# STATISTICS FOR BUSINESS AND ECONOMICS by xiaopangnv

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```									    Chapter 3-PART I
QUANTITATIVE
METHODS FOR
DISCREETE PROBABILITY
DISTRIBUTION
John Kooti

Slide 1
Chapter 3, Part I
Discrete Probability Distributions

   Random Variables
   Discrete Random Variables
   Expected Value and Variance
   Binomial Probability Distribution
   Poisson Probability Distribution

Slide 2
Random Variables

   A random variable is a numerical description of the
outcome of an experiment.
   A random variable can be classified as being either
discrete or continuous depending on the numerical
values it assumes.
   A discrete random variable may assume either a
finite number of values or an infinite sequence of
values.
   A continuous random variable may assume any
numerical value in an interval or collection of
intervals.

Slide 3
Example: JSL Appliances

   Discrete random variable with a finite number of
values:
Let x = number of TV sets sold at the store in one day
where x can take on 5 values (0, 1, 2, 3, 4)

   Discrete random variable with an infinite sequence of
values:
Let x = number of customers arriving in one day
where x can take on the values 0, 1, 2, . . .
We can count the customers arriving, but there is no
finite upper limit on the number that might arrive.

Slide 4
Discrete Probability Distributions

   The probability distribution for a random variable
describes how probabilities are distributed over the
values of the random variable.
   The probability distribution is defined by a
probability function, denoted by f(x), which provides
the probability for each value of the random variable.
   The required conditions for a discrete probability
function are:
f(x) > 0
Sf(x) = 1
   We can describe a discrete probability distribution
with a table, graph, or equation.

Slide 5
Example: JSL Appliances

   Using past data on TV sales (below left), a tabular
representation of the probability distribution for TV
sales (below right) was developed.
Number
Units Sold of Days              x      f(x)
0          80              0      .40
1          50              1      .25
2          40              2      .20
3          10              3      .05
4          20              4      .10
200                    1.00

Slide 6
Example: JSL Appliances

   A graphical representation of the probability
distribution for TV sales in one day
.50

.40
Probability

.30

.20

.10

0    1   2    3    4
Values of Random Variable x (TV sales)

Slide 7
Expected Value and Variance

   The expected value, or mean, of a random variable is
a measure of its central location.
• Expected value of a discrete random variable:
E(x) = m = Sxf(x)
   The variance summarizes the variability in the values
of a random variable.
• Variance of a discrete random variable:
Var(x) = s2 = S(x - m)2f(x)
   The standard deviation, s, is defined as the positive
square root of the variance.

Slide 8
Example: JSL Appliances

   Expected Value of a Discrete Random Variable

x      f(x)    xf(x)
0      .40      .00
1      .25      .25
2      .20      .40
3      .05      .15
4      .10      .40
1.20 = E(x)

The expected number of TV sets sold in a day is 1.2

Slide 9
Example: JSL Appliances

   Variance and Standard Deviation
of a Discrete Random Variable
x     x-m     (x - m)2   f(x)           (x - m)2f(x)
_____   _________ ___________   _______   _______________

0      -1.2        1.44        .40     .576
1      -0.2        0.04        .25     .010
2       0.8        0.64        .20     .128
3       1.8        3.24        .05     .162
4       2.8        7.84        .10     .784
1.660 = s 2
The variance of daily sales is 1.66 TV sets squared.
The standard deviation of sales is 1.29 TV sets.

Slide 10
Example: JSL Appliances

   Formula Spreadsheet for Computing Expected Value
and Variance

A        B           C                  D
1     x       f(x)       xf(x)          (x- m ) 2 f(x)
2     0       0.40      =A2*B2        =(A2-C\$7)^2*B2
3     1       0.25      =A3*B3        =(A3-C\$7)^2*B3
4     2       0.20      =A4*B4        =(A4-C\$7)^2*B4
5     3       0.05      =A5*B5        =(A5-C\$7)^2*B5
6     4       0.10      =A6*B6        =(A6-C\$7)^2*B6
7                     =SUM(C2:C6)      =SUM(D2:D6)
8                    Expected Value      Variance

Slide 11
Binomial Probability Distribution

   Properties of a Binomial Experiment
• The experiment consists of a sequence of n identical
trials.
• Two outcomes, success and failure, are possible on
each trial.
• The probability of a success, denoted by p, does not
change from trial to trial.
• The trials are independent.

