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Lecture 9

Lecture 9 (week 10)

8 Pages
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Department
Quantitative Methods
Course Code
QMS 102
Professor
Clare Chua

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11/9/2010
1
Quiz 2 covered materials in
Chapter 5
Chapter 6
Cha
p
ter 7
p
Normal Probability
Distribution
Chapter 6
Continuous Probability
Distribution
Normal Distribution Uniform Distribution
Normal Distribution is symmetrical and bell
shaped, implying that most values tend to
cluster around the mean, equal to median.
The uniform distribution is symmetrical
and therefore the mean = median.
Normal Probability Distribution
The focus is on a
continuous,
bell-shaped
distribution
distribution
Discuss the
properties
Properties
1. The shape is determines by mean, Pand
standard deviation, V
2. The highest point on the normal curve is
located at the mean, which is also the
median and the mode of the distribution.
3. The normal distribution is symmetrical: the
curve’s shape to the left of the mean is the
mirror image of its shape to the right of the
mean. Thus, mean = median
4The tails of the normalcurveextend toinfinity
4
.
The
tails
of
the
normal
curve
extend
to
infinity
in both directions and never touch the
horizontal axis. Thus, it has an infinite range.
-f< X < +f
5. However, the tails get close enough to the
horizontal axis quickly enough to ensure that
the total area under the normal curve equals
1.
6. Since the normal curve is symmetrical, the
area under the normal curve to the right of
the mean (P) equals the area under the
normal curve to the left of the mean, and
each of these areas equals 0.5.
Normal Probability Distribution
When a mathematical
expression is available to
represent a continuous variable,
the probability that various
values occur
within certain ranges
or intervalscan be calculated.
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11/9/2010
2
Normal Probability Distribution
However, the EXACT
probability of a
particular value from
a Normal
Distribution is
ZERO.
e.g. P(X=140) = 0
P=95 140
Normal Probability Distribution
b
a
What information do we need to calculate P(a x b)?
The normal probability distribution
is defined by the equation
linereal the on x all or
e
21
1
=f(x)
2
1
-x
2
1
-¸
¸
¹
·
¨
¨
©
§
Normal Probability Distribution
b
a
What information do we need to calculate P(a x b)?
Pand V
Normal Probability Distribution
b
a
To calculateTo calculate
P(a x b):P(a x b):
Pand V
Normal
Random
Variable
X=measured”
e.g. X= a person’s weight
Parameters
Presentation of Solution
Follow the “normal” template – page 269
X = ____________________
For example: X= number of .... WRONG
1=
µ = Normal
P ( ) = P ( X symbol # ) = Ncd ( L, U, 1, µ ) = 0.xxxx
(words) ( <, >, , ) 4 decimals
Must draw a diagram
(Mean, 1, value, shading)
Suppose that the weights of adults are normally distributed
with a mean of 170.0lbs and a standard deviation of 25.0
lbs.
What is the probability that a person weighs from 150 to
190 lbs?
V= 25.0
Note:
Calculating “normal” probability =
Finding area under a portion
of the normal curve
P=170 190
150
X= a person’s weight
P= 170.0 lbs
V= 25.0 lbs
P( 150 x 190) = Ncd (150,190, 25, 170)
Lower upper VP
Refer to the calculator Lesson 5:
Page 169
normal
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11/9/2010
3
Normal Probability Exercises
Page 269 to 270
Q6.1 to Q6.6
Inverse Normal
Normal Probability Distribution
Given the value X, find
the probability (or
the area under the
)
The Inverse Normal
Probability means:
Given the probability
curve
)
(area), find the x
value?
What
is the
area?
Given X µ
1
Area=
0.3
X=? µ
1
Inverse Normal Probability
The weights of adults are normally distributed with a mean of 170.0 lbs
and a standard deviation of 25.0 lbs.
1) What is the maximum weight of the lightest 30% of the population?
Present your solution as follows(use the template)
X = person’s weight
P= 170.0
V= 25.0
P( X
d
?)
=
0.30
Area= V= 25.0
Normal
P(
X
d
?
)
0.30
? = =
CASIO Calculator: select STATmode F5 (DIST) F1 (NORM) F3 (InvN)
X = InvN(0.3, 25, 170) =
Written statement: The lightest 30% weigh at most 156.9 lbs
0.3
X=?P= 170.0
»
»
»
¼
º
«
«
«
¬
ª
P
V
