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Chapter 6
Continuous Random Variables
McGraw-Hill/Irwin
Copyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved.
Continuous Random Variables
6.1 Continuous Probability Distributions
6.2 The Uniform Distribution (Optional)
6.3 The Normal Probability Distribution
6.4 Approximating the Binomial Distribution
by Using the Normal Distribution
(Optional)
6.5 The Exponential Distribution (Optional)
6.6 The Normal Probability Plot (Optional)
6-2
Two Types of Random Variables
• Random variable: a variable that assumes
numerical values that are determined by the
outcome of an experiment
– Discrete
– Continuous
• Discrete random variable: Possible values
can be counted or listed; doesn’t take values
on an interval of the real line.
– The number of defective units in a batch of 20
– Toss a coin. Let x=1 if we have a head, x=0 if we
have a tail
6-3
Random Variables
Continued
• Continuous random variable: May
assume any numerical value in one or
more intervals
– The waiting time for a credit card
authorization
– The interest rate charged on a business
loan
6-4
Continuous Probability Distributions
• A continuous random variable may
assume any numerical value in one or
more intervals
• To compute probabilities about the
range of values that a continuous
random variable x might attain, we use
continuous probability distribution (or
called probability density function),
denoted by f(x).
6-5
Properties of Continuous Probability
Distributions
•
Properties of probability density
function f(x): f(x) is a continuous
function such that
1. f(x)  0 for all x
2. The total area under the curve of f(x) is
equal to 1
3. The probability of X to be between a and
b, P(a<X<b), is the area under the curve
between a and b.
6-6
Properties of Density Function
P(a ≤ x ≤ b) is the area colored in blue
under the curve and between the values
x = a and x = b
Note: P(a ≤ x ≤ b) =P(a<x<b)
6-7
Area and Probability
• The blue area under the curve f(x) from x = a to x = b
is the probability that x could take any value in the
range a to b
– Symbolized as P(a  X  b)
– Note for continuous variable x , P(X=a)=0; P(a  X 
b)=P(a<X<b), Or as P(a < X < b), because each of the
interval endpoints has a probability of 0
– P(X<b) is the total area under curve to the left of b; and
P(X>b) is the total area under curve to the right of b.
6-8
The Normal Probability Distribution
• A random variable with normal distribution is uniquely
determined by its mean m and standard deviation s.
Or equivalently a normal density function is uniquely
determined by its mean m and standard deviation s.
• The normal curve is bell-shaped symmetrical about
its mean m
– So m is also the median
– The curve of f(x) is tallest over its m.
• The area under the entire normal curve is 1
– What is the total area under the normal curve to
the right of m? What does it mean?
6-9
The Normal Probability Distribution
Continued
6-10
Gauss
6-11
Means and Standard Deviations
• A normal distribution can have any mean and any positive standard
deviation.
• The mean gives the location of the line of symmetry.
• The standard deviation describes the spread of the data.
μ = 3.5
σ = 1.5
μ = 3.5
σ = 0.7
μ = 1.5
σ = 0.7
6-12
Example: Understanding Mean and
Standard Deviation
1. Which curve has a greater mean?
Solution:
Curve A has the greater mean (The line of symmetry of curve A occurs at
x = 15. The line of symmetry of curve B occurs at x = 12.)
6-13
How to decide the Position and Shape of
the Normal Curve
(a) The mean m positions the peak of the normal curve over the real axis
(b) The variance s2 measures the width or spread of the normal curve
6-14
Properties of Normal Distributions
1. The mean, median, and mode are equal, µ
2. The normal curve is bell-shaped and symmetric about
the mean µ.
3. The total area under the curve is equal to one.
4. The normal curve approaches, but never touches the xaxis as it extends farther and farther away from the
mean.
Total area = 1
f( x) =
μ
1
σ 2π
1  x m 
 

2 s 
e
2
x
6-15
Properties of the Normal Distribution
• There are an infinite number of normal
curves
– The shape of any individual normal curve
depends on its specific mean m and
standard deviation s
– The highest point is over the mean
• Any normal distribution can always be
converted to a standard normal distribution.
All problems concerning a normal random
variable can always be solved by a standard
normal random variable.
6-16
The Standard Normal Distribution
Standard normal distribution
• A normal distribution with a mean of 0 and a standard
deviation of 1.
Area = 1
3
2
1
z
0
1
2
3
6-17
The Standard Normal Table
• The standard normal table is a table
that lists the area under the standard
normal curve to the left of the z (z
score) value of interest, that is P(Z<z)
– See Table 6.1, Table A.3 in Appendix A,
and the table on the back of the front cover
• This table is so important that it is repeated 3
times in the textbook!
• Always look at the accompanying figure for
guidance on how to use the table
6-18
Example: Using The Standard
Normal Table
Find the area to the left of 1.15.
Solution:
μ =0
1.15
Find 1.1 in the left hand column.
P(Z<1.15)=? The area to the left of z = 1.15 is 0.8749.
This means the probability that for a standard normal random
variable Z the probability P(Z 1.15)=P(Z<1.15)=0.8749
Question: What is P(Z>1.15)?1-0.8749
6-19
The Standard Normal Table
Continued
• In this course to find out P(Z<z), always
round z to the nearest one hundredth.
• Then find the one hundredth digit in the
first row of the table, and then find the
remaining digits in the first column.
• At the intersection of the column and row you
find in the last step, you find out P(Z<z) or
P(Z≤z), which is equivalent to the areas
under the normal curve from the leftmost part
up to any value of z are given in the body of
the table.
6-20
Example: Using The Standard
Normal Table
Find the cumulative area that corresponds to a z-score of
-0.24.
P(Z<-0.24)=0.4052
P(Z>-0.24)=1-0.4052?
6-21
Z
• P( a<Z <b)
=P(Z<b)-P(Z<a)
• P(-1<Z<2)=P(Z<2)-P(Z<-1)
=0.9772-0.1577
• P(Z<2)=0.9772
• P(Z<-1)=0.1577
6-22
The Non-Standard Normal
Distribution
• If X is normally distributed with mean m
and standard deviation s, then the
random variable z
xm
z
s
is normally distributed with mean 0 and
standard deviation 1
• Then P(X  x)=P(Z (x-µ)/σ)
6-23
Probability and Normal
Distributions
• Find out P(X<600) for a normal random
variable with m500 and standard
deviation s100.
μ = 500
σ = 100
P(x < 600) = Area
μ =500 600
P(X < x)=P(Z (x-µ)/σ) ;
P(X < 600) = P(Z < 1)
x
6-24
Probability and Normal
Distributions
X has Normal Distribution
μ = 500 σ = 100
xm
z
s
P(x < 600)
x  m 600  500
z