Slide 12
Example: Evans Electronics

   Binomial Probability Distribution
Evans is concerned about a low retention rate for
employees. On the basis of past experience,
management has seen a turnover of 10% of the
hourly employees annually. Thus, for any hourly
employees chosen at random, management estimates
a probability of 0.1 that the person will not be with
the company next year.
Choosing 3 hourly employees a random, what is
the probability that 1 of them will leave the company
this year?
Let:          p = .10, n = 3, x = 1

Slide 13
Binomial Probability Distribution

   Binomial Probability Function

n!
f ( x)                p x (1  p ) (n  x )
x !( n  x )!
where
f(x) = the probability of x successes in n trials
n = the number of trials
p = the probability of success on any one trial

Slide 14
Example: Evans Electronics

   Using the Binomial Probability Function
n!
f ( x)                p x (1  p ) (n  x )
x !( n  x )!
3!
f (1)                ( 0.1)1 ( 0. 9) 2
1!( 3  1)!
= (3)(0.1)(0.81)

= .243

Slide 15
Example: Evans Electronics

   Using the Tables of Binomial Probabilities

p
n    x    .10     .15     .20     .25     .30     .35     .40     .45     .50
3    0   .7290   .6141   .5120   .4219   .3430   .2746   .2160   .1664   .1250
1   .2430   .3251   .3840   .4219   .4410   .4436   .4320   .4084   .3750
2   .0270   .0574   .0960   .1406   .1890   .2389   .2880   .3341   .3750
3   .0010   .0034   .0080   .0156   .0270   .0429   .0640   .0911   .1250

Slide 16
Example: Evans Electronics

   Using a Tree Diagram
First           Second       Third         Value
Worker           Worker       Worker         of x   Probab.
L (.1)              3     .0010
Leaves (.1)
S (.9)              2     .0090
Leaves (.1)
L (.1)
2     .0090
Stays (.9)
1     .0810
S (.9)
L (.1)              2     .0090
Leaves (.1)
S (.9)             1      .0810
Stays (.9)                            L (.1)
1      .0810
Stays (.9)
0     .7290
S (.9)
Slide 17
Example: Evans Electronics

• Step 1: Select a cell in the worksheet where you
want the binomial probabilities 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 BINOMDIST from the Function
Name box.
Select OK.
continued

Slide 18
Example: Evans Electronics

   Using an Excel Spreadsheet (continued)
• Step 5: When the BINOMDIST dialog box appears:
Enter 1 in the Number_s box (value of x).
Enter 3 in the Trials box (value of n).
Enter .1 in the Probability_s box (value of p).
Enter false in the Cumulative box.
[Note: At this point the desired binomial
probability of .243 is automatically computed
and appears in the right center of the dialog
box.]
Select OK (and .243 will appear in the
worksheet cell requested in Step 1).

Slide 19
The Binomial Probability Distribution

   Expected Value
E(x) = m = np
   Variance
Var(x) = s 2 = np(1 - p)
   Standard Deviation
SD( x )  s  np(1  p)
   Example: Evans Electronics
E(x) = m = 3(.1) = .3 employees out of 3
Var(x) = s 2 = 3(.1)(.9) = .27
SD( x)  s  3(.1)(.9)  .52 employees

Slide 20
MINITAB APPLICATION

   Minitab Program

Slide 21
Poisson Probability Distribution

   Properties of a Poisson Experiment
• The probability of an occurrence is the same for
any two intervals of equal length.
• The occurrence or nonoccurrence in any interval is
independent of the occurrence or nonoccurrence
in any other interval.

