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                                                                                4-6 NORMAL DISTRIBUTION   115







                                                                       x – 10
                                                                    z =    = 2.05
                                                                         2
                 Figure 4-16  Deter-
                                                                          0.98
                 mining the value of x
                 to meet a specified
                 probability.                              10           x


                                   Therefore, (x   10) 2   2.05, and the standardizing transformation is used in reverse to solve
                                   for x. The result is

                                                         x   212.052 
 10   14.1 milliamperes


                 EXAMPLE 4-15      Assume that in the detection of a digital signal the background noise follows a normal distri-
                                   bution with a mean of 0 volt and standard deviation of 0.45 volt. The system assumes a digi-
                                   tal 1 has been transmitted when the voltage exceeds 0.9. What is the probability of detecting
                                   a digital 1 when none was sent?
                                       Let the random variable N denote the voltage of noise. The requested probability is

                                                           N     0.9
                                           P1N   0.92   P  a        b   P1Z   22   1   0.97725   0.02275
                                                          0.45   0.45

                                   This probability can be described as the probability of a false detection.
                                       Determine symmetric bounds about 0 that include 99% of all noise readings. The question
                                   requires us to find x such that P1 x   N   x2   0.99 . A graph is shown in Fig. 4-17. Now,

                                                     P1 x   N   x2   P1 x	0.45   N	0.45   x	0.452
                                                                   P1 x	0.45   Z   x	0.452   0.99

                                   From Appendix Table II

                                                            P 1 2.58   Z   2.582   0.99










                                                                              Distribution of N
                                                      N
                 Figure 4-17  Deter-  Standardized distribution of  0.45
                 mining the value of x
                 to meet a specified
                 probability.                          – z    0      z                  – x  0  x
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