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Ch 7 & 8 實習
Random Variables and Probability
Distributions

A random variable is a function or rule that
assigns a numerical value to each simple event
in a sample space.

為了降低分析的複雜性，將所有可能結果加以數值化
例如投銅板十次，正面出現次數的事件就是random
variable

短期不知道是什麼，長期下來會呈現某種分配


There are two types of random variables:


2
Discrete random variable
Continuous random variable
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Discrete Probability Distribution

A table, formula, or graph that lists all possible
values a discrete random variable can assume,
together with associated probabilities, is called
a discrete probability distribution.

To calculate the probability that the random variable
X assumes the value x, P(X = x),



3
add the probabilities of all the simple events for which X is
equal to x, or
Use probability calculation tools (tree diagram),
Apply probability definitions
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Example 1


The number of cars a dealer is selling daily were
recorded in the last 100 days. This data was
summarized in the table below.
Estimate the probability distribution, and determine
the probability of selling more than 2 cars a day.
Daily sales Frequency
0
5
1
15
2
35
3
25
4
20
100
4
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Solution

From the table of frequencies we can calculate
the relative frequencies, which becomes our
estimated probability distribution
Daily sales Relative Frequency
0
5/100=.05
1
15/100=.15
2
35/100=.35
3
25/100=.25
4
20/100=.20
1.00
5
.35
.25
.15
.20
.05
0
1
2
3
4
X
P(X>2) = P(X=3) + P(X=4) =
.25 + .20 = .45
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Describing the Population/
Probability Distribution



6
The probability distribution represents a population
We’re interested in describing the population by
computing various parameters.
Specifically, we calculate the population mean and
population variance.
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Population Mean (Expected Value) and
Population Variance

Given a discrete random variable X with values xi, that
occur with probabilities p(xi), the population mean of X is.

加權平均概念(權數是機率)
E(X)  m   x i  p( x i )
all xi

Let X be a discrete random variable with possible values
xi that occur with probabilities p(xi), and let E(xi) = m. The
variance of X is defined by
V ( X )   2  E[( X - m) 2 ]  ( x i - m) 2 p( x i )
all xi
7
The s tan dard deviation is
  2
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The Mean and the Variance

The variance can also be calculated as follows:
V ( X )   2  E( X 2 ) - m 2 
2
2
x
p
(
x
)
m
 i i
all xi

Proof
E[( X - m ) 2 ]
 E[ X 2 - 2 X m  m 2 ]
 E ( X 2 ) - 2m E ( X )  E ( m 2 )
 E( X 2 ) - m 2
8
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Laws of Expected Value and Variance

Laws of Expected Value




Laws of Variance



9
E(c) = c
E(X + c) = E(X) + c
E(cX) = cE(X)
V(c) = 0
V(X + c) = V(X)
V(cX) = c2V(X)
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Example 2

We are given the following probability distribution:
x
0
P(x) 0.4




10
1
2
3
0.3
0.2
0.1
a. Calculate the mean, variance, and standard deviation
b. Suppose that Y=3X+2. For each value of X, determine the value
of Y. What is the probability distribution of Y?
c. Calculate the mean, variance, and standard deviation from the
probability distribution of Y.
d. Use the laws of expected value and variance to calculate the
mean, variance, and standard deviation of Y from the mean,
variance, and standard deviation of X. Compare your answers in
Parts c and d. Are they the same (except for rounding)
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Solution

a.
E(X) = 0(0.4) +1(0.3) + 2(0.2) +3(0.1) = 1
V(X)=E(X2)-{E(X)}2


=(0)2(0.4)+(1)2(0.3)+(2)2(0.2)+(3)2(0.1)-(1)2 =1
 (X ) 1


b.
x
y
P(y)
11
0
2
0.4
1
5
0.3
2
8
0.2
3
11
0.1
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Solution

c.


E(Y) = 2(0.4) + 5(0.3) + 8(0.2) + 11(0.1) = 5
V(Y)=E(Y2)-{E(Y)}2
=(2)2(0.4)+(5)2(0.3)+(8)2(0.2)+(11)2(0.1)-(5)2 =9


d.


12
 (Y )  3
E(Y) = E(3X+2) = 3E(X)+2 = 5
V(Y) = V(3X+2) = 9V(X) = 9
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Bivariate Distributions

The bivariate (or joint) distribution is used when
the relationship between two random variables is
studied.


