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Statistics for
Business and Economics
8th Edition
Chapter 5
Continuous Random Variables
and Probability Distributions
Continuous Probability Distributions
 A continuous random variable is a variable that
can assume any value in an interval




thickness of an item
time required to complete a task
temperature of a solution
height, in inches
 These can potentially take on any value,
depending only on the ability to measure
accurately.
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-2
Continuous Probability Distributions
Two characteristics
1. The probability that x assumes a value in any
interval lies in the range 0 to 1
2. The total probability of all the (mutually
exclusive) intervals within which x can
assume a value of 1.0
Figure 6.3 Area under a curve between
two points.
Figure 6.4 Total area under a
probability distribution curve.
Figure 6.5 Area under the curve as
probability.
Figure 6.6 Probability that x lies in the
interval 65 to 68 inches.
Expectations for Continuous
Random Variables

The mean of X, denoted μX , is defined as the
expected value of X
μX  E(X)

The variance of X, denoted σX2 , is defined as the
expectation of the squared deviation, (X - μX)2, of a
random variable from its mean
σ 2X  E[(X  μX )2 ]
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-8
Linear Functions of Variables
 Let W = a + bX , where X has mean μX and
variance σX2 , and a and b are constants
 Then the mean of W is
μW  E(a  bX)  a  bμX
 the variance is
σ  Var(a  bX)  b σ
2
W
2
2
X
 the standard deviation of W is
σW  b σX
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-9
Example 5.2: (b.p-185)
home heating costs
 A homeowner estimates that within the range of
likely temperatures her January heating bill, Y,
in dollars, will be
 Y = 290 – 5 T
 Where T is the average temperature for the
month, in degrees Fahrenheit.
 If the average January temperature can be
represented by a random variable with mean 24
and S.D. 4, find the mean and S.D. of this home
owner’s January heating bill.
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-10
Exercise 5.10
 The profit for a production process is equal to
$1000 minus 2 times the number of units
produced. The mean and variance for the
number of units produced are 50 and 90,
respectively. Find the mean and variance of the
profit.
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-11
THE NORMAL DISTRIBUTION
Normal Probability Distribution
A normal probability distribution , when
plotted, gives a bell-shaped curve such that:
1.
2.
3.
The total area under the curve is 1.0.
The curve is symmetric about the mean.
The two tails of the curve extend indefinitely.
Figure 6.11 Normal distribution with
mean μ and standard deviation σ.
The Normal Distribution
(continued)
‘Bell Shaped’
 Symmetrical
f(x)
 Mean, Median and Mode
are Equal
Location is determined by the
mean, μ

Spread is determined by the
standard deviation, σ
The random variable has an
infinite theoretical range:
+  to  
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
σ
μ
x
Mean
= Median
= Mode
Chap 6-14
Properties of Normal
Distribution
 Suppose that the Random variable X follows
a normal distribution with parameters μ and
σ2. Then the following properties hold:
1. The mean of random variable is μ: E(X) = μ
2. The variance of random variable is σ2:
Var(X) = E [(X-μ)2] = σ2
3. The shape of the probability density function is a
symmetric bell-shaped curve centered on the
mean (see Fig. 6.8)
4. We define the normal distribution using the
notation X~N(μ,σ2).
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-15
The Normal Distribution
(continued)

The normal distribution closely approximates the
probability distributions of a wide range of random
variables

Distributions of sample means approach a normal
distribution given a “large” sample size

Computations of probabilities are direct and elegant

The normal probability distribution has led to good
business decisions for a number of applications
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-16
The Normal Distribution Shape
By varying the parameters μ and σ, we obtain
different normal distributions
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-17
The Normal Distribution
Shape
f(x)
Changing μ shifts the
distribution left or right.
σ
Changing σ increases
or decreases the
spread.
μ
x
Given the mean μ and variance σ we define the normal
distribution using the notation
X ~ N(μ,σ 2 )
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-18
Cumulative Normal Distribution
 For a normal random variable X with mean μ and
variance σ2 , i.e., X~N(μ, σ2), the cumulative
distribution function is
F(x 0 )  P(X  x 0 )
f(x)
P(X  x 0 )
0
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
x0
x
Chap 6-19
Finding Normal Probabilities
The probability for a range of values is
measured by the area under the curve
P(a  X  b)  F(b)  F(a)
a
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
μ
b
x
Chap 6-20
Finding Normal Probabilities
(continued)
F(b)  P(X  b)
a
μ
b
x
a
μ
b
x
a
μ
b
x
F(a)  P(X  a)
P(a  X  b)  F(b)  F(a)
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-21
THE STANDARD NORMAL
DISTRIBTUION
z Values or z Scores
Definition
The units marked on the horizontal axis of the
standard normal curve are denoted by z and are
called the z values or z scores. A specific value of z
gives the distance between the mean and the point
represented by z in terms of the standard deviation.
The Standardized Normal

