Continuous Probability Distributions

Chapter 7

McGraw-Hill/Irwin Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved.
 GOALS

1. Understand the difference between discrete and continuous

distributions.
2. Compute the mean and the standard deviation for a uniform
distribution.
3. Compute probabilities by using the uniform distribution.
4. List the characteristics of the normal probability distribution.
5. Define and calculate z values.
6. Determine the probability an observation is between two
points on a normal probability distribution.
7. Determine the probability an observation is above (or below)
a point on a normal probability distribution.
8. Use the normal probability distribution to approximate the
binomial distribution.

7-2
 Continuous Probability
Distributions

 The Uniform Distribution

a  b
 The Normal Distribution

 The Exponential Distribution 

7-3

 Continuous Probability
Distributions

 A continuous random variable can assume any value in

an interval on the real line or in a collection of intervals.

 It is not possible to talk about the probability of the

random variable assuming a particular value.

 Instead, we talk about the probability of the random

variable assuming a value within a given interval.

7-4
 Continuous Probability
Distributions
 The probability of the continuous random variable
assuming a specific value is 0.

 The probability of the random variable assuming a value

within some given interval from x1 to x2 is defined to be
the area under the graph of the probability density
function between x1 and x2.

7-5
 Continuous Probability
Distributions

a  b a x1  b
x 1 x2
P(x1 ≤ x≤ x2) P(x≤ x1)

P(x≥ x1)= 1- P(x<x1)

a x1  b

P(x≥ x1)
7-6
 The Uniform Distribution

The uniform probability

distribution is perhaps
the simplest distribution
for a continuous
random variable.

This distribution is
rectangular in shape
and is defined by
minimum and maximum
values.

7-7
 The Uniform Distribution – Mean and
Standard Deviation

7-8
 The Uniform Distribution - Example

Southwest Arizona State University provides bus service to students while

they are on campus. A bus arrives at the North Main Street and
College Drive stop every 30 minutes between 6 A.M. and 11 P.M.
during weekdays. Students arrive at the bus stop at random times.
The time that a student waits is uniformly distributed from 0 to 30
minutes.

1. Draw a graph of this distribution.

2. Show that the area of this uniform distribution is 1.00.
3. How long will a student “typically” have to wait for a bus? In other
words what is the mean waiting time? What is the standard deviation
of the waiting times?
4. What is the probability a student will wait more than 25 minutes
5. What is the probability a student will wait between 10 and 20
minutes?

7-9
 The Uniform Distribution - Example

1. Draw a graph of this distribution.

7-10
 The Uniform Distribution - Example

2. Show that the area of this distribution is 1.00

7-11
 The Uniform Distribution - Example

3. How long will a student

“typically” have to wait for a
bus? In other words what is
the mean waiting time?

What is the standard

deviation of the waiting
times?

7-12
 The Uniform Distribution - Example

4. What is the P (25  Wait Time  30)  (height)(b ase)

probability a
1
student will wait  (5)
(30  0)
more than 25
minutes?  0.1667

7-13
 The Uniform Distribution - Example

5. What is the P (10  Wait Time  20)  (height)(b ase)

probability a
1
student will wait  (10)
(30  0)
between 10 and 20
minutes?  0.3333

7-14
 Example: Slater's Buffet

Slater customers are charged for the amount of salad

they take. Sampling suggests that the amount of
salad taken is uniformly distributed between 5 ounces
and 15 ounces.

f (x ) = 1/10 for 5 < x < 15

= 0 elsewhere
where
x = salad plate filling weight

7-15
 Example: Slater's Buffet

What is the probability that a customer will take

between 12 and 15 ounces of salad?
F (x )

P(12 < x < 15) = (1/10)*(3) = .3

1/10

x
5 10 12 15
Salad Weight (oz.)

