Download P(Z < -2.23)

Chapter 8
Continuous Probability
Distributions
8.1 Continuous Probability Distributions
 Unlike a discrete random variable which we studied in Chapter
7, a continuous random variable（連續型隨機變數） is
one that can assume an uncountable number of values in the
interval (a,b).
 We cannot list the possible values because there is an infinite
number of them.
 Because there is an infinite number of values, the probability
that a continuous variable X will assume any particular value is
0. Why?
The probability of each value
1/4
1/3
1/2
0
2007會計資訊系統計學(一)上課投影片
+
1/4
+
1/3
+
1/3
+
1/4
1/2
2/3
+
+
1/4 = 1
1/3 = 1
1/2 = 1
1
8-2
8.1 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
2007會計資訊系統計學(一)上課投影片
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
8-3
Point Probabilities are Zero
 Because there is an infinite number of values, the
probability of each individual value is virtually 0.
 Thus, we can determine the probability of a range of
values only.


E.g. with a discrete random variable like tossing a die, it
is meaningful to talk about P(X=5), say.
In a continuous setting (e.g. with time as a random
variable), the probability the random variable of interest,
say task length, takes exactly 5 minutes is infinitesimally
small, hence P(X=5) = 0.
 It is meaningful to talk about P(X ≤ 5).
2007會計資訊系統計學(一)上課投影片
8-4
Probability Density Function
 To calculate probabilities we define a probability density
function f(x).
 A function f(x) is called a probability density function（機
率密度函數） (over the range a ≤ x ≤ b if it meets the
following requirements:
1) f(x) ≥ 0 for all x between a and b, and
2) The total area under the curve between a and b is 1.0
Area = 1
x1
P(x1≤X≤x2)
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.
2007會計資訊系統計學(一)上課投影片
8-5
Uniform Distribution（均等分配）
 Consider the uniform probability distribution
(sometimes called the rectangular probability
distribution（矩形分配）).
 It is described by the function:
1
f(x)
, where a  x  b
ba
f(x)
area  width  height
1
 ( b  a )
1
ba
1
ba
a
2007會計資訊系統計學(一)上課投影片
b
x
8-6
Uniform Distribution
 A random variable X is said to be uniformly
distributed if its density function is
1
f(x)
, where a  x  b
ba
 The expected value and the variance are
ab
E(X) 
2
( b  a )2
V( X ) 
12
2007會計資訊系統計學(一)上課投影片
8-7
Example 8.1
 The amount of gasoline sold daily at a service
station is uniformly distributed with a minimum of
2,000 gallons and a maximum of 5,000 gallons.
Find the probability that sales are:



Between 2,500 and 3,000 gallons
More than 4,000 gallons
Exactly 2,500 gallons
2007會計資訊系統計學(一)上課投影片
8-8
Example 8.1
 The daily sale of gasoline is uniformly distributed
between 2,000 and 5,000 gallons.
 Find the probability that sales are between 2,500 and
3,000 gallons.
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(2500X3000) = (3000-2500)(1/3000) = .1667
1/3000
2000 2500 3000
2007會計資訊系統計學(一)上課投影片
5000
x
There is about
a 17% chance
that between
2,500 and 3,000
gallons of gas
will be sold on
a given day.
8-9
Example 8.1
 The daily sale of gasoline is uniformly distributed
between 2,000 and 5,000 gallons.
 Find the probability that sales are more than 4,000
gallons.
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(X4000) = (5000-4000)(1/3000) = .333
1/3000
2000
2007會計資訊系統計學(一)上課投影片
4000
5000
x
There is a 33%
chance the
gas station will
sell more than
4,000 gallons
on any given
day.
8-10
Example 8.1
 The daily sale of gasoline is uniformly distributed
between 2,000 and 5,000 gallons.
 Find the probability that sales are exactly 2,500 gallons.
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(X=2500) = (2500-2500)(1/3000) = 0
1/3000
2000 2500
2007會計資訊系統計學(一)上課投影片
5000
x
The
probability
that the gas
station will
sell exactly
2,500 gallons
is zero.
8-11
8.2 Normal Distribution（常態分配）
 This is the most important continuous distribution.


