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202                        Computational Statistics Handbook with MATLAB

                                                                                           ⁄
                             normal curve of size  α 2⁄   to the left of it. In other words, we use  z ( α 2)   to
                             denote that value such that

                                                             ⁄
                                                     PZ <(  z ( α 2) ) =  α  . 2 ⁄
                                                        ⁄
                                                      ( α 2)    ( 1 –  α 2)
                                                                   ⁄
                             Thus, the area between  z     and  z     is  1 –  α.   This is shown in
                             Figure 6.3.


                                     0.45
                                      0.4

                                     0.35
                                      0.3

                                     0.25
                                      0.2
                                     0.15

                                      0.1
                                     0.05

                                        0
                                        −4   −3    −2   −1     0     1    2     3    4

                                                          ⁄
                                                                                       ⁄
                              FI F F F U URE GU 6.  RE RE RE 6. 6. 6. 3  3  ( α 2)  ( 1 –  α 2)
                               IG
                               GU
                               G
                                     3
                               II
                                     3
                              The left vertical line corresponds to  z  , and the right vertical line is at  z  . So, the
                              non-shaded areas in the tails each have an area of  α 2⁄  , and the shaded area in the middle
                              is 1 –  α  .
                              We can see from this that the shaded area has probability 1 –  α  ,  and
                                                      ⁄
                                                                 ⁄
                                                 Pz (  ( α 2) < Z <  z (  1 –  α 2) ) =  1 – α  ,  (6.5)
                             where
                                                              X – µ
                                                         Z =  --------------- .             (6.6)
                                                              σ ⁄  n

                             If we substitute this into Equation 6.5, then we have






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