Gerstman_PP07.ppt

7: Normal Probability Distributions 1
September 22
Chapter 7:
Normal Probability
Distributions

In Chapter 7:
7.1 Normal Distributions
7.2 Determining Normal Probabilities
7.3 Finding Values That Correspond to
Normal Probabilities
7.4 Assessing Departures from Normality

§7.1: Normal Distributions
• This pdf is the most popular distribution
for continuous random variables
• First described de Moivre in 1733
• Elaborated in 1812 by Laplace
• Describes some natural phenomena
• More importantly, describes sampling
characteristics of totals and means

Normal Probability Density
Function
• Recall: continuous
random variables are
described with
probability density
function (pdfs)
curves
• Normal pdfs are
recognized by their
typical bell-shape
Figure: Age distribution
of a pediatric population
with overlying Normal
pdf

Area Under the Curve
• pdfs should be viewed
almost like a histogram
• Top Figure: The darker
bars of the histogram
correspond to ages ≤ 9
(~40% of distribution)
• Bottom Figure: shaded
area under the curve
(AUC) corresponds to
ages ≤ 9 (~40% of area)
2
2
1
2
1
)
(





 

 



x
e
x
f

Parameters μ and σ
• Normal pdfs have two parameters
μ - expected value (mean “mu”)
σ - standard deviation (sigma)
σ controls spread
μ controls location

Mean and Standard Deviation
of Normal Density
μ
σ

Standard Deviation σ
• Points of inflections
one σ below and
above μ
• Practice sketching
Normal curves
• Feel inflection points
(where slopes change)
• Label horizontal axis
with σ landmarks

Two types of means and standard
deviations
• The mean and standard deviation from
the pdf (denoted μ and σ) are
parameters
• The mean and standard deviation from
a sample (“xbar” and s) are statistics
• Statistics and parameters are related,
but are not the same thing!

68-95-99.7 Rule for
Normal Distributions
• 68% of the AUC within ±1σ of μ
• 95% of the AUC within ±2σ of μ
• 99.7% of the AUC within ±3σ of μ

Example: 68-95-99.7 Rule
Wechsler adult
intelligence scores:
Normally distributed
with μ = 100 and σ = 15;
X ~ N(100, 15)
• 68% of scores within
μ ± σ
= 100 ± 15
= 85 to 115
• 95% of scores within
μ ± 2σ
= 100 ± (2)(15)
= 70 to 130
• 99.7% of scores in
μ ± 3σ =
100 ± (3)(15)
= 55 to 145

Symmetry in the Tails
… we can easily
determine the AUC in
tails
95%
Because the Normal
curve is symmetrical
and the total AUC is
exactly 1…

Example: Male Height
• Male height: Normal with μ = 70.0˝ and σ = 2.8˝
• 68% within μ ± σ = 70.0  2.8 = 67.2 to 72.8
• 32% in tails (below 67.2˝ and above 72.8˝)
• 16% below 67.2˝ and 16% above 72.8˝ (symmetry)

Reexpression of Non-Normal
Random Variables
• Many variables are not Normal but can be
reexpressed with a mathematical
transformation to be Normal
• Example of mathematical transforms used
for this purpose:
– logarithmic
– exponential
– square roots
• Review logarithmic transformations…

Logarithms
• Logarithms are exponents of their base
• Common log
(base 10)
– log(100) = 0
– log(101) = 1
– log(102) = 2
• Natural ln (base e)
– ln(e0) = 0
– ln(e1) = 1
Base 10 log function

Example: Logarithmic Reexpression
• Prostate Specific Antigen
(PSA) is used to screen
for prostate cancer
• In non-diseased
populations, it is not
Normally distributed, but
its logarithm is:
• ln(PSA) ~N(−0.3, 0.8)
• 95% of ln(PSA) within
= μ ± 2σ
= −0.3 ± (2)(0.8)
= −1.9 to 1.3
Take exponents of “95% range”
 e−1.9,1.3 = 0.15 and 3.67
 Thus, 2.5% of non-diseased
population have values greater
than 3.67  use 3.67 as
screening cutoff

§7.2: Determining Normal
Probabilities
When value do not fall directly on σ
landmarks:
1. State the problem
2. Standardize the value(s) (z score)
3. Sketch, label, and shade the curve
4. Use Table B

Step 1: State the Problem
• What percentage of gestations are
less than 40 weeks?
• Let X ≡ gestational length
• We know from prior research:
X ~ N(39, 2) weeks
• Pr(X ≤ 40) = ?

Step 2: Standardize
• Standard Normal
variable ≡ “Z” ≡ a
Normal random
variable with μ = 0
and σ = 1,
• Z ~ N(0,1)
• Use Table B to look
up cumulative
probabilities for Z

Example: A Z variable
of 1.96 has cumulative
probability 0.9750.





x
z
Step 2 (cont.)
5
.
0
2
39
40
has
)
2
,
39
(
~
from
40
value
the
example,
For



z
N
X
z-score = no. of σ-units above (positive z) or below
(negative z) distribution mean μ
Turn value into z score:

3. Sketch
4. Use Table B to lookup Pr(Z ≤ 0.5) = 0.6915
Steps 3 & 4: Sketch & Table B

a represents a lower boundary
b represents an upper boundary
Pr(a ≤ Z ≤ b) = Pr(Z ≤ b) − Pr(Z ≤ a)
Probabilities Between Points

Pr(-2 ≤ Z ≤ 0.5) = Pr(Z ≤ 0.5) − Pr(Z ≤ -2)
.6687 = .6915 − .0228
Between Two Points
See p. 144 in text
.6687 .6915
.0228
-2 0.5 0.5 -2

§7.3 Values Corresponding to
Normal Probabilities
1. State the problem
2. Find Z-score corresponding to
percentile (Table B)
3. Sketch
4. Unstandardize:

 p
z
x 


z percentiles
 zp ≡ the Normal z variable with
cumulative probability p
 Use Table B to look up the value of zp
 Look inside the table for the closest
cumulative probability entry
 Trace the z score to row and column

Notation: Let zp
represents the z score
with cumulative
probability p,
e.g., z.975 = 1.96
e.g., What is the 97.5th
percentile on the Standard
Normal curve?
z.975 = 1.96

Step 1: State Problem
Question: What gestational length is
smaller than 97.5% of gestations?
• Let X represent gestations length
• We know from prior research that
X ~ N(39, 2)
• A value that is smaller than .975 of
gestations has a cumulative probability
of.025

Step 2 (z percentile)
Less than 97.5%
(right tail) = greater
than 2.5% (left tail)
z lookup:
z.025 = −1.96
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
–1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233

35
)
2
)(
96
.
1
(
39 




 
 p
z
x
The 2.5th percentile is 35 weeks
Unstandardize and sketch

7.4 Assessing Departures
from Normality
Same distribution on
Normal “Q-Q” Plot
Approximately
Normal histogram
Normal distributions adhere to diagonal line on Q-Q
plot

Negative Skew
Negative skew shows upward curve on Q-Q plot

Positive Skew
Positive skew shows downward curve on Q-Q plot

Same data as prior slide with
logarithmic transformation
The log transform Normalize the skew

Leptokurtotic
Leptokurtotic distribution show S-shape on Q-Q plot

Gerstman_PP07.ppt

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