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Important continuous distributions


                                                            f(x)

                                                           0.4

                                                           0.3
                                               Φ(a)
                                                           0.2

                                                           0.1
                                                                       a
                                                                                             x
                                   −4    −3     −2     −1     0     1      2      3     4
                                                            Φ(x)
                                                            1
                                                          Φ(a)


                                                           0.5



                                                                       a
                                                                                             x
                                   −4    −3     −2     −1           1      2      3     4

               Figure 6.10 – On the above, the standard Gaussian distribution f(x); the shaded area gives P(X <
               a) = Φ(a). On the below, the cumulative probability function Φ(x) for a standard Gaussian
               distribution f(x).

                                                 ( (         )      (        ))
                                                      x 2 − a         x 1 − a
                                              =   Φ            − Φ              .                         (6.23)
                                                         σ              σ
               In the special case when the interval [x 1 , x 2 ] is symmetrical with respect to the expectation a, x 1 =
               a − ε, x 2 = a + ε, the formula (6.23) gives
                                                                                ε           ε
                                                                           ( ( )        (    ))
                              P(a − ε ≤ X ≤ a + ε) = P(|X − a| ≤ ε) = Φ            − Φ −        .         (6.24)
                                                                                σ          σ
               Hence taking into account the oddness of the function Φ(x) we obtain

                                                                         ε
                                                                        ( )
                                                  P(|X − a| ≤ ε) = 2Φ        .                            (6.25)
                                                                         σ
               Example 6.8. If X is described by a Gaussian distribution of mean µ and variance
                 2
               σ , compute the probabilities that X lies within 1σ, 2σ and 3σ of the mean.                    ,
               Solution. From (6.24)

                               P(µ − nσ < X ≤ µ + nσ) = Φ(n) − Φ(−n) = Φ(n) − [1 − Φ(n)],

               and so from tabular valued of Φ we obtain

                                   P(µ − σ < X ≤ µ + σ) = 2Φ(1) − 1 = 0.6826 ≈ 68.3%,

                                  P(µ − 2σ < X ≤ µ + 2σ) = 2Φ(2) − 1 = 0.9544 ≈ 95.4%,
                                  P(µ − 3σ < X ≤ µ + 3σ) = 2Φ(3) − 1 = 0.9974 ≈ 99.7%.

               Thus we expect X to be distributed in such a way that about two thirds of the values will lie
               between µ − σ and µ + σ, 95% will lie within 2σ of the mean and 99.7% will lie within 3σ
               of the mean. These limits are called the one-, two- and three-sigma limits respectively; it is
               particularly important to note that they are independent of the actual values of the mean and
               variance.


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