Slide 22
Poisson Probability Distribution

   Poisson Probability Function

m x e m
f ( x) 
x!
where
f(x) = probability of x occurrences in an interval
m = mean number of occurrences in an interval
e = 2.71828

Slide 23
Example: Mercy Hospital

Patients arrive at the emergency room of Mercy
Hospital at the average rate of 6 per hour on weekend
evenings. What is the probability of 4 arrivals in 30
minutes on a weekend evening?

   Using the Poisson Probability Function
m = 6/hour = 3/half-hour, x = 4
34 ( 2. 71828) 3
f ( 4)                    .1680
4!

Slide 24
Example: Mercy Hospital

    Using the Tables of Poisson Probabilities
m
x      2.1      2.2      2.3      2.4       2.5      2.6      2.7      2.8      2.9     3.0
0    .1225    .1108    .1003    .0907     .0821    .0743    .0672    .0608    .0550    .0498
1    .2572    .2438    .2306    .2177     .2052    .1931    .1815    .1703    .1596    .1494
2    .2700    .2681    .2652    .2613     .2565    .2510    .2450    .2384    .2314    .2240
3    .1890    .1966    .2033    .2090     .2138    .2176    .2205    .2225    .2237    .2240
4     .0992    .1082    .1169    .1254     .1336    .1414    .1488    .1557    .1622   .1680
5     .0417    .0476    .0538    .0602   ..0668     .0735    .0804    .0872    .0940   .1008
6     .0146    .0174    .0206    .0241     .0278    .0319    .0362    .0407    .0455   .0504
7     .0044    .0055    .0068    .0083     .0099    .0118    .0139    .0163    .0188   .0216
8     .0011    .0015    .0019    .0025     .0031    .0038    .0047    .0057    .0068   .0081
9     .0003    .0004    .0005    .0007     .0009    .0011    .0014    .0018    .0022   .0027
10    .0001    .0001    .0001    .0002     .0002    .0003    .0004    .0005    .0006   .0008
11    .0000    .0000    .0000    .0000     .0000    .0001    .0001    .0001    .0002   .0002
12    .0000    .0000    .0000    .0000     .0000    .0000    .0000    .0000    .0000   .0001

Slide 25
Poisson Probability Distribution

The Poisson probability distribution can be used as
an approximation of the binomial probability
distribution when p, the probability of success, is small
and n, the number of trials, is large.
• Approximation is good when p < .05 and n > 20
• Set m = np and use the Poisson tables.

Slide 26
MINIAB APPLICATION

   Minitab Program

Slide 27
The End of Discrete Probability Distribution

Slide 28
Chapter 3, Part II
Continuous Random Variables

   Continuous Random Variables
   Normal Probability Distribution
   Exponential Probability Distribution

Slide 29
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.
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 30
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 31
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 32
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

Slide 33
MINITAB APPLICATION

   Minitab Program

Slide 34
The Normal Probability Distribution

   Graph of the Normal Probability Density Function

f(x)

x
m

Slide 35
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 36
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 37
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

xm
z
                               s
We can think of z as a measure of the number of
standard deviations x is from m.

Slide 38
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 39
Example: Pep Zone

   Standard Normal
Distribution                         Area = .2967

z = (x - m)/s                       Area = .2033
= (20 - 15)/6
= .83           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 40
Example: Pep Zone

    Using the Standard Normal Probability Table
z      .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 41
Example: Pep Zone

• 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 42
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 43
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
cutting the z.05 area of .05.
Let z.05 represent the z value 0           tail

Slide 44
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 45
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 46
Exponential Probability Distribution

   Exponential Probability Density Function
1
f ( x)        e x/m   for x > 0, m > 0
m
where              m = mean
e = 2.71828
   Cumulative Exponential Distribution Function
P ( x  x0 )  1  e  xo / m
where      x0 = some specific value of x

Slide 47
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 48
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 49
Relationship Between the
Poisson and Exponential Distributions
   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 50
The End of Chapter 3

Slide 51

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