Area
Inv
Inverse Normal Probability
The weights of adults are normally distributed with a mean of 170.0 lbs
and a standard deviation of 25.0 lbs.
2) What is the minimum weight to be in the heaviest 30% of the
population? Present your solution as follows(use the template)
X = person’s weight
P= 170.0
V= 25.0
P( X
?) 030
V= 25.0
NormalArea = 0.3
P(
X
?
)
=
0
.
30
? = =
CASIO Calculator: select STATmode F5 (DIST) F1 (NORM) F3 (InvN)
X = InvN((1-0.3), 25, 170) = InvN(0.7, 25, 170)
Written statement: The heaviest 30% weigh at least 183.1 lbs.
X=?
P= 170.0
»
»
»
¼
º
«
«
«
¬
ª
P
V
Area
Inv
NOTE
When using the InvN function,
the Area value that is required is
LEFT
of the
LEFT
of
the
desired X value.
Area=
0.3
Area=
0.7
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Description
11/9/2010 Quiz 2 covered materials in Chapter 5 Chapter 6 Normal Probability Chap pter 7 Distribution Chapter 6 Continuous Probability Distribution Normal Probability Distribution The focus is on a continuous, bell-shaped diisttrbuuttionn Discuss the properties Normal Distribution Uniform Distribution Normal Distribution is symmetricaThe uniform distribution is symmetrical shaped, implying that most valuesand therefore the mean = median. cluster around the mean, equal to median. Properties Normal Probability Distribution 1. The shape is determines by mean, 2 and standard deviation, 8 2. The highest point on the normal curve is median and the mode of the distribution. 3. The normal distribution is symmetrical: the curves shape to the left of the mean is the mirror image of its shape to the right of the mean. Thus, mean = median 4.Theaiioffhenomaalure exend onfniittyy in both directions and never touch the When a mathematical horizontal axis. Thus, it has an infinite range. expression is available to - B < X < +B represent a continuous variable, 5. However, the tails get close enough to the theprobability that various the total area under the normal curve equalst within certain ranges 1. or intervals can be calculated. 6. Since the normal curve is symmetrical, the area under the normal curve to the right of the mean (2) equals the area under the normal curve to the left of the mean, and each of these areas equals 0.5. 1 www.notesolution.com 11/9/2010 Normal Probability Distribution Normal Probability Distribution However, the EXACT probability of a particular value from a Normal Distribution is ZERO. a b e.g. P(X=140) = 0 What information do we need to calculate P(a x b)? The normal probability distribution 2 is defined by the equation - x- 2=95 140 f(x)= 1 e 2 2 or all x on the realline Normal Probability Distribution Normal Probability Distribution a b a b What information do we need to calculate P(a x b)? Tocaculaete Normal Pa x b)):: Random X= measured Variable e.g. X= a persons weight 2 and 8 Parameters 2 and 8 Suppose that the weights of adults are normally distributed Presentation of Solution with a mean of 170.0lbs and a standard deviation of 25.0 lbs. Follow the normal template page 269 What is the probability that a person weighs from 150 to 190 lbs? X = ____________________ 8 = 25.0 Note: For example: X= number of .... WRONG Calculating normal probability = Findingarea under a portion of the normal curve = Normal = 150 2=170190 X= apersons weight Refer to the calculator Lesson 5: P ( ) = P ( X symbol # ) = Ncd ( L, U, , ) = 0.xxxx Page 169 (words) ( , , ) 4 decimals 2 = 170.0 lbs 8 = 25.0 lbs normal Must draw a diagram P( 150 x 190) = Ncd (150,190, 25, 170) Lower upper 8 (Mean, , value, shading) 2 2 www.notesolution.com11/9/2010 Normal Probability Exercises Inverse Normal Page 269 to 270 Q6.1 to Q6.6 Inverse Normal Probability The weights of adults are normally distributed with a mean of 170.0 lbs Normal Probability Distribution and a standard deviation of 25.0 lbs. 1) What is the maximum weight of the lightest 30% of the population? Present your solution as follows (use the template) Given the value X, find The Inverse Normal X = persons weight the probability (or Probability means: 2 = 170.0 8 = 25.0 Normal the area under the Given the probability Area= 8 = 25.0 curve)) P(X @? ))0.30 0.3 (area), find the x Area value?
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