Many distributions can be approximated by a
normal distribution.
The normal distribution is the cornerstone
distribution of statistical inference.
2007會計資訊系統計學(一)上課投影片
8-12
Normal Distribution
 A random variable X with mean m and variance
s2 is normally distributed if its probability density
function is given by
(
1 x μ 2
)(
)
2
σ
1
f(x) 
e
  x  
σ 2π
where π  3.14159  and e  2.71828 
Unlike the range of the uniform distribution (a ≤ x ≤ b),
Normal distributions range from minus infinity to plus infinity.
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8-13
The Normal Distribution
 Important things to note:
The normal distribution is fully defined by two
parameters: its standard deviation and mean
1 x μ 2
( )(
)
2
σ
1
f(x) 
e
  x  
σ 2π
where π  3.14159  and e  2.71828 
The normal distribution is bell shaped
and symmetrical about the mean.
2007會計資訊系統計學(一)上課投影片
8-14
The Shape of the Normal Distribution
The normal distribution is bell shaped, and
symmetrical around m.
90
m
Why symmetrical? Let m = 100. Suppose x = 110.
f (110) 
1
s 2
 110 100 
(1/ 2)

s 

e
2007會計資訊系統計學(一)上課投影片
2

1
s 2
 10 
(1/ 2) 
s
e
110
Now suppose x = 90
2
f (90) 
1
s 2
 90 100 
(1/ 2)

s 

e
2

1
s 2
 10 
(1/ 2)

s 

e
2
8-15
The effects of m and s
How does the standard deviation affect the shape of f(x)?
s= 2
s =3
s =4
Increasing the
standard deviation
“flattens” the curve.
How does the expected value affect the location of f(x)?
m = 10 m = 11 m = 12
Increasing the mean shifts
the curve to the right.
2007會計資訊系統計學(一)上課投影片
8-16
Standard Normal Distribution
 A normal distribution whose mean is 0 and standard
deviation is 1 is called the standard normal
distribution（標準常態分配）.
z 2 1 x  μ0 2
 ( )(
)
2 2
1σ
11
f(z) 
ee
   z x  
f(x)
σ1 2π2π
where π  3.14159  and ee  2.71828
2.71828

 As we shall see shortly, any normal distribution can
be converted to a standard normal distribution with
simple algebra. This makes calculations much
easier.
2007會計資訊系統計學(一)上課投影片
8-17
Finding Normal Probabilities
 Two facts help calculate normal probabilities:


The normal distribution is symmetrical.
Any normal distribution can be transformed into a
specific normal distribution called
“STANDARD NORMAL DISTRIBUTION”
X  mX
Z
sX
 Example
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?
2007會計資訊系統計學(一)上課投影片
8-18
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.
This shifts the mean of X to zero…
Every normal variable
with some m and s,
can be transformed
into this Z.
Z
E(Z) = m = 0, V(Z) = s2 = 1
2007會計資訊系統計學(一)上課投影片
X  mX
sX
Therefore, once probabilities
for Z are calculated,
probabilities of any normal
variable can be found.
This changes the shape of the curve…
8-19
Finding Normal Probabilities
 Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-0.5 < Z < 1)
To complete the calculation we need to compute
the probability under the standard normal distribution
2007會計資訊系統計學(一)上課投影片
8-20
Using the Standard Normal Table
Standard normal probabilities have been
calculated and are provided in a table .
P(0<Z<z0)
The tabulated probabilities correspond
to the area between Z=0 and some Z = z0 >0
z
0.0
0.1
.
.
1.0
.
.
1.2
.
.
2007會計資訊系統計學(一)上課投影片
0
0.0000
0.0398
.
.
0.3413
.
.
0.3849
.
.
0.01
0.0040
0.0438
.
.
0.3438
.
.
0.3869
.
.
…….
…….
.
.
Z=0
0.05
0.0199
0.0596
.
.
0.3531
.
.
0.3944
.
.
0.06
0.0239
0.0636
.
.
0.3554
.
.
0.3962
.
.
Z = z0
8-21
Finding Normal Probabilities
 Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-.5 < Z < 1)
We need to find the shaded area
z0 = -.5
2007會計資訊系統計學(一)上課投影片
z0 = 1
8-22
Finding Normal Probabilities
 Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1
P(0<Z<1)
z
0.0
0.1
.
.
1.0
.
0
0.0000
0.0398
.
.
0.3413
.
0.1
0.0040
0.0438
.
.
0.3438
.
…….
0.05
0.0199
0.0596
.
.3413
.
0.3531
.
0.06
0.0239
0.636
.
.
0.3554
.
z0 =-.5z=0 z0 = 1
2007會計資訊系統計學(一)上課投影片
8-23
Finding Normal Probabilities
 The symmetry of the normal distribution makes it
possible to calculate probabilities for negative
values of Z using the table as follows:
-z0
0
+z0
P(-z0<Z<0) = P(0<Z<z0)
2007會計資訊系統計學(一)上課投影片
8-24
Finding Normal Probabilities
 Example - continued
0
0.0000
0.0398
.
.
0.1915
.
0.1
0.0040
0.0438
.
.
….
.
…….
.3413
.1915
z
0.0
0.1
.
.
0.5
.
-.5
2007會計資訊系統計學(一)上課投影片
0.05
0.0199
0.0596
.
.
….
.
0.06
0.0239
0.636
.
.
….
.
.5
8-25
Finding Normal Probabilities
 Example - continued
0
0.0000
0.0398
.
.
0.1915
.
0.1
0.0040
0.0438
.
.
….
.
…….
.3413
.1915
.1915
.1915
z
0.0
0.1
.
.
0.5
.
-.5
0.05
0.0199
0.0596
.
.
….
.
0.06
0.0239
0.636
.
.
….
.
.5 1.0
P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328
“Just over half the time, 53% or so, a computer will have an
assembly time between 45 minutes and 1 hour”
2007會計資訊系統計學(一)上課投影片
8-26
Using the Normal Table (Table 3)
 What is P(Z > 1.6) ?
P(0 < Z < 1.6) = .4452
z
0
1.6
P(Z > 1.6) = .5 – P(0 < Z < 1.6)
= .5 – .4452
= .0548
2007會計資訊系統計學(一)上課投影片
8-27
Using the Normal Table (Table 3)
 What is P(Z < -2.23) ?
P(0 < Z < 2.23)
P(Z < -2.23)
P(Z > 2.23)
z
-2.23
0
2.23
P(Z < -2.23) = P(Z > 2.23)
= .5 – P(0 < Z < 2.23)
= .0129
2007會計資訊系統計學(一)上課投影片
8-28
Using the Normal Table (Table 3)
 What is P(Z < 1.52) ?
P(0 < Z < 1.52)
P(Z < 0) = .5
z
0
1.52
P(Z < 1.52) = .5 + P(0 < Z < 1.52)
= .5 + .4357
= .9357
2007會計資訊系統計學(一)上課投影片
8-29
Using the Normal Table (Table 3)
 What is P(0.9 < Z < 1.9) ?
P(0 < Z < 0.9)
P(0.9 < Z < 1.9)
z
0
0.9
1.9
P(0.9 < Z < 1.9) = P(0 < Z < 1.9) – P(0 < Z < 0.9)
=.4713 – .3159
= .1554
2007會計資訊系統計學(一)上課投影片
8-30
Finding Normal Probabilities
 Additional Example - 1

