Download 05 Normal distribution and binomial distri

Lecture 5 probability model
normal distribution & binomial distribution
[email protected]
Contents
 Normal distribution for continuous data
 Binomial distribution for binary categorical
data
2
The Normal Distribution
The most important distribution in statistics.
Normal distribution
 Introduction to normal distribution




History
Parameters and shape
standard normal distribution and Z score
Area under the curve
 Application
 Estimate of frequency distribution
 Reference interval (range) in health_related
field.
4
histroy-Normal Distribution





Johann Carl Friedrich Gauss
Germany
One of the greatest mathematician
Applied in physics, astronomy
Gaussian distribution
(1777~1855)
Mark and Stamp in memory of Gauss.
5
The Most Important Distribution
 Many real life distributions are
approximately normal. such as height,
EFV1,weight, IQ, and so on.
 Many other distributions can be almost
normalized by appropriate data
transformation (e.g. taking the log). When
log X has a normal distribution, X is said to
have a lognormal distribution.
6
Frequency distributions of heights of adult men.
(a)
(b)
(c)
(d)
7
Sample & Population
 Histogram the area of the bars
 Cumulative relative
frequency
 in the sample, the
proportion of the boys
of age 12 that are lower
than a specified height.





normal distribution curve
The area under the curve
The cumulative probability.
In the population.
Generally speaking, the
chance that a boy of aged
12 is lower than a
specified height if he grow
normally
8
Definition of Normal distribution
 X ～ N(,2), X is distributed as normal
distribution with mean  and variance 2.
 The probability density function (PDF)
f (x) for a normal distribution is given by
f (X) 
1
 2

e
( X   )2
2 2
(- < X < +)
Where: e = 2.7182818285, base of natural logarithm
 = 3.1415926536, ratio of the circumference
of a circle to the diameter.
9
The shape of a normal distribution
.4
f (X) 
f(x)
1
 2

e
( X   )2
2 2
.3
.2
.1
0
x
10
The normal distributions with the equal variance
but different means
3
1
2
11
The normal distributions with the same mean
but different variances
2
1
3

12
Properties Of Normal Distribution
 & completely determine the characterization of the
normal distribution.
 Mean, median , mode are equal
 The curve is symmetric about mean.
 The relationship between  and the area under the
normal curve provides another main characteristic of the
normal distribution.
13
Areas under the Standard Normal Curve
 A variable that has a normal distribution with mean 0
and variance 1 is called the standard normal variate
and is commonly designated by the letter Z.
 N(0,1)
 As with any continuous variable, probability
calculations here are always concerned with finding
the probability that the variable assumes any value in
an interval between two specific points a and b.
14
Cumulative distribution Function (
 the area under the curve)
from -∞ to x, cumulative
Probability
S(-, )=1
 Example: What is the probability
of obtaining a z value of 0.5 or
less?
 We have
15
Area under standard normal distribution (Z)
Z
-3.0
-2.5
-2.0
-1.9
-1.6
-1.0
-0.5
0
0.00
0.0013
0.0062
0.0228
0.0287
0.0548
0.1587
0.3085
0.5000
-0.02
0.0013
0.0059
0.0217
0.0274
0.0526
0.1539
0.3015
0.4920
-0.04
0.0012
0.0055
0.0207
0.0262
0.0505
0.1492
0.2946
0.4840
-0.06
0.0011
0.0052
0.0197
0.0250
0.0485
0.1446
0.2877
0.4761
-0.08
0.0010
0.0049
0.0188
0.0239
0.0465
0.1401
0.2810
0.4681
Z 0
Z is the standard score, that is the units of standard deviation.
16
Figure Standard normal curve and some
important divisions.
•P(-1 < z < 1)=0.6826
•P(-2 < z < 2)=0.9545
•P(-3 < z < 3)=0.9974
17
Find probability in Excel
 Using an electronic table, find the area under the
standard normal density to the left of 2.824.
 We use the excel2007 function NORMSDIST
evaluated at 2.824 [NORMSDIST(2.824)]with the
result as follows:
18
EXAMPLE
 What is the probability of obtaining a z
value between 1.0 and 1.58?
 We have
19
CUMULATIVE PROBABILITY FOR
X~N(μ,σ2)
 Z=(X-μ)/σ
-3 -2 -
X= μ+Zσ

+ +2 +3
x
20
Areas under the Normal Curve
S(-, )=1
+1
+3)=0.6587
+2
)=0.9987
)=0.9772
S(-, )=0.5
-3)=0.1587
-2
-1
)=0.0013
)=0.0228
21
-3 -2 -
-4
-3
-2
-1

