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7: Normal Probability Distributions 1
June 23
Chapter 7:
Normal Probability
Distributions
7: Normal Probability Distributions 2
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: Normal Probability Distributions 3
§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
7: Normal Probability Distributions 4
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
7: Normal Probability Distributions 5
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
7: Normal Probability Distributions 6
Parameters μ and σ
• Normal pdfs have two parameters
μ - expected value (mean “mu”)
σ - standard deviation (sigma)
σ controls spread
μ controls location
7: Normal Probability Distributions 7
Mean and Standard Deviation
of Normal Density
μ
σ
7: Normal Probability Distributions 8
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
7: Normal Probability Distributions 9
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!
7: Normal Probability Distributions 10
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
7: Normal Probability Distributions 11
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
7: Normal Probability Distributions 12
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…
7: Normal Probability Distributions 13
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)
7: Normal Probability Distributions 14
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…
7: Normal Probability Distributions 15
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
7: Normal Probability Distributions 16
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: Normal Probability Distributions 17
§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
7: Normal Probability Distributions 18
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) = ?
7: Normal Probability Distributions 19
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
7: Normal Probability Distributions 20
Example: A Z variable
of 1.96 has cumulative
probability 0.9750.
7: Normal Probability Distributions 21




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:
7: Normal Probability Distributions 22
3. Sketch
4. Use Table B to lookup Pr(Z ≤ 0.5) = 0.6915
Steps 3 & 4: Sketch & Table B
7: Normal Probability Distributions 23
a represents a lower boundary
b represents an upper boundary
Pr(a ≤ Z ≤ b) = Pr(Z ≤ b) − Pr(Z ≤ a)
Probabilities Between Points
7: Normal Probability Distributions 24
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: Normal Probability Distributions 25
§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 

7: Normal Probability Distributions 26
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
7: Normal Probability Distributions 27
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
7: Normal Probability Distributions 28
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
7: Normal Probability Distributions 29
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
7: Normal Probability Distributions 30
35
)
2
)(
96
.
1
(
39 




 
 p
z
x
The 2.5th percentile is 35 weeks
Unstandardize and sketch
7: Normal Probability Distributions 31
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
7: Normal Probability Distributions 32
Negative Skew
Negative skew shows upward curve on Q-Q plot
7: Normal Probability Distributions 33
Positive Skew
Positive skew shows downward curve on Q-Q plot
7: Normal Probability Distributions 34
Same data as prior slide with
logarithmic transformation
The log transform Normalize the skew
7: Normal Probability Distributions 35
Leptokurtotic
Leptokurtotic distribution show S-shape on Q-Q plot

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Understanding Normal Probability Distributions

  • 1. 7: Normal Probability Distributions 1 June 23 Chapter 7: Normal Probability Distributions
  • 2. 7: Normal Probability Distributions 2 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
  • 3. 7: Normal Probability Distributions 3 §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
  • 4. 7: Normal Probability Distributions 4 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
  • 5. 7: Normal Probability Distributions 5 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
  • 6. 7: Normal Probability Distributions 6 Parameters μ and σ • Normal pdfs have two parameters μ - expected value (mean “mu”) σ - standard deviation (sigma) σ controls spread μ controls location
  • 7. 7: Normal Probability Distributions 7 Mean and Standard Deviation of Normal Density μ σ
  • 8. 7: Normal Probability Distributions 8 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
  • 9. 7: Normal Probability Distributions 9 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!
  • 10. 7: Normal Probability Distributions 10 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
  • 11. 7: Normal Probability Distributions 11 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
  • 12. 7: Normal Probability Distributions 12 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…
  • 13. 7: Normal Probability Distributions 13 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)
  • 14. 7: Normal Probability Distributions 14 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…
  • 15. 7: Normal Probability Distributions 15 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
  • 16. 7: Normal Probability Distributions 16 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
  • 17. 7: Normal Probability Distributions 17 §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
  • 18. 7: Normal Probability Distributions 18 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) = ?
  • 19. 7: Normal Probability Distributions 19 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
  • 20. 7: Normal Probability Distributions 20 Example: A Z variable of 1.96 has cumulative probability 0.9750.
  • 21. 7: Normal Probability Distributions 21     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:
  • 22. 7: Normal Probability Distributions 22 3. Sketch 4. Use Table B to lookup Pr(Z ≤ 0.5) = 0.6915 Steps 3 & 4: Sketch & Table B
  • 23. 7: Normal Probability Distributions 23 a represents a lower boundary b represents an upper boundary Pr(a ≤ Z ≤ b) = Pr(Z ≤ b) − Pr(Z ≤ a) Probabilities Between Points
  • 24. 7: Normal Probability Distributions 24 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
  • 25. 7: Normal Probability Distributions 25 §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  
  • 26. 7: Normal Probability Distributions 26 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
  • 27. 7: Normal Probability Distributions 27 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
  • 28. 7: Normal Probability Distributions 28 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
  • 29. 7: Normal Probability Distributions 29 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
  • 30. 7: Normal Probability Distributions 30 35 ) 2 )( 96 . 1 ( 39         p z x The 2.5th percentile is 35 weeks Unstandardize and sketch
  • 31. 7: Normal Probability Distributions 31 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
  • 32. 7: Normal Probability Distributions 32 Negative Skew Negative skew shows upward curve on Q-Q plot
  • 33. 7: Normal Probability Distributions 33 Positive Skew Positive skew shows downward curve on Q-Q plot
  • 34. 7: Normal Probability Distributions 34 Same data as prior slide with logarithmic transformation The log transform Normalize the skew
  • 35. 7: Normal Probability Distributions 35 Leptokurtotic Leptokurtotic distribution show S-shape on Q-Q plot

Notes de l'éditeur

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