7-16
 The Uniform Probability
Distribution
f (x )

P(8<x < 12) = ? 1/10

5 8 12 15 x

P(8<x < 12) = (1/10)*(12-8) = .4

7-17
 The Uniform Probability
Distribution
f (x )

1/10
P(0<x < 12) = ?

5 12 15 x

P(0<x < 12) = P(5<x < 12)=

= (1/10)(12-5) = .7

7-18
 Continuous Probability

McGraw-Hill/Irwin Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved.
 Normal Probability Distribution

7-20
 Characteristics of a Normal
Probability Distribution
1. It is bell-shaped and has a single peak at the center of the
distribution.
2. It is symmetrical about the mean
3. It is asymptotic: The curve gets closer and closer to the X-axis
but never actually touches it. To put it another way, the tails of
the curve extend indefinitely in both directions.
4. The location of a normal distribution is determined by the
mean,, the dispersion or spread of the distribution is
determined by the standard deviation,σ .
5. The arithmetic mean, median, and mode are equal
6. The total area under the curve is 1.00; half the area under the
normal curve is to the right of this center point and the other
half to the left of it

7-21
 The Normal Distribution - Graphically

7-22
 The Family of Normal Distribution

Equal Means and Different Different Means and

Standard Deviations Standard Deviations

Different Means and Equal Standard Deviations

7-23
 The Standard Normal Probability Distribution

 The standard normal distribution is a normal

distribution with a mean of 0 and a standard
deviation of 1.
 It is also called the z distribution.
 A z-value is the signed distance between a
selected value, designated X, and the population
mean , divided by the population standard
deviation, σ.
 The formula is:

7-24
 Areas Under the Normal Curve

7-25
 Using Standard Normal Distribution Table

P(0<Z<1.57)
= 0.4418

7-26
 P(Z>0.94)=

0.50-0.3264

=0.1736

7-27
 The Normal Distribution – Example

The weekly incomes of shift

foremen in the glass
industry follow the
normal probability
distribution with a mean
of $1,000 and a
standard deviation of
$100.
What is the z value for the
income, let’s call it X, of
a foreman who earns
$1,100 per week? For a
foreman who earns
$900 per week?
7-28
 Finding the Area Under Standard Normal
Probability Distribution:

 Four Situations to Find Area:

1. Area Between 0 and z (or –z), is looked up directly from the
table.

2. Find Area Beyond +z or less than –z, locate the probability of z in

the table and subtract the probability from 0.50

3. To find the area between two points on different sides of the

mean, determine the z values and add the corresponding
probabilities.

4. To find the area between two points on the same side of the
mean determine the z values and subtract the smaller probability
7-29
from the larger
 The Empirical Rule

 About 68 percent of
the area under the
normal curve is
within one standard
deviation of the
mean.
 About 95 percent is
within two standard
deviations of the
mean.
 Practically all is
within three
standard deviations
of the mean.
7-30
 The Empirical Rule - Example

As part of its quality assurance

program, the Autolite Battery
Company conducts tests on
battery life. For a particular
D-cell alkaline battery, the
mean life is 19 hours. The
useful life of the battery
follows a normal distribution
with a standard deviation of
1.2 hours.

Answer the following questions.

1. About 68 percent of the
batteries failed between
what two values?
2. About 95 percent of the
batteries failed between
what two values?
3. Virtually all of the batteries
failed between what two
values?

7-31
 Normal Distribution – Finding Probabilities

In an earlier example we
reported that the
mean weekly income
of a shift foreman in
the glass industry is
normally distributed
with a mean of $1,000
and a standard
deviation of $100.

What is the likelihood of

selecting a foreman
whose weekly income
is between $1,000
and $1,100?

7-34
 Normal Distribution – Finding Probabilities

7-35
 Finding Areas for Z Using Excel

The Excel function

=NORMDIST(x,Mean,Standard_dev,Cumu)
=NORMDIST(1100,1000,100,true)
generates area (probability) from
Z=1 and below

7-36
 Normal Distribution – Finding Probabilities
(Example 2)

Refer to the information

regarding the weekly income
of shift foremen in the glass
industry. The distribution of
weekly incomes follows the
normal probability
distribution with a mean of
$1,000 and a standard
deviation of $100.
What is the probability of
selecting a shift foreman in
the glass industry whose
income is:
Between $790 and $1,000?