也就是第六章所看到的聯合機率分配
The probability that X assumes the value x, and Y
assumes the value y is denoted
p(x,y) = P(X=x and Y = y)
The joint probabilit y function
satisfies the following conditions :
1. 0  p(x,y)  1
2.   p(x,y)  1
13
all x all y
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Bivariate Distributions

14
Example 7.5

Xavier and Yvette are two real estate agents.
Let X and Y denote the number of houses that
Xavier and Yvette will sell next week,
respectively.

The bivariate probability distribution is
presented next.
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Bivariate Distributions
0.42
p(x,y)
Example 7.5 –
continued
Y
0
1
2
0.21
X
1
.42
.06
.02
0
.12
.21
.07
2
.06
.03
.01
0.12
0.06
0.06
0.07
0.02
0.01
Y
15
X=0
y=0
X
0.03
y=1
y=2
X=1
X=2
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Marginal Probabilities

Example 7.5 – continued

Sum across rows and down columns
p(0,0)
p(0,1)
p(0,2)
Y
0
1
2
p(x)
0
.12
.21
.07
.40
X
1
.42
.06
.02
.50
2
.06
.03
.01
.10
p(y)
.60
.30
.10
1.00
P(Y=1), the
marginal
probability.
The marginal probability P(X=0)
16
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Describing the Bivariate Distribution


The joint distribution can be described by the
mean, variance, and standard deviation of
each variable.
This is done using the marginal distributions.
x
0
1
2
p(x)
.4
.5
.1
E(X) = .7
V(X) = .41
17
y
0
1
2
p(y)
.6
.3
.1
E(Y) = .5
V(Y) = .45
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Describing the Bivariate Distribution

To describe the relationship between the
two variables we compute the covariance
and the coefficient of correlation

Covariance:


Coefficient of Correlation

18
COV(X,Y) = S(X – mx)(Y- my)p(x,y)=E(XY)-E(X)E(Y)
COV(X,Y)
xy
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Describing the Bivariate Distribution

Example 7.6


Calculate the covariance and coefficient of correlation between
the number of houses sold by the two agents in Example 7.5
Solution
= S(x-mx)(y-my)p(x,y) =
(0-.7)(0-.5)p(0,0)+…(2-.7)(2-.5)p(2,2) = -.15
 r=COV(X,Y)/xy = - .15/(.64)(.67) = -.35
 COV(X,Y)
Y
0
1
2
p(x)
19
0
.12
.21
.07
.40
X
1
.42
.06
.02
.50
2
.06
.03
.01
.10
p(y)
.60
.30
.10
1.00
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Sum of Two Variables

The probability distribution of X + Y is determined by



Example 7.5 - continued


20
Determining all the possible values that X+Y can assume
For every possible value C of X+Y, adding the probabilities of all
the combinations of X and Y for which X+Y = C
Find the probability distribution of the total number of houses sold
per week by Xavier and Yvette.
Solution
 X+Y is the total number of houses sold. X+Y can have the
values 0, 1, 2, 3, 4.
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The Probability Distribution of X+Y
P(X+Y=0) = P(X=0 and Y=0) = .12
P(X+Y=1) = P(X=0 and Y=1)+ P(X=1 and Y=0) =.21 + .42 = .63
P(X+Y=2) = P(X=0 and Y=2)+ P(X=1 and Y=1)+ P(X=2 and Y=0)
= .07 + .06 + .06 = .19
Y
0
1
2
p(x)
0
.12
.21
.07
.40
X
1
.42
.06
.02
.50
2
.06
.03
.01
.10
p(y)
.60
.30
.10
1.00
The probabilities P(X+Y)=3 and P(X+Y) =4 are
calculated the same way. The distribution follows
21
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The Expected Value and Variance of X+Y

The distribution of X+Y
x+y
p(x+y)

1
.63
2
.19
3
.05
4
.01
The expected value and variance of X+Y can
be calculated from the distribution of X+Y.


22
0
.12
E(X+Y)=0(.12)+ 1(63)+2(.19)+3(.05)+4(.01)=1.2
V(X+Y)=(0-1.2)2(.12)+(1-1.2)2(.63)+… =.56
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The Expected Value and Variance of X+Y

The following relationship can assist in calculating
E(X+Y) and V(X+Y)
E(X+Y) =E(X) + E(Y);
 V(X+Y) = V(X) +V(Y) +2COV(X,Y)
 When X and Y are independent COV(X,Y)
= 0, and V(X+Y) = V(X)+V(Y).
 Proof

Var ( X  Y )  E[( X  Y - m x - m y ) 2 ]
 E[( X - m x ) 2  (Y - m y ) 2  2( X - m x )(Y - m y )]
 Var ( X )  Var (Y )  2Cov( X , Y )
23
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Example 3

