Page 172 - Applied statistics and probability for engineers
P. 172

150     Chapter 4/Continuous Random Variables and Probability Distributions


              Example 4-28     The time to complete a task in a large project is modeled as a generalized beta distribution with
                               minimum and maximum times a = 8  and b = 20  days, respectively, along with mode of m = 16
                                                                                      /
                               days. Also, assume that the mean completion time is μ = (a  + m  + ) 6. Determine the parameters
                                                                                    b
                                                                               4
               α and β of the generalized beta distribution with these properties.
                              ,
                           ,
                 The values (a m b ) specify the minimum, mode, and maximum times, but the mode value alone does not uniquely deter-
                                                                                                          /
               mine the two parameters α and β. Consequently, the mean completion time, μ, is assumed to equal μ = (a  + m  + ) 6.
                                                                                                        b
                                                                                                    4
                                                                −
                                                           a
                                                                  )
                 Here the generalized beta random variable is W = + ( b a X, where X is a beta random variable. Because the
               minimum and maximum values for W are 8 and 20, respectively, a = 8 and b = 20. The mean of W is
                                                                          α
                                                      −
                                                                     −
                                                                a
                                                 a
                                                                  (
                                                           (
                                              μ = + b a E X)  = + b a)
                                                    (
                                                         )
                                                                          +
                                                                        ( α β)
                                            +
                                         16
                                       4
               The assumed mean is μ = (8 + ( ) 20 )  / = 15 .333. The mode of W is
                                                 6
                                                                  α  −1
                                                             −
                                                        a
                                                    m = + ( b a)
                                                                α β
                                                                  + − 2
               with m = 16. These equations can be solved for α and β to obtain
                                                            a
                                                              2
                                                    α =  ( μ − )( m  − − )a b
                                                          (m −  μ)(b  − )a
                                                     β =  α(b −  μ)
                                                         μ − a
               Therefore,
                                                  (15 .333  − )( ( )8 2 16  − − 20 )
                                                                  8
                                              α =                       = . 3 665
                                                    (16  −15 .333 )(20  − )8
                                                            .
                                              β =  . 3 665 (20  −115 333)  = .
                                                                   2 333
                                                       .
                                                           −
                                                     15 333 8
                 Practical Interpretation: The program evaluation and review technique (PERT) widely uses the distribution of W to
               model the duration of tasks. Therefore, W is said to have a PERT distribution. Notice that we need only specify the
               minimum, maximum, and mode (most likely time) for a task to specify the distribution. The model assumes that the
               mean is the function of these three values and allows the α and β parameters to be computed.
               Exercises            FOR SECTION 4-12
                  Problem available in WileyPLUS at instructor’s discretion.
                           Tutoring problem available in WileyPLUS at instructor’s discretion.
               4-184.  Suppose that X has a beta distribution with  parameters   (a)  Calculate the mode, mean, and variance of the distribution
               α = 2.5 and β = 2.5. Sketch an approximate graph of the prob-  for α = 3 and β = 1 4. .
               ability density function. Is the density symmetric?  (b) Calculate the mode, mean, and variance of the distribution
                                                                                 .
               4-185.  Suppose that x has a beta distribution with parameters   for α = 10 and β = 6 25.
               α = 2.5 and β = 1. Determine the following:      (c)  Comment on the difference in dispersion in the distribution
                                   . (
                   (
               (a)  P X < 0 25)  (b) P 0 25 <  X < 0.75)           from parts (a) and (b).
                        .
               (c)  Mean and variance                           4-188.  The length of stay at a hospital emergency department
                                                                is the sum of the waiting and service times. Let X denote the
               4-186.  Suppose that X has a beta distribution with parameters   proportion of time spent waiting and assume a beta distribution
               α = 1 and β = 4 2. Determine the following:      with α = 10 and β = 1. Determine the following:
                          .
                   (
                                   . (
               (a)  P X < 0 25)  (b) P 0 5 < )  (c)  Mean and variance  (a)  P X > 0 9)  (b) P X < 0 5)  (c)  Mean and variance
                                                                    (
                                                                                   (
                        .
                                      X
                                                                         .
                                                                                       .
               4-187.  A European standard value for a low-emission win-  4-189.  The maximum time to complete a task in a project is 2.5
               dow glazing uses 0.59 as the proportion of solar energy that   days. Suppose that the completion time as a proportion of this max-
               enters a room. Suppose that the distribution of the proportion   imum is a beta random variable with α = 2 and β = 3. What is the
               of solar energy that enters a room is a beta random variable.  probability that the task requires more than two days to complete?
   167   168   169   170   171   172   173   174   175   176   177