7-37
 Normal Distribution – Finding Probabilities
(Example 3)

Refer to the information

regarding the weekly income
of shift foremen in the glass
industry. The distribution of
weekly incomes follows the
normal probability
distribution with a mean of
$1,000 and a standard
deviation of $100.
What is the probability of
selecting a shift foreman in
the glass industry whose
income is:
Less than $790?

7-38
 Normal Distribution – Finding Probabilities
(Example 4)

Refer to the information

regarding the weekly income
of shift foremen in the glass
industry. The distribution of
weekly incomes follows the
normal probability
distribution with a mean of
$1,000 and a standard
deviation of $100.
What is the probability of
selecting a shift foreman in
the glass industry whose
income is:
Between $840 and $1,200?

7-39
 Normal Distribution – Finding
Probabilities (Example 5)
Refer to the information
regarding the weekly income
of shift foremen in the glass
industry. The distribution of
weekly incomes follows the
normal probability
distribution with a mean of
$1,000 and a standard
deviation of $100.
What is the probability of
selecting a shift foreman in
the glass industry whose
income is:
Between $1,150 and $1,250

7-40
 Finding the Area Under Standard Normal
Probability Distribution:

 Four Situations to Find Area:

1. Area Between 0 and z (or –z), is looked up directly from the
table.

2. Find Area Beyond +z or less than –z, locate the probability of z in

the table and subtract the probability from 0.50

3. To find the area between two points on different sides of the

mean, determine the z values and add the corresponding
probabilities.

4. To find the area between two points on the same side of the
mean determine the z values and subtract the smaller probability
7-41
from the larger
 Finding Probabilities

 Find P(-1.32<Z<2.14)

 Find P(Z>0.82)

 Find P(1.56<Z<2.22)

 Find P(-0.95<Z<-0.43)

 Find P(Z<1.65)
7-42
 Using Z in Finding X Given Area - Example

Layton Tire and Rubber Company

wishes to set a minimum
mileage guarantee on its new
MX100 tire. Tests reveal the
mean mileage is 67,900 with a
standard deviation of 2,050
miles and that the distribution of
miles follows the normal
probability distribution. Layton
wants to set the minimum
guaranteed mileage so that no
more than 4 percent of the tires
will have to be replaced.
What minimum guaranteed
mileage should Layton
announce?

7-43
 Using Z in Finding X Given Area - Example

Solve X using the formula :

x -  x  67,900
z 
 2,050

The value of z is found using the 4% informatio n

The area between 67,900 and x is 0.4600, found by 0.5000 - 0.0400
Using Appendix B.1, the area closest to 0.4600 is 0.4599, which
gives a z alue of - 1.75. Then substituti ng into the equation :

x - 67,900
- 1.75  , then solving for x
2,050

- 1.75(2,050)  x - 67,900

x  67,900 - 1.75(2,050)

x  64,312
7-44
 Normal Approximation to Binomial

McGraw-Hill/Irwin Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved.
 Binomial Probability Experiment

1. An outcome on each trial of an experiment is

classified into one of two mutually exclusive
categories—a success or a failure.
2. The random variable counts the number of successes
in a fixed number of trials.
3. The probability of success and failure stay the same
for each trial.
4. The trials are independent, meaning that the outcome
of one trial does not affect the outcome of any
other trial.

7-46
 Normal Approximation to the Binomial

 The normal distribution (a continuous

distribution) yields a good approximation of the
binomial distribution (a discrete distribution) for
large values of n.

 The normal probability distribution is generally

a good approximation to the binomial
probability distribution when n and n(1- )
are both greater than 5.