The bivariate distribution of X and Y is described here.
x





24
y
1
2
1
0.28
0.42
2
0.12
0.18
a. Find the marginal probability distribution of X.
b. Find the marginal probability distribution of Y
c. Compute the mean and variance of X
d. Compute the mean and variance of Y
e. Compute the covariance and variance of Y
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Solution



25
a
x
P(x)
1
.4
2
.6
b y
P(y)
1
.7
2
.3
c E(X) = 1(.4) + 2(.6) = 1.6
V(X) = (1–1.6)^2*(.4) + (2–1.6)^2*(.6) = .24
or (1^2)*0.4+(2^2)*0.6-(1.6)^2=.24
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Solution



d E(Y) = 1(.7) + 2(.3) = 1.3
V(Y) = (1–1.3)^2(.7) + (2–1.3)^2(.3) = .21
e E(XY) = (1)(1)(.28) + (1)(2)(.12) +
(2)(1)(.42) + (2)(2)(.18) = 2.08
COV(X, Y) = E(XY)–E(X)E(Y)
= 2.08 – (1.6)(1.3) = 0
 r
26
COV (X, Y )
xy
=0
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The Binomial Distribution


The binomial experiment can result in only one
of two possible outcomes.
Binomial Experiment





Binomial Random Variable


27
There are n trials (n is finite and fixed).
Each trial can result in a success or a failure.
The probability p of success is the same for all the trials.
All the trials of the experiment are independent
The binomial random variable counts the number of successes
in n trials of the binomial experiment.
By definition, this is a discrete random variable.
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Calculating the Binomial Probability
In general, The binomial probability is calculated by:
P( X  x)  p( x)  C p (1 - p)
n
x
w here C nx
x
n- x
n!

x! (n - x )!
Mean and Variance of Binomial Variable
E(X)  m  np
V(X)  s 2  np(1 - p)
28
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Example 4

a.
b.
29
In the game of blackjack as played in casinos in
Las Vegas, Atlantic City, Niagara Falls, as well as
many other cities, the dealer has the advantages.
Most players do not play very well. As a result, the
probability that the average player wins a about
45%.Find the probability that an average player
wins
twice in 5 hands
ten or more times in 25 hands
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Solution
5!
2
5- 2
(.45)
(1
.45)
 a P(X = 2) =
= .3369
2!(5 - 2)!

30
b Excel with n = 25 and p = .45:
P(X  10) = 1 – P(X 9) = 1 – .2424 = .7576
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Poisson Distribution
31

The Poisson experiment typically fits cases of
rare events that occur over a fixed amount of
time or within a specified region

Typical cases
 The number of errors a typist makes per page
 The number of customers entering a service station
per hour
 The number of telephone calls received by a
switchboard per hour.
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Properties of the Poisson Experiment

The number of successes (events) that occur in a certain
time interval is independent of the number of successes that
occur in another time interval.

The probability of a success in a certain time interval is
 the same for all time intervals of the same size,
 proportional to the length of the interval.
The probability that two or more successes will occur in an
interval approaches zero as the interval becomes smaller.

32
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The Poisson Variable and Distribution

The Poisson Random Variable
 The Poisson variable indicates the number of
successes that occur during a given time interval
or in a specific region in a Poisson experiment

Probability Distribution of the Poisson Random
Variable.
e -mm x
P( X  x)  p( x) 
x!
E( X )  V( X )  m
33
x  0,1, 2...
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Example 5

a.
b.
34
The number of students who seek assistance with
their statistics assignments is Poisson distributed
with a mean of three per day.
What is the probability that no student seek
assistance tomorrow?
Find the probability that 10 students seek
assistance in a week.
Jia-Ying Chen
Solution

a. P(X = 0 with m = 3) =
e -m m x
x!
=
= .0498
e
 b. P(X = 10 with m = 21) =
-m
mx
x!
e-3 (3)0
0!
e -21 (21)10
10!
=
= .0035
35
Jia-Ying Chen
Continuous Probability Distributions

A continuous random variable has an uncountably
infinite number of values in the interval (a,b).