Any normal distribution (with any mean and variance
combination) can be transformed into the
standardized normal distribution (Z), with mean 0
and variance 1
f(Z)
Z ~ N(0 ,1)
1
0

Z
Need to transform X units into Z units by subtracting the
mean of X and dividing by its standard deviation
X μ
Z
σ
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-23
Example

If X is distributed normally with mean of 100
and standard deviation of 50, the Z value for
X = 200 is
X  μ 200  100
Z

 2.0
σ
50

This says that X = 200 is two standard
deviations (2 increments of 50 units) above
the mean of 100.
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-24
Comparing X and Z units
100
0
200
2.0
X
Z
(μ = 100, σ = 50)
(μ = 0, σ = 1)
Note that the distribution is the same, only the
scale has changed. We can express the problem in
original units (X) or in standardized units (Z)
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-25
Example 5.3 (b.p-191):
investment portfolio value
 A client has an investment portfolio whose
mean value is equal to $500,000 with a S.D. of
$15,000. She has asked you to determine the
probability that the value of her portfolio is
between $485,000 and $530,000. Also draw the
normal distribution diagram for this problem.
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-26
Example 5.4 (b.p-192):
normal probabilities
If X~N(15,16), find the probability that X is larger than 18.
Show relevant diagram
What will be the probability that X is less than 18?
Use Table 1 given in the Appendix (will be given in the exam)
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-27
Probability as
Area Under the Curve
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below
f(X) P(  X  μ)  0.5
0.5
P(μ  X  )  0.5
0.5
μ
X
P(  X  )  1.0
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-28
Appendix Table 1
 The Standardized Normal table in the textbook
(Appendix Table 1) shows values of the
cumulative normal distribution function
 For a given Z-value a , the table shows F(a)
(the area under the curve from negative infinity to a )
F(a)  P(Z  a)
0
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
a
Z
Chap 6-29
The Standardized Normal Table
 Appendix Table 1 gives the probability F(a) for
any value a
.9772
Example:
P(Z < 2.00) = .9772
0
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
2.00
Z
Chap 6-30
The Standardized Normal Table
(continued)
 For negative Z-values, use the fact that the
distribution is symmetric to find the needed
probability:
.9772
.0228
Example:
P(Z < -2.00) = 1 – 0.9772
= 0.0228
0
2.00
Z
.9772
.0228
-2.00
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
0
Z
Chap 6-31
General Procedure for
Finding Probabilities
To find P(a < X < b) when X is
distributed normally:

Draw the normal curve for the problem in
terms of X
 Translate X-values to Z-values
 Use the Cumulative Normal Table
Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chap 6-32
Finding an x Value for a Normal
Distribution
For a normal curve, with known values of μ and σ
and for a given area under the curve to the left of
x, the x value is calculated as
x = μ + zσ
Example 6-19
(similar to Ex-5.7 in book pg-194)
Almost all high school students who intend to go to college
take the SAT test. In a recent test, the mean SAT score (in
verbal and mathematics) of all students was 1020. Debbie
is planning to take this test soon. Suppose the SAT scores
of all students who take this test with Debbie will have a
normal distribution with a mean of 1020 and a standard
deviation of 153. What should her score be on this test so
that only 10% of all examinees score higher than she
does?
Example 6-19: Solution
 Area to the left of the x value
= 1.0 - .10 = .9000
 Look for .9000 in the body of the normal distribution
table. The value closest to .9000 in Table IV is .8997,
and the z value is 1.28.
 x = μ + zσ = 1020 + 1.28(153)
= 1020 + 195.84 = 1215.84 ≈ 1216
 Thus, if Debbie scores 1216 on the SAT, only about 10%
of the examinees are expected to score higher than she
does.
Finding an x value.