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                                                                     7-5 SAMPLING DISTRIBUTIONS OF MEANS  241


                                   shape of the population. If n   30, the central limit theorem will work if the distribution of the
                                   population is not severely nonnormal.

                 EXAMPLE 7-13      An electronics company manufactures resistors that have a mean resistance of 100 ohms and
                                   a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability
                                   that a random sample of n   25 resistors will have an average resistance less than 95 ohms.
                                       Note that the sampling distribution of  X  is normal, with mean     100 ohms  and a
                                                                                             X
                                   standard deviation of

                                                                           10
                                                                                 2
                                                                X
                                                                    1n    125
                                   Therefore, the desired probability corresponds to the shaded area in Fig. 7-7. Standardizing
                                   the point X   95  in Fig. 7-7, we find that


                                                                   95   100
                                                               z              2.5
                                                                      2
                                   and therefore,


                                                             P 1X   952   P1Z   2.52
                                                                         0.0062

                                       The following example makes use of the central limit theorem.

                 EXAMPLE 7-14      Suppose that a random variable X has a continuous uniform distribution

                                                                      1 2,  4   x   6
                                                              f  1x2   e
                                                                       0,  otherwise
                                   Find the distribution of the sample mean of a random sample of size n   40.
                                                                                      2
                                                                            2
                                       The mean and variance of X are    5 and     16   42  12   1 3 . The central limit
                                   theorem indicates that the distribution of  X  is approximately normal with mean     5  and
                                                                                                     X
                                                2
                                   variance    2      n   1  3314024   1 120 . The distributions of X and  are shown in Fig. 7-8.
                                                                                           X
                                            X
                                       Now consider the case in which we have two independent populations. Let the first pop-
                                                                 2
                                   ulation have mean   and variance   1  and the second population have mean   and variance
                                                                                                  2
                                                    1
                                     2
                                     . Suppose that both populations are normally distributed. Then, using the fact that linear
                                     2
                                                                             4          5          6   x
                                                              σ   = 2                       σ  2  = 1/120
                                                               X                             X



                                             95        100            x      4          5          6   x
                                         Figure 7-7 Probability for Example 7-13.  Figure 7-8 The distributions of X and
                                                                          X for Example 7-14.
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