7
Theoretical Probability
Distributions
Random variables (RV)
Represented by X,Y, or Z
Discrete or continuous RV
Discrete RV  martial status: single, married, divorced
Continuous RV  weight, height
7.1 Probability distributions
Every RV has a corresponding
probability distribution.
X = the birth order of each child born to
a woman residing in US
X = 1, 2  first-born, second-born child
Let X = the RV, x = the outcome of a
particular child
P(X=4) = 0.058
P(X=1 or X=2) = 0.416 + 0.330 = 0.746
Chapter7 p163
7.1 Probability distributions
Probability distribution of
Table 7.1 data.
Probabilities that are
calculated (from a finite
amount of data, based on
theoretical consideration) are
called (empirical, theoretical)
probabilities.
Chapter7 p164
7.2 The binomial distribution
Dichotomous RV, Y = life and death, male and female, sickness and health
Also known as Bernoulli RV
Example
Y denotes smoking status, Y=0,1  non-smoking, smoking
In 1987, 29% of the adults in the US smoked cigar, cigarettes or pipes
P(Y=1) = p = 0.29  P(Y=0) = 0.71
X denotes the number of persons selected from the population of adults in the
US  X can take on three possible values: 0, 1, 2
P(X=0) = (1-p)(1-p) = 0.504
P(X=1) = p(1-p) + (1-p)p = 0.412
P(X=2) = p*p = 0.084
P(X=0) + P(X=1) + P(X=2) = 1
7.2 The binomial distribution
X would be a binomial RV with parameters n=3 and p=0.29
P(X=0) = (1-p) (1-p) (1-p) = 0.358
P(X=1) = p(1-p) (1-p) + (1-p)p (1-p) + (1-p) (1-p)p = 0.439
P(X=2) = p*p (1-p) + p (1-p)p + (1-p)p*p = 0.179
P(X=3) = p*p*p = 0.024
In case X=n
P( X  x)  C xn p x (1  p) n  x
(mean,variance) of X = (np, np(1-p))
For n=10, (np, np(1-p)) = (10*0.29, 10*0.29*(0.71) = (2.9, 2.059)
Skew to right
=
=
Chapter7 p171
symmetric
=
=
Chapter7 p172
7.3 The Poisson distribution
When n>>1, and p is very small, such as p = the probability of a
person involved in a motor vehicle accident each year in the US =
0.00024
The Poisson distribution is used to model disctete events that occur
infrequently in time or space.
e  l lx
P( X  x) 
x!
X is said to have a Poisson distribution with parameter l
7.3 The Poisson distribution
Binomial distribution, np, np(1-p), if p <<1
np, np  mean = variance
Example
Determine the number of people in a population of 10000 who
will be involved in a motor vehicle accident per year
l = 10000*0.00024 = 2.4
e 2.4 (2.4) 0
P( X  0) 
 0.091
0!
e  2.4 (2.4)1
P( X  1) 
 0.218
1!
e  2.4 (2.4) 2
P ( X  2) 
 0.261
2!
l
Chapter7 p175
The Poisson distribution is highly skewed for small l, as l
increases, the distribution becomes more symmetric.
Chapter7 p175
The Poisson distribution is highly skewed for small l, as l increases,
the distribution becomes more symmetric.
Chapter7 p175
7.4 The Normal distribution
Discrete binomial or Poisson distribution
as n increases  Normal distribution
f ( x) 
1
e
2 
1  x 
 

2  
2
where -∞＜x＜ ∞
=
=
Chapter7 p177
7.4 The Normal distribution
Change of variable  standard normal distribution
Z
x

1  12 z 2
 f ( z) 
e
2
With mean =0, variance 2= 1
Chapter7 p177
=
-
Chapter7 p179
=
Figure 7.10 The standard normal curve, area between z = -2.00 and z = 2.00
Chapter7 p180
=
Chapter7 p181
Normal
distribution
table
NORMDIST - Area under the curve start from left
hand side
Z=0
Z=2
X 2
Z
0 .5
Chapter7 p181
Let X = systolic blood pressure. For the population of 18- to 74-year-old males in the US,
systolic 收縮的 blood pressure is distributed with a mean 129 mm Hg and standard
deviation 19.8 mm Hg.
Find the value of x that cuts off the upper 2.5% of the curve of systolic blood pressure,
Find P(X>x) = 0.025
for the upper 2.5%
z = 1.96 = (x – 129)/ 19.8
 x = 167.8 mm Hg
Symmetric (the lower 2.5%)
z = -1.96
 x = 90.2 mm Hg
Chapter7 p182
Comparison of two normal distributions (ND)
Not taking corrective medication, diastolic 舒張 blood
pressure is approximately ND with
mean = 80.7 mm Hg, s.d = 9.2 mm Hg
For the men using antihypertensive drugs, with
mean = 94.9 mm Hg, s.d = 11.5 mm Hg
Example
Identify 90% of the persons who are currently taking
medication, what value of diastolic blood pressure
should be designated as the lower cutoff point ?
From Table, lower 10%  z = -1.28  x = 80.2 mm Hg
Below 80.2 mm Hg represent FN
Person who is taking medication are not identified as
such
Other probability distributions
Negative binomial distribution, multi-nomial distribution,
hypergeometric distribution
Negative binomial distribution
When X=x, among the previous x-1 test, r-1 times are success, x-r
times are failure
f ( x)  C xxr1 p r (1  p) x r , x  r , r  1,...
  r / p, _ and _  2  rq / p 2
Example
A telegraph system has a probability of 0.1sending wrong message.
What is the probability that the 10th message is the third error ?
1
3
103
f ( x  10)  C 10
0
.
1
(
1

0
.
1
)
 0.0172
x103
Multi-nomial distribution
n independent tests, each test has r types of outcome, where each
type has a probability of occurrence p1, ….., pr. Let the RV be
X=(X1, ….Xr).
f ( x1 ,....., xr )  C xn1 p1x1 C xn2 x1 p2x2 ....C xxrr prxr 
n!
p1x1 ..... prxr
x1! x2 !......xn !
i  npi _ and _  i2  npi ( pi )
Example
A dice is thrown 10 times, what is the probabilities that number
1,3 and 5 occur 2,3,and 5 times respectively ?
2
f (2,0,3,0,5,0) 
3
5
10!  1   1   1 
5
       4.1676 10
2!3!5!  6   6   6 
Hypergeometric distribution
N balls, R red color balls, N-R white color balls,
RV, X = n balls are drawn without replacement  X is said to have
hypergeometric distribution - the probability of having x red ball
from R red balls, and n-x white ball for N-R white balls.
C xN CnNxR
f ( x) 
CnN
  nR / N , _ and _  2 
nR  R  N  n
1  
N  N  N 1
Example
A cargo of 50 goods, 5 are defected and 45 are good. Five pieces
are drawn, what is the probability of identify defected goods ?
P(X≧1) = 1 – P(X≦0) = 1-f(0)
C05C5505
 0.423
50
C5
7.5 Further applications
Chapter7 p189
Chapter7 p190
Chapter7 p171