1
s
100
x
μ =500 600
P(X < x)=P(Z (x-µ)/σ) ;
P(X < 600) = P(Z < 1)=0.8413
6-25
Solution: Finding Probabilities for
Normal Distributions
Normal Distribution
μ = 45 σ = 12
P(24 < x < 54)
x
24
z1 
x-m
s
45 54

24 - 45
 -1.75
12
z2 
x-m
s

54 - 45
 0.75
12
P(x1  X  x2)=P(x2-µ)/σ Z (x1-µ)/σ) ;
P(24 < X < 54) = P(-1.75 < Z < 0.75)
= 0.7734 – 0.0401 = 0.7333
6-26
The Properties of z-score
• Z-score, (x-µ)/σ, measures the number of
standard deviations that x is from the mean m
– The algebraic sign on z indicates on which side of
m is x
– z is positive if x > m (x is to the right of m on the
number line)
– z is negative if x < m (x is to the left of m on the
number line)
6-27
Finding a z-Score Given an area or probability (the
inverse problem or percentile problem)
Find the z-score that such that
P(Z<z)=0.025. z=?
0.025
z
0
z
The areas closest to 0.025 in the table is 0.025 with z =
-1.96 . Then z-score is -1.96.
Find the z score which gives you the given probability; if you
can not find the exact value, find the closest; if you have a
tie, take the average of the two z scores.
6-28
Example: Finding a z-Score Given
an Area
Find the z-score that corresponds to a
cumulative area of 0.3632.
0.3632
z
z 0
6-29
Solution: Finding a z-Score Given
an Area
• Locate 0.3632 in the body of the Standard Normal Table.
The z-score
is -0.35.
• The values at the beginning of the corresponding row and at the
top of the column give the z-score.
• If you cannot find the exact given area, find the closest value in
the table; in case of tie, take the average of the two z-scores.
6-30
Solution: Finding a z-Score Given
an Area
• Find z such that P(Z<z)=0.8923
• The closest number in the table to 0.8923 is 0.8925.
The z-score
is 1.24.
6-31
• Find the z-score that such that
P(Z<z)=0.01. z=?
0.01
z
0
z
6-32
Finding a z-Score Given an area or probability
(the inverse problem or percentile problem)
Find the z-score that corresponds to
P(Z<z)=0.05. z=?
The z-score corresponds to an area of 0.05.
0.05
z
z
0
The areas closest to 0.05 in the table are 0.0495 (z = -1.65)
and 0.0505 (z = -1.64). Because 0.05 is halfway between
the two areas in the table, 0.0495 and 0.0505, and we have
a tie, so the z score is the average of -1.64 and -1.65.
6-33
Finding Percentiles for Non-Standard Normal
Random Variable with mean m sd s
1. Formulate the problem in terms of x and
write is as P(X<x)=p
2. Calculate the corresponding z values such
that P(Z<z)=p
3. x = μ + zσ
Note: It is always useful to draw a picture
showing the required areas before using the
normal table
6-34
Example: Finding a Specific Data
Value
Scores for a civil service exam are normally distributed, with a
mean of 75 and a standard deviation of 6.5. To be eligible for civil
service employment, you must score in the top 5%. What is the
lowest score you can earn and still be eligible for employment?
P(X<x)=0.95
1 – 0.05
= 0.95
75
5%
?
x
An exam score in the top 5% is
any score above the 95th
percentile. Find the z-score that
corresponds to a cumulative area
of 0.95.
6-35
Solution: Finding a Specific Data
Value
From the Standard Normal Table, the areas closest to 0.95
are 0.9495 (z = 1.64) and 0.9505 (z = 1.65). Because 0.95 is
halfway between the two areas in the table, use the z-score
that is halfway between 1.64 and 1.65. That is, z = 1.645.
1.P(X<x)=0.95, X normal with mean 75 standard deviation
6.5
2.Find z such that P(Z<z)=0.95, z=1.645
3. Using the equation x = μ + zσ,
x = 75 + 1.645(6.5) ≈ 85.69
5%
z
0
1.645
6-36
Solution: Finding a Specific Data
Value
Using the equation x = μ + zσ
x = 75 + 1.645(6.5) ≈ 85.69
5%
z
75 85.69
x
The lowest score you can earn and still be eligible for employment
is 86.
6-37