Determine the following probability:
P(Z>1.47)
0.5 - P(0<Z<1.47) = P(Z>1.47)
0
1.47
P(Z>1.47) = 0.5 - 0.4292 = 0.0708
2007會計資訊系統計學(一)上課投影片
8-31
Finding Normal Probabilities
• Additional Example - 2
– Determine the following probability: P(-2.25<Z<1.85)
P(-2.25<Z<0) ==.4878
?
-2.25
P(0<Z<1.85) = .4678
P(0<Z<2.25) = .4878
0
1.85 2.25
P(-2.25<Z<1.85) = 0.4878 + 0.4678 = 0.9556
2007會計資訊系統計學(一)上課投影片
8-32
Finding Normal Probabilities
• Additional Example - 3
– Determine the following probability:
P(.65<Z<1.36)
P(0<Z<1.36) = .4131
P(0<Z<.65) = .2422
0 .65 1.36
P(.65<Z<1.36) = .4131 - .2422 = .1709
2007會計資訊系統計學(一)上課投影片
8-33
Example 8.2
 The rate of return (X) on an investment is normally
distributed with mean of 10% and standard deviation of
(i) 5%, (ii) 10%.
 What is the probability of losing money? Some advice: always
draw a picture!
0%
10%
0 - 10
(i) P(X< 0 ) = P(Z<
) = P(Z< - 2)
5
X
.4772
-2
=P(Z>2) = 0.5 - P(0<Z<2) = 0.5 - .4772 = .0228
2007會計資訊系統計學(一)上課投影片
0
2
Z
8-34
Example 8.2
 The rate of return (X) on an investment is normally
distributed with mean of 10% and standard deviation of
(i) 5%, (ii) 10%.
 What is the probability of losing money?
X
0%
10%
0 - 10
(ii) P(X< 0 ) = P(Z< 10 )
.3413
-1
1
Z
= P(Z< - 1) = P(Z>1) = 0.5 - P(0<Z<1) = 0.5 - .3413 = .1587（>.0228）
Increasing the standard deviation increases the probability of losing money,
which is why the standard deviation (or the variance) is a measure of risk.
2007會計資訊系統計學(一)上課投影片
8-35
Finding Values of Z
 Sometimes we need to find the value of Z for a given
probability
 We use the notation zA to express a Z value for which
P(Z > zA) = A
A
zA
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8-36
Finding Values of Z
 Example 8.3 & 8.4


Determine z exceeded by 5% of the population
Determine z such that 5% of the population is below
 Solution
z.05 is defined as the z value for which the area on its right under the
standard normal curve is .05.
0.45
0.05
0.05
-Z0.05
2007會計資訊系統計學(一)上課投影片
0
Z0.05
1.645
8-37
Finding Values of Z
 Other Z values are

Z.05 = 1.645

Z.01 = 2.33
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8-38
Using the values of Z
 Because z.025 = 1.96 and - z.025= -1.96, it follows that
we can state
P(-1.96 < Z < 1.96) = .95
 Similarly
P(-1.645 < Z < 1.645) = .90
2007會計資訊系統計學(一)上課投影片
8-39
8.3 Exponential Distribution（指數分配）
 The exponential distribution can be used to model



the length of time between telephone calls
the length of time between arrivals at a service station
the life-time of electronic components.
 When the number of occurrences of an event
follows the Poisson distribution, the time between
occurrences follows the exponential distribution.
2007會計資訊系統計學(一)上課投影片
8-40
Exponential Distribution
 A random variable is exponentially distributed if its
probability density function is given by
f ( x )   e  x , x  0
where   0 is the distribution parameter.
1
1
E(X) 
and V ( X )  2


Note that x ≥ 0. Time (for example) is a non-negative quantity; the
exponential distribution is often used for time related phenomena such as the
length of time between phone calls or between parts arriving at an assembly
station. Note also that the mean and standard deviation are equal to each
other and to the inverse of the parameter of the distribution (lambda)
2007會計資訊系統計學(一)上課投影片
8-41
Exponential Distribution
 The exponential distribution depends upon the value of .
 Smaller values of  “flatten” the curve.
 E.g. exponential distributions for  = .5, 1, 2
2007會計資訊系統計學(一)上課投影片
8-42
Exponential Distribution
 If X is an exponential random variable, then we can
calculate probabilities by:



P(X > a) = e–a
P(X < a) = 1 – e –a
P(a1 < X < a2) = e – (a1) – e – (a2)
2007會計資訊系統計學(一)上課投影片
8-43
Exponential Distribution
 Example 8.6



The lifetime of an alkaline battery is exponentially
distributed with  = .05 per hour.
What is the mean and standard deviation of the battery’s
lifetime?
Find the following probabilities:
• The battery will last between 10 and 15 hours.
• The battery will last for more than 20 hours.
2007會計資訊系統計學(一)上課投影片
8-44
Exponential Distribution
 Solution


