Download Ch 3B

Anderson
Sweeney
Williams
QUANTITATIVE
METHODS FOR
BUSINESS 8e
Slides Prepared by JOHN LOUCKS
© 2001 South-Western College Publishing/Thomson Learning
Slide 1
Chapter 3, Part II
Continuous Random Variables



Continuous Random Variables
Normal Probability Distribution
Exponential Probability Distribution
Slide 2
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.
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.
Slide 3
Uniform Probability Distribution
A random variable is uniformly distributed whenever
the probability is proportional to the length of the
interval.
 Uniform Probability Density Function
f(x) = 1/(b - a) for a < x < b
= 0 elsewhere
 Expected Value of x
E(x) = (a + b)/2
 Variance of x
Var(x) = (b - a)2/12
where: a = smallest value the variable can assume
b = largest value the variable can assume
Slide 4
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.
 Probability Density Function
f(x) = 1/10 for 5 < x < 15
= 0 elsewhere
where
x = salad plate filling weight
Slide 5
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
15
10 12
Salad Weight (oz.)
Slide 6
The Normal Probability Distribution

Graph of the Normal Probability Density Function
f(x)
m
x
Slide 7
Normal Probability Distribution

The Normal Curve
• The shape of the normal curve is often illustrated
as a bell-shaped curve.
• The highest point on the normal curve is at the
mean, which is also the median and mode of the
distribution.
• The normal curve is symmetric.
• The standard deviation determines the width of
the curve.
• The total area under the curve is 1.
• Probabilities for the normal random variable are
given by areas under the curve.
Slide 8
Normal Probability Distribution

Normal Probability Density Function
1
( x  m )2 / 2s 2
f (x) 
e
s 2p
where
m = mean
s = standard deviation
p = 3.14159
e = 2.71828
Slide 9
Standard Normal Probability Distribution



A random variable that has a normal distribution
with a mean of zero and a standard deviation of one
is said to have a standard normal probability
distribution.
The letter z is commonly used to designate this
normal random variable.
Converting to the Standard Normal Distribution
z

xm
s
We can think of z as a measure of the number of
standard deviations x is from m.
Slide 10
Example: Pep Zone
Pep Zone sells auto parts and supplies including a
popular multi-grade motor oil. When the stock of this
oil drops to 20 gallons, a replenishment order is placed.
The store manager is concerned that sales are being
lost due to stockouts while waiting for an order. It has
been determined that leadtime demand is normally
distributed with a mean of 15 gallons and a standard
deviation of 6 gallons.
The manager would like to know the probability of a
stockout, P(x > 20).
Slide 11
Example: Pep Zone

Standard Normal
Distribution
z = (x - m)/s
= (20 - 15)/6
= .83
Area = .2967
Area = .2033
Area = .5
z
0 .83
The Standard Normal table shows an area of .2967 for
the region between the z = 0 line and the z = .83 line
above. The shaded tail area is .5 - .2967 = .2033. The
probability of a stockout is .2033.
Slide 12
Example: Pep Zone

z
Using the Standard Normal Probability Table
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359
.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753
.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141
.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517
.4
.1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879
.5
.1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224
.6
.2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549
.7
.2580 .2612 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852
.8
.2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133
.9
.3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389
Slide 13
Example: Pep Zone

Using an Excel Spreadsheet
• Step 1: Select a cell in the worksheet where you
want the normal probability to appear.
• Step 2: Select the Insert pull-down menu.
• Step 3: Choose the Function option.
• Step 4: When the Paste Function dialog box
appears:
Choose Statistical from the Function
Category box. Choose NORMDIST from the
Function Name box. Select OK.
continue
Slide 14
Example: Pep Zone

Using an Excel Spreadsheet (continued)
• Step 5: When the NORMDIST dialog box appears:
Enter 20 in the x box.
Enter 15 in the mean box.
Enter 6 in the standard deviation box.
Enter true in the cumulative box.
Select OK.
At this point, .7967 will appear in the cell selected in
Step 1, indicating that the probability of demand during
lead time being less than or equal to 20 gallons is .7967.
The probability that demand will exceed 20 gallons is
1 - .7967 = .2033.
Slide 15
Example: Pep Zone
If the manager of Pep Zone wants the probability of a
stockout to be no more than .05, what should the
reorder point be?
Area = .05
Area = .5 Area = .45
z.05
0
Let z.05 represent the z value cutting the tail area of .05.
Slide 16
Example: Pep Zone

Using the Standard Normal Probability Table
We now look-up the .4500 area in the Standard
Normal Probability table to find the corresponding
z.05 value.
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441
1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545
1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633
1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706
1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767
.
z.05 = 1.645 is a reasonable estimate.
Slide 17
Example: Pep Zone
The corresponding value of x is given by
x = m + z.05s
= 15 + 1.645(6)
= 24.87
A reorder point of 24.87 gallons will place the
probability of a stockout during leadtime at .05.
Perhaps Pep Zone should set the reorder point at 25
gallons to keep the probability under .05.
Slide 18
Exponential Probability Distribution

Exponential Probability Density Function
f ( x) 
where

1
m
e x /m
for x > 0, m > 0
m = mean
e = 2.71828
Cumulative Exponential Distribution Function
P ( x  x0 )  1  e  xo / m
where
x0 = some specific value of x
Slide 19
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.
P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866
Slide 20
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
Slide 21
Relationship Between the
Poisson and Exponential Distributions
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

The continuous exponential probability distribution is
related to the discrete Poisson distribution.
The Poisson distribution provides an appropriate
description of the number of occurrences per interval.
The exponential distribution provides a description of
the length of the interval between occurrences.
Slide 22
The End of Chapter 3, Part II
Slide 23