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STATISTICS FOR BUSINESS AND ECONOMICS

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					    Chapter 3-PART I
    QUANTITATIVE
    METHODS FOR
      BUSINESS
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

   Using an Excel Spreadsheet
    • 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.
   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 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
                   Salad Weight (oz.)


                                                    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

   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 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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