0
+ +2 +3
1
2
3
4
x
Z
Area Under Normal Curve
S(-, -3)=0.0013
S(-3, -2)=0.0115
S(-, -2)=0.0228
S(-2, -1)=0.1359
S(-, -1)=0.1587
S(-1, 
)=0.3413
S(-, -0)=0.5
-3
-3
-2
-

-3
-
-2
+
+2
-2
-1
0
1
2
 +
+3
+2
+3
3
22
Area Under Normal Curve
95%
2.5%
2.5%

-1.96
+1.96
23
-3
-2
-1
0
1
2
3
Area Under Normal Curve
90%
5%
5%

-1.64
+1.64
24
-3
-2
-1
0
1
2
3
Area Under Normal Curve
99%
0.5%
0.5%
-2.58
+2.58

25
-3
-2
-1
0
1
2
3
Area Under Normal Curve
95%
2.5%
2.5%

-1.96
+1.96
26
-3
-2
-1
0
1
2
3
95% heights of females will fall in the range between
mean -1.96SD and mean +1.96SD and
Z score, Standard Score
 Transform N(,2) to N(0,1z is refer to as
Standard Normal score
 How many SD’s the observation from the
mean?
 Transformation of a normal distribution such that
the units are in SD’s. (z score, Standard Score)
 By the units of SD, we can compare the
observations from diff population.
A female with height 172 cm
a male with height 172 cm
28
Values of variable
& area under curve
Observation distributed as
AUC
Standard normal score (Z)
normal (x)
（probability）
μ-1σ～μ＋1σ
-1～＋1
68.27%
μ-1.96σ～μ＋1.96σ
-1.96~1.96
95.00%
μ-2.58σ～μ＋2.58σ
-2.58~2.58
99.00%
The area that falls in the interval under the
nonstandard normal curve is the same as that
under the standard normal curve within the
corresponding u-boundaries.
29
The Most Important Distribution
 In practice Many real life distributions are
approximately normal, such as height, weight, IQ,
GB and so on
 In theory Many other distributions can be almost
normalized by appropriate data transformation (e.g.
taking the log);
•30
30
Summarizing
 The fundamental probability distribution of
statistics.
 A very important distribution both in theory and in
practice.
 The normal distribution has a set of curves.
Defined by mean and SD. (infinite)
 N(0,1) is unique.
 The areas under normal curve are equal when
measured by standard deviation.
31
Applications of Normal distribution
 Estimate frequency distribution
 Estimate Reference Range
32
Estimate frequency distribution
Example:
 IF the distribution of birth weights follows a
normal distribution with mean 3150g, and
standard deviation is 350g。
 To estimate what proportion of infants whose
birth weight are less than 2500g?
33
Solve for the Example:
 The standard normal deviate if x=2500:
Z=(x-3150)/350=-1.86
 The probability when Z<-1.86 under the standard
normal distribution :
ϕ(-1.86)=P(z<-1.86)=0.0314
 Result: there are about 3.14% infants whose birth
weight are less than 2500g.
34
Estimate Frequency Distribution
0.0314
2500
3150

35
Using Normal Distribution
 For any variables distributed as normal
distribution, 95% individuals assume values
between μ-1.96σ～μ＋1.96σ;
 99% between -2.58～ +2.58 ;
 And so on.
36
Reference Interval( Range)
 In health-related fields, a reference range or reference
interval usually describes the variations of a measurement or
value in healthy individuals.
 It is a basis for a physician or other health professional to
interpret a set of results for a particular patient.
 The standard definition of a reference range (usually referred
to if not otherwise specified) basically originates in what is
most prevalent in a reference group taken from the
population. However, there are also optimal health ranges that
are those that appear to have the optimal health impact on
people.
Reference Interval( Range)
 What is ?
 A range of values within which majority of measurements from “normal” subjects will lie.
 Majority: 90%，95%，99%, etc.。
 Usage：
 Used as the basis for assessing the result of diagnostic
tests in clinic. (normal? abnormal?)
 Definitions of “Normal subject”:
 Normal  Healthy
 maybe suffer from other diseases, but do not influence
the variable we studied.
38
How to estimate a reference interval?