The probability that a continuous variable X will
assume any particular value is zero. Why?
1/4
1/3
1/2
0
36
The probability of each value
+
1/4
+
1/4
+
+
1/3
+
+
1/3
1/2
2/3
1/4 = 1
1/3 = 1
1/2 = 1
1
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Continuous Probability Distributions
As the number of values increases the probability of each
value decreases. This is so because the sum of all the
probabilities remains 1.
When the number of values approaches infinity (because X
is continuous) the probability of each value approaches 0.
1/4
1/3
1/2
0
37
The probability of each value
+
1/4
+
1/4
+
+
1/3
+
+
1/3
1/2
2/3
1/4 = 1
1/3 = 1
1/2 = 1
1
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Probability Density Function


To calculate probabilities we define a
probability density function f(x).
The density function satisfies the following
conditions


Area = 1
f(x) is non-negative,
P(x1<=X<=x2)
The total area under the curve representing f(x) equals 1.
x1 x2
• The probability that X falls between x1 and x2 is
found by calculating the area under the graph of f(x)
between x1 and x2.
38
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Uniform Distribution

A random variable X is said to be uniformly
distributed if its density function is
1
f ( x) 
a  x  b.
b-a

The expected value and the variance are
ab
E(X) 
2
39
(b - a) 2
V( X ) 
12
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Example 6

a.
b.
40
The weekly output of a steel mill is a uniformly
distributed random variable that lies between 110
and 175 metric tons.
Compute the probability that the steel mill will
produce more than 150 metric tons next week.
Deter the probability that the steel mill will produce
between 120 and 160 metric tons next week.
Jia-Ying Chen
Solution
1
1

 f(x) =
,110
(175 - 110 ) 65


41
a P(X ≧
≦ x ≦ 175
1
150) = (175 - 150 ) 65 = 0.3846
b P(120 ≦ X ≦ 160) =
1
(160 - 120 ) =
65
0.6154
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Normal Distribution

A random variable X with mean m and variance 2 is normally
distributed if its probability density function is given by
 x-m 
- (1 / 2 ) 




2
1
f ( x) 
e
-  x  
 2
where   3.14159... and e  2.71828...
N(m ,  2 )
f
1
2 
42
Jia-Ying Chen
Finding Normal Probabilities

Two facts help calculate normal probabilities:



“Standard Normal Distribution”

Example:

43
The normal distribution is symmetrical.
Any normal distribution can be transformed into
a specific normal distribution called…
The amount of time it takes to assemble a computer is
normally distributed, with a mean of 50 minutes and a
standard deviation of 10 minutes. What is the
probability that a computer is assembled in a time
between 45 and 60 minutes?
Jia-Ying Chen
Finding Normal Probabilities

Solution


If X denotes the assembly time of a computer,
we seek the probability P(45 ≦ X ≦ 60).
This probability can be calculated by creating a
new normal variable the standard normal
variable.
Every normal variable
with some m and , can
be transformed into this Z.
44
X - mx
Z
x
E(Z) = m = 0
Therefore, once probabilities for Z
are calculated, probabilities of any
normal variable can be found.
V(Z) = 2 = 1
Jia-Ying Chen
Finding Normal Probabilities

Example - continued
45 - 50
P(45 ≦ X ≦ 60) = P(
≦
10
X- m
≦

60 - 50
)
10
= P(-0.5 ≦ Z ≦ 1)
To complete the calculation we need to compute
the probability under the standard normal distribution
45
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46
Jia-Ying Chen
Finding Normal Probabilities

Example - continued
45 - 50
P(45 ≦ X ≦ 60) = P(
≦
10
X- m

≦
60 - 50
)
10
= P(-.5 ≦ Z ≦ 1)=0.8413-(1-0.6915)=0.5328
We need to find the shaded area
z0 = -.5
47
z0 = 1
Jia-Ying Chen
Example 7



48
X is normally distributed with mean 300 and
standard deviation 40. What value of X does
only the top 15 % exceed?
P(0 < Z < z.15 ) = 1-0.15 = 0.85
z .15 = 1.04; z  x - m
.15

x - 300
1.04 
40
 x  341.6
Jia-Ying Chen
Example 8




49
The long-distance calls made by the
employees of a company are normally
distributed with a mean of 7.2 minutes and a
standard deviation of 1.9 minutes. Find the
probability that a call
a. Last between 5 and 10 minutes
b. Last more than 7 minutes
c. Last less than 4 minutes
Jia-Ying Chen
Solution



 5 - 7.2 X - m 10 - 7.2 
P




1.9 
 1.9
a P(5 < X < 10) =
= P(–1.16 < Z < 1.47) = 0.9292-(1-0.8770)
= .8062
b. P(X > 7) =
c. P(X < 4)
 X - m 7 - 7.2 
P


1.9 
 
= P(Z > –.11) = 0.5438
 X - m 4 - 7.2

P


1.68

=  
1.9

=1-0.9535= .0465
50
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