The mean = standard deviation = 1/  1/.05 = 20 hours.
Let X denote the lifetime.
P( 10  X  15 )
 e .05( 10 )  e .05( 15 )
 e .5  e .75
 .6065  .4724
 .1341
“There is about a 13%
• P(X > 20) =
2007會計資訊系統計學(一)上課投影片
e-.05(20)
= .3679
chance a battery will only
last 10 to 15 hours”
8-45
Exponential Distribution
 Example 8.7


The service rate at a supermarket checkout is 6
customers per hour.
If the service time is exponential, find the following
probabilities:
• A service is completed in 5 minutes,
• A customer leaves the counter more than 10 minutes
after arriving
• A service is completed between 5 and 8 minutes.
2007會計資訊系統計學(一)上課投影片
8-46
Exponential Distribution
 Solution




A service rate of 6 per hour =
A service rate of .1 per minute ( = .1/minute).
P(X < 5) = 1-e-x = 1 – e-.1(5) = .3935
P(X >10) = e-x = e-.1(10) = .3679
P(5 < X < 8) = e-.1(5) – e-.1(8) = .1572
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8-47
Exponential Distribution
 Additional Example


Cars arrive randomly and independently to a
tollbooth at an average of 360 cars per hour.
Use the exponential distribution to find the probability
that the next car will not arrive within half a minute.
 Solution

Let X denote the time (in minutes) that elapses
before the next car arrives. X is exponentially
distributed with  = 360/60 = 6 cars per minute.
P(X>.5) = e-6(.5) = .0498.
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8-48
Exponential Distribution vs. Poisson Distribution


What is the probability that no car will arrive within the
next half minute?
Solution
• If Y counts the number of cars that will arrive in the
next half minute, then Y is a Poisson variable with m =
(.5)(6) = 3 cars per half a minute.
P(Y = 0) = e-3(30)/0! = .0498.
Comment:
If the first car will not arrive within the next half a minute
then no car will arrives within the next half minute.
Therefore, not surprisingly, the probability found here is
the exact same probability found in the previous question.
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8-49
8.4 Other Continuous Distribution
 Three other important continuous distributions which
will be used extensively in later sections are
introduced here:
 Student t Distribution
 Chi-Squared Distribution
 F Distribution
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8-50
Student t Distribution（t分配）
 Here the letter t is used to represent the random
variable, hence the name. The density function for the
Student t distribution is as follows:
 [(  1 ) / 2 ]  t 2 
f (t ) 
1  
  ( / 2 )   
 1

2
for    t  
  (nu) is called the degrees of freedom（自由度）
 (Gamma function) is (k)=(k-1)(k-2)…(2)(1)
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8-51
Student t Distribution
 Much like the standard normal distribution, the
Student t distribution is “mound” shaped and
symmetrical about its mean of zero:
 The mean and variance of a Student t random
variable are

for ν  2
E(t) = 0 and V ( t ) 
 2
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Student t Distribution
 In much the same way that m and s define the normal
distribution,  , the degrees of freedom, defines the Student
t Distribution:
 As the number of degrees of freedom increases, the t
distribution approaches the standard normal distribution.
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8-53
Determining Student t Values
 The student t distribution is used extensively in statistical
inference.
 Thus, it is important to determine values of tA associated
with a given number of degrees of freedom.
 We can do this using
 t tables
 Excel
 Minitab
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Determining Student t Values
 The student t distribution is used extensively in statistical
inference. Table 4 in Appendix B lists values of t A , .
 That is, values of a Student t random variable with degrees
of freedom  such that:
P( t  t A , )  A
 The values for A are pre-determined “critical” values,
typically in the 10%, 5%, 2.5%, 1% and 0.5% range.
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Using the t Table
 The table provides the t values (tA) for which P(t > tA) = A
A = .05
A = .05
The t distribution is
symmetrical around 0
t.100
t.05
t.025
t.01
t.005
3.078
1.886
.
.
1.372
6.314
2.92
.
.
1.812
12.706
4.303
.
.
2.228
31.821
6.965
.
.
2.764
.
.
.
.
.
.
.
.
.
.
200
1.286
1.282
1.653
1.645
1.972
1.96
2.345
2.326
63.657
9.925
.
.
3.169
.
.
2.601
2.576
Degrees of Freedom
1
2
.
.
10