Homogeneity of normal subjects. 100
Measurement errors are controlled
One side? Two sides?
Majority? 90%,95%?
Is it necessary to estimated RI in subgroups?
(considerations of partitioning based on age, sex etc)
 Determine the suspect range if necessary
39
Two-side or One-side
 Determined by medical professional.
 Two-side:
 WBC, BP, serum total cholesterol, ……
 One-side:
 Upper Limit : urine Ld, hair Hg, …Normal as long
as lower than

Low Limit:
Vital Capacity, IQ,
FEV1 (forced expiratory volume in one second)
 Normal as long as great than
40
Overlap distributed of observations for
Normal and Abnormal (one-side)
Normal Subject
False-negative
rate
False-positive
rate
Abnormal
界值
41
Overlap distributed of observations for
Normal and Abnormal (one-side)
Normal Subject
False-negative
rate
False-positive
rate
Abnormal
42
Overlap distributed of observations for
Normal and Abnormal (two-side)
False-negative
rate
False-positive
rate
Normal Subject
Abnormal
Abnormal
43
Normal approximate method
 For normally distributed
data
 A 95% reference interval
 Two-side:
 One-side:
X  1.96 s
For upper limit:
For low limit:
X  1.64 s
X  1.64 s
Percentile Method
 For non-normally distributed data
 A 95% reference interval
 Two-side:
 One-side:
P2.5 ~ P97.5
For upper limit:
For low limit:
<P95
>P5
45
Example
 Hb (hemoglobin) for 360 normal male.
 The mean is 13.45 g/100ml;
 The standard deviation is 0.71 g/100ml;
 Hb is normally distributed.
 Estimate the 95% reference range
and the 90% reference range.
46
Example (cont.)
 Two side
X  1.96 s
X  1.96 s  13.45  1.96  0.71  12.06 (g/100ml)
X  1.96 s  13.45  1.96  0.71  14.84 (g/100ml )
 The 95% reference range is
12.06～14.84 (g/100ml)
47
Example (cont.)
 Two side
X  1.64 s
X  1.64 s  13.45  1.64  0.71  12.29 (g/100ml)
X  1.64 s  13.45  1.64  0.71  14.61 (g/100ml)
The 90% reference range is 12.29～14.61 (g/100ml)
The 95% reference range is 12.06～14.84 (g/100ml)
48
Two methods for reference intervals.
Method
two-side
One-side
Low
Normal
Percentile
X  u / 2 s X  u s
P2.5～P97.5
>P5
Upper
X  u s
<P95

49
Central Limit Theorem
 As a sample size increased, the means of
samples drawn from a population of and
distribution will approach the normal
distribution. This theorem is known as the
central limit theorem (CLT).
 That is Sampling distributions
 Probability and the central limit theorem
50
Sampling distribution
 A sampling distribution is the probability
distribution of a sample statistic that is
formed when samples of size n are
repeatedly taken from a population.
 The sampling distribution of sample means
51
Binomial Distribution
Probability Model for discrete data
52
Review
 binary qualitative data
 rate-incidence /proportion-prevalence
53
Tossing coin
 What’s the probability that you flip exactly 3
heads in 5 coin tosses?
54
•P(3 heads & 2 tails) =5C3*P(heads)3*P(tails)2
•=10*(0.5)3(0.5)2=31.25%
•
 
 
 3
5
• ways to
arrange 3
heads in
5 trials
Outcome
Probability
THHHT
(1/2)3 * (1/2)2
HHHTT
(1/2)3 * (1/2)2
TTHHH
(1/2)3 * (1/2)2
HTTHH
(1/2)3 * (1/2)2
HHTTH
(1/2)3 * (1/2)2
HTHHT
(1/2)3 * (1/2)2
THTHH
(1/2)3 * (1/2)2
HTHTH
(1/2)3 * (1/2)2
HHTHT
(1/2)3 * (1/2)2
THHTH
(1/2)3 * (1/2)2
•5C3 = 5!/3!2! = 10 10 arrangements
•The
probability of
each unique
outcome
(note: they
are all equal)
(1/2)3 *(1/2)2
•Factorial review: n! = n(n-1)(n-2)…
55
Binomial distribution function:
X= the number of heads tossed in 5 coin tosses
•p(x)
•0 •1 •2
•3 •4 •5
•x
•number of heads
56
Example for side effect of drug
 if a certain drug is known to cause a side effect
10% of the time and if five patients are given this
drug, what is the probability that four or more
experience the side effect?
 Let S denote a side-effect outcome and N an outcome
without side effects.
57
Table
58
Solution to example
 The probability of obtaining
an outcome with four S’s
and one N is
 The probability of obtaining
all five S’s is
 the probability of the
compound event that ‘‘four
or more have side effects is
59
probability density function(PDF)
 The model is concerned with the total
number of successes in n trials as a
 random variable, denoted by X. Its
probability density function is given by
the number of combinations of x objects
selected from a set of n
60
Assumptions for Binomial Distribution
 The experiment consists of n repeated trials
satisfying these assumptions:
 1. The n trials are all independent.
 2. The parameter p of one in 2 is the same for
each trial.
61
The mean and variance of the
binomial distribution
 x  n
 x  n (1   )
2
 x  n (1   )
when the number of trials n is from moderate to
large (n > 25, say), we approximate the binomial
distribution by a normal distribution and answer
probability questions by first converting to a
standard normal score:
 where π is the probability of having a positive
outcome from a single trial
62
Solution to Example
 For π =0.1 and n =30, we have
63
PDF
•n••=20•
••=0.5•
•0•.4•
•n••=5•
••=0.3•
•n••=10•
••=0.3•
•n••=30•
••=0.3•
P•(•X•)•
•0•.3•
•0•.2•
•0•.1•
•0•.0•
•4•
•8•
•12•
•16•
•0• •2• •4•
•0• •2• •4• •6•
•X•
•4•
•8•
•12•
•16•
65
Review – experiment & survey
 2 type of researches_ experimental and
observational research
 Clinical trial (4 phases)
 Statistical consideration in clinical trial
 Controlled /Randomization/blindness/
replication (appropriate sample size).
 probabilistic sampling techniques
66
Review on idea of probability
67
Idea of probability
 Definitions of probability



