Y (or ln(x)) distributed as
normal with mean  and variance 2.
The geometric mean is e, the backtransformed population mean in the ln
scale.
If we have the difference between two
means in the ln scale then backtransforming give us
e
1  2

e
e
1
2
= ratio of geometric means.
About geometric means

A fact is that if the variances of both
populations are the same, then the
ratio of the population geometric
means is the same as the ratio of
the population means.

Question: Why not just use the means
in the original scale?

Answer: Means are best when
populations are normal. Using the ratio
of the geometric means will give us a
more precise estimate of the true ratio
than using the ratio of the means in the
original scale.

A similar fact explains why we use means
rather than medians.

For a normal population the mean is the
same as the median. We could use either
the sample mean or the sample median to
estimate .

BUT, the mean will be a more precise guess
(estimate of) the true value, .

It would take us roughly 50% more values
(larger n) using the median as our guess at
 to accomplish the same degree of precision
as we get using the mean as our guess at .
9.4 Application of the normal
distribution

1960-62 Public Health Service
Health Examination Survey
6,672 Americans 18-79 years old
The woman’s heights were approximately
normal with 63 and standard deviation
2.5 .
What percentage of women were over 68
tall?
Solution:

X=height
P(X>68)=P(Z>(68-63)/2.5))
=P(Z>2)
=0.5-0.4772
=0.0228
Continuity Correction for a Better
Approximation

Sometimes only integer values are
possible for x.
x=score of LSAT
x=# of heads in 10 tosses of a fair coin
A normal approximation is more accurate
with a “continuity correction”

1976 LSAT
Approximately normal
mean 650, st. dev 60
P(X≥680)P(Z>(679.5-650)/60)
=P(Z>0.49)
=0.5-0.1879
=0.3121
9.5 Normal Approximation
to Binomial
A binomial distribution:
n=10, p=0.5
μ=np=5
σ2=np(1-p)=2.5  σ=1.58
1.
P(X≥7)=0.172 from Binomial
2.
P(X≥7)= P(Z>(6.5-5)/1.58)
3. =P(Z>0.95)
=0.5-0.3289=0.1711
from normal approximation

Dots: Binomial Probabilities
0.00
0.05
0.10
fx
0.15
0.20
0.25
Smoot Line: Normal Curve With Same Mean and Variance
0
2
4
6
x
8
10
Normal Approximation Is Good If

The normal curve has the same
mean and standard deviation as
binomial

np>5 and n(1-p)>5

Continuity correction is made
Example

1.
2.
3.
4.
Records show that 60% of the
customers of a service station pay with
a credit card. Use normal approximation
to find the probabilities that among 100
customers
At most 65 will pay with a credit
At least 55 will pay with a credit
Between 55 and 65 will pay with a credit
card
Exactly 65 will pay with a credit card
Solution:
X=# of customers who pay with a
credit card
μ=np=60,
σ2=np(1-p)=24  σ=4.8990

65.5  60
P ( X  65)  P ( Z 
)  P ( Z  1.12)
4.899
 0.5  0.3686  0.8686
54.5  60
P ( X  55)  P ( Z 
)  P ( Z  1.12)
4.899
 P ( Z  1.12)  0.5  0.3686  0.8686
Normal Approximation
3.
54.5  60
65.5  60
P(55  X  65)  P(
Z 
)
4.899
4.899
 P(1.12  Z  1.12)  2(0.3686)  0.7372
4. P( X  65)  P(65  X  65)
64.5  60
65.5  60
 P(
Z
)
4.899
4.899
 P(0.92  Z  1.12)  0.3686  0.3212
 0.0474