2007會計資訊系統計學(一)上課投影片
tA =1.812
-tA=-1.812
8-56
Student t Probabilities and Values
 Excel can calculate Student distribution probabilities
and values.
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Chi – Squared Distribution（卡方分配）
 The chi-squared density function is given by:
1
1
2 (  / 2 )1   2 / 2
f( )
( )
e
/2
 ( / 2 ) 2
2
for χ 2  0
 As before, the parameter  is the number of degrees
of freedom.
E(  )  
2
V (  )  2
2
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The Chi – Squared Distribution
 The chi-squared distribution is not symmetrical
2
 The square, i.e.  , forces non-negative values
(e.g. finding P(  2  0 ) is illogical).
 Chi squared values can be found from the chi
squared table, from Excel, or from Minitab.
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The Chi – Squared Distribution
 Table 5 in Appendix B makes it easy to look-up the
right hand tail probability (A), for which
P(    )  A
2
2
A
 A2
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The Chi – Squared Distribution
 For probabilities of this sort:

2
1 A
 We use 1–A,
2
2
i.e. we are determining P(   1 A )  A
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Using the Chi-Squared Table
To find 2 for which
P(2<2)=.01, lookup
the column labeled
21-.01 or 2.99
A =.99
0
5
Degrees of 2
freedom  .995
1
0.0000393
.
.
10
2.15585
.
.
.
.
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A =.05
10
20
22.05
A
15
2.990
0.0001571
.
2.55821
.
.
.
25
30
35
2.05 2.010 2.005
3.84146
6.6349
7.87944
.
18.307
.
.
23.2093
.
.
25.1882
.
.
8-62
For Example
 To find the point in a chi-squared distribution with 8 degrees
of freedom, such that the area to the right is .05,
 .205
 Look up the intersection of the 8 d.f. row with the 0.05
column, yielding a value of 15.5073
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For Example
 To find the point in a chi-squared distribution with 8 degrees
of freedom, such that the area to the left is .05,
 .295
 Look up the intersection of the 8 d.f. row with the 0.95
column, yielding a value of 2.73264
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Chi-Squared Probabilities and Values
 Excel can calculate chi-squared distribution
probabilities and values.
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F Distribution（F分配）
 The F density function is given by:
(
 1  2
)
1
 1 2
2

F
2
f(F )
( 1 )2

1
2 2
 1F
 ( ) ( )
(1
)
2
2
2
1  2
2
for F  0
 Two parameters define this distribution, and like we’ve already
seen these are again degrees of freedom.
  1 is the “numerator” degrees of freedom（分子的自由度）
  2 is the “denominator” degrees of freedom （分母的自由度）
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F Distribution
 The mean and variance of an F random variable are given
by:
E( F ) 
2
2  2
ν2  2
 and
2 ( 1  2  2 )
V( F ) 
 1 ( 2  2 )2 ( 2  4 )
2
2
ν2  4
 The F distribution is similar to the  distribution in that its
starts at zero (is non-negative) and is not symmetrical.
2
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Determining Values of F
 The values of the F variable can be found in the F
table, Excel, or from Minitab.
 The entries in the table are the values of the F
variable of the right hand tail probability (A), for
which
P( F 1 , 2  FA )  A
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Determining Values of F
 For example, what is the value of F for 5% of the area under
the right hand “tail” of the curve, with a numerator degree of
freedom of 3 and a denominator degree of freedom of 7?
 Solution: use the F look-up (Table 6)
There are different tables
for different values of A.
Make sure you start with
the correct table!!
F.05,3,7
F.05 ,3 ,7  4.35
Denominator Degrees of Freedom : ROW
Numerator Degrees of Freedom : COLUMN
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Determining Values of F
 For areas under the curve on the left hand side of
the curve, we can leverage the following relationship:
F1 A , 1 , 2 
1
FA , 2 , 1
Pay close attention to the order of the terms!
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F Probabilities and Values
 Excel can calculate F distribution probabilities and
values.
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