Classic probability- If a random experiment can result in n possible
mutually exclusive and equally likely outcomes and if nA of these
outcomes have an attribute A, then the probability (Pr) of A is
written as nA /n
Statistical probability-If an experiment is performed n times and if
nA of these result in the outcome A, then the probability of A
occurring is defined as the limiting ratio: P(A)=nA/n
Subjective probability-Probability represents one’s belief regarding
the likelihood of an outcome A occurring
 Probability of Event = p
0 <= p <= 1
68
Rule for Computing
 If A and B have no outcomes in common, they can not
occur simultaneously, they are Mutually Exclusive
events
P(A or B) = P(A) + P(B)
 if events A & B are independent
then P(A&B) = P(A)*P(B)
69
Conditional Probability
 Concern the odds of one event occurring,
given that another event has occurred
 P(A|B)=Prob of A, given B
 if A and B are independent, then
P(B|A) = P(A)*P(B)/P(A)
P(B|A) = P(B)
70
Percentile calculation
71
Quartiles

Quartiles divide data into four equal parts
 First quartile—Q1


25% of observations are below Q1 and 75% above Q1
Also called the lower quartile
 Second quartile—Q2


50% of observations are below Q2 and 50% above Q2
This is also the median
 Third quartile—Q3


75% of observations are below Q3 and 25% above Q3
Also called the upper quartile
Calculating percentiles
Example
The sorted observations are:
2,5, 9, 12, 14,15,18,24,60,find the median
and P20.
Solution
The number of observations n = 9
12-73
P50  X ( n1)50%  X 5  14
PX  X ( n 1) X %
P20  X ( n1)20%  X 2  5
i  (n  1)  X %
Calculating percentiles
The sorted observations are:
4, 9, 10, 12,14,20,24,61, Find the median
and P20.
 (n+1)*20%=1.8
PX  X ( n 1) X %
PX  X j  ( X j 1  X j )  (i  j )
P20  X 1  ( X 2  X 1 )  (1.8  1)  4  (9  4)  0.8  8
12-74
•Calculation of percentile from a
grouped frequency table
•Example: The frequency distribution for the systolic blood pressure readings (in
mm or mercury) of 200 randomly selected college students is shown here.
Boundaries
Frequency
cumulative
frequency
cumulative percent(%)
89.5-
24
24
12
104.5-
62
86
43
119.5-
72
158
79
134.5-
26
184
92
149.5-
12
196
95
164.5-
4
200
100
The class interval that contains the relevant quartile is called the
quartile class
75
Calculation of quartiles from a grouped
frequency table
n

 C
4
 i 
Q1  L  
f
 3n

  C
4
 i 
Q3  L  
f
where:
L = the real lower limit of the quartile class (containing Q1 or Q3)
n = Σf = the total number of observations in the entire data set
C = the cumulative frequency in the class immediately before the
quartile class
f = the frequency of the relevant quartile class
i = the length of the real class interval of the relevant quartile class
Q1  P25  104.8 
 111.09
12-76
50  24 15 
62
•Calculation of percentile from a
grouped frequency table
class
Frequency
(f)
cumulative
frequency(C)
cumulative
percent(%)
89.5-
24
24
12
104.5-
62
86
43
119.5-
72
158
79
134.5-
26
184
92
149.5-
12
196
95
164.5-
4
200
100
n

 C
4
 i 
Q1  L  
f
The class interval that contains the relevant quartile is called the
quartile class
Q1  P25  104.8 
50  24 15   111.09
62
77
•Calculation of quartiles from a grouped frequency table
PX

n X %  C
i 
 L
f
78