Class Notes
(811,457)

Canada
(494,752)

Ryerson University
(28,350)

Quantitative Methods
(263)

QMS 102
(188)

Clare Chua
(18)

Lecture 9

# Lecture 9 (week 10)

Unlock Document

Ryerson University

Quantitative Methods

QMS 102

Clare Chua

Fall

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?

More
Less
Related notes for QMS 102