7-47
 Normal Approximation to the Binomial
Using the normal distribution (a continuous distribution) as a substitute
for a binomial distribution (a discrete distribution) for large values of n
seems reasonable because, as n increases, a binomial distribution gets
closer and closer to a normal distribution.

7-48
 Continuity Correction Factor

The value .5 subtracted or added, depending on the

problem, to a selected value when a binomial probability
distribution (a discrete probability distribution) is being
approximated by a continuous probability distribution (the
normal distribution).

7-49
 How to Apply the Correction Factor

Only one of four cases may arise:

1. For the probability at least X occurs, use the
area above (X -0.5).
2. For the probability that more than X occurs, use
the area above (X+0.5).
3. For the probability that X or fewer occurs, use
the area below (X +0.5).
4. For the probability that fewer than X occurs, use
the area below (X-0.5).
7-50
 Normal Approximation to the Binomial - Example

Suppose the management

of the Santoni Pizza
Restaurant found that 70
percent of its new
customers return for
another meal. For a week
in which 80 new (first-
time) customers dined at
Santoni’s, what is the
probability that 60 or
more will return for
another meal?

7-51
 Normal Approximation to the Binomial -
Example

Binomial distribution solution:

P(X ≥ 60) = 0.063+0.048+ … + 0.001) = 0.197

7-52
 Normal Approximation to the Binomial -
Example

Step 1. Find the mean

and the variance of a
binomial distribution
and find the z
corresponding to an
X of 59.5 (x-.5, the
correction factor)
Step 2: Determine the
area from 59.5 and
beyond

7-53
 Exponential Distribution
 Continuous Distribution

 Usually Describes Times Between Events in

a Sequence

 Actions Occur Independently at a Constant

Rate Per Unit of Time.

 Exponential Random Variable is Always

Positive. 54
 Exponential Distributions
 Distribution is positively skewed

 Defined by only one parameter, which is λ.

 λ is the rate parameter.

 λ =1/ (closely related to Poisson distribution).

 As we decrease λ, the shape of the distribution becomes less

skewed.

55
 Examples of Exponential Distributions

 Service Times in A System. Time It Takes to serve a

customer.

 Time Between “Hits” on a website

 The lifetime of an electrical component.

 The time until the next phone call arrives in a

customer service center.

56
 The Exponential Probability Distribution
 Exponential Probability Density Function

xx> 0,  > 0
f ( x )  e
for

where  = mean and λ=1/

e = 2.71828
 Cumulative Exponential Distribution Function

 x0
P ( x  x0 )  1  e
where x0 = some specific value of x
57
 Area Greater Than A Particular “x” Value

P( x  x0 )  1  P ( x  x0 )  1  (1  e  x0 )  e  x0

58
 Example: Al’s Carwash
The time between arrivals of cars at Al’s Carwash
follows an exponential probability distribution with a
mean time between arrivals of 3 minutes. Al would like
to know the probability that the time between two
successive arrivals will be 2 minutes or less.

λ=1/3
Find P(x < 2)
We know that:
 x0
P ( x  x0 )  1  e
P(x < 2) = 1 - e-(1/3)(2) = 1 - .5134 = .4866
59
 Example: Al’s Carwash
 Graph of the Probability Density Function

F (x )

.4
.3 P(x < 2) = area = .4866

.2
.1
x
1 2 3 4 5 6 7 8 9 10

60
 EXAMPLE
 Compton Computers wishes to set a minimum lifetime guarantee on its
new power supply unit. Quality testing shows the time to failure
follows an exponential distribution with a mean of 4,000 hours.
Compton wants a warranty period such that only 5 percent of the power
supply units fail during that period. What value should they set for the
warranty period?
 λ= 1/ =1/4000
 Warranty such that only 5% of units fail during this period.

 Hence, P(Arrival Time<x)=0.05

 1- e-(1/4000)(x) =0.05
 Solve for x:

61