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Unit 7: Mean Deviation and Standard Deviation


                        The average marks obtained by P are higher. Hence his performance is better. The  Notes
                        coefficient of variation is lower in case of  P hence he is a more consistent student.
            Example 15:  Suppose that samples of polythene bags two manufactures, A and B are tested by
                        prospective buyer for bursting pressure, with the following results:
                   Bursting                                                                              Number of bags
                 Pressure (lb.)                    A                            B

                    5.0–9.9                        2                            9
                   10.0–14.9                       9                           11
                   15.0–19.9                      29                           18
                   20.9–24.9                      54                           32
                   25.0–29.9                      11                           27
                   30.0–34.9                       5                           13
                                                 110                           110

                        Which set of the bags has the highest average bursting pressure ? Which has more
                        uniform pressure ? If prices are the same, which manufacture’s bags would be preferred
                        by the buyer ? Why ?
            Solution:   For determining the set of bags having average bursting pressure, calculate arithmetic
                        mean and for finding out set of bags having more uniform pressure compute coefficient
                        of variation.
            Manufacturer A

                                 Calculation of Mean and Standard Deviation

                                                       ⎛  – 17.45 ⎞ m
                     Bursting         m         f      ⎜        ⎟      fd         fd 2
                                                       ⎝   5    ⎠
                   pressure (lb.)                          d
                     4.95–9.95       7.45       2          – 2         – 4         8
                     9.95–14.95      12.45      9          – 1         – 9         9
                    14.95–19.95      17.45      29         0            0          0
                    19.95–24.95      22.45      54        + 1         + 54         54
                    24.95–29.95      27.45      11        + 2         + 22         44
                    29.95–34.95      32.45      5         + 3         + 15         45

                                                                                  2
                                    N = 110                          Σfd  = 78    Σfd  = 160

                                                Σfd
                                        X =  A  +  ×  i
                                                N
                        Here            A = 17.45,  fdΣ  = 78, N = 110, i = 5

                                                   78
                                        X = 17.45 +   110  × 5  = 17.45 + 3.55 = 21.


                                              Σ  fd 2  ⎛  Σ  fd ⎞  2  160 ⎛  78 ⎞  2
                                        σ =       –  ⎜  ⎟  i ×  =   –  ⎜  ⎟  × 5
                                               N   ⎝  N ⎠      110 ⎝  110 ⎠




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