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198   •  Chapter 6    /    Mechanical Properties of Metals

                                where n is the number of observations or measurements and x i  is the value of a discrete
                                measurement.
                                   Furthermore, the standard deviation s is determined using the following expression:

                                                              n         1/2
                                                              a (x i - x) 2
            Computation of                                  £ i=1      §
            standard deviation                          s =                                        (6.22)
                                                                n - 1

                                where x i , x, and n were defined earlier. A large value of the standard deviation corre-
                                sponds to a high degree of scatter.


                       EXAMPLE PROBLEM 6.6

                         Average and Standard Deviation Computations
                         The following tensile strengths were measured for four specimens of the same steel alloy:

                                           Sample Number        Tensile Strength (MPa)
                                                1                       520
                                                2                       512
                                                3                       515
                                                4                       522


                        (a)  Compute the average tensile strength.
                        (b)  Determine the standard deviation.

                         Solution
                        (a)  The average tensile strength (TS) is computed using Equation 6.21 with n = 4:
                                                        4
                                                       a  (TS) i
                                                       i=1
                                                  TS =
                                                          4
                                                       520 + 512 + 515 + 522
                                                      =
                                                                 4
                                                      = 517 MPa
                        (b)  For the standard deviation, using Equation 6.22, we obtain
                                        4             1/2
                                       a {(TS) i - TS} 2
                                      £ i=1          §
                                  s =
                                           4 - 1
                                                                            2
                                                 2
                                                              2
                                       (520 - 517) + (512 - 517) + (515 - 517) + (522 - 517) 2 1/2
                                    = c                                                   d
                                                              4 - 1
                                   = 4.6 MPa
                            Figure 6.20 presents the tensile strength by specimen number for this example problem
                            and also how the data may be represented in graphical form. The tensile strength data
                            point (Figure 6.20b) corresponds to the average value TS, and scatter is depicted by error
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