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110     Chapter 4/Continuous Random Variables and Probability Distributions



                                      If X is a continuous random variable, for any x 1  and x 2 ,
                                          (
                                                       P x 1 <
                                                                     P x 1 ≤
                                         P x 1 ≤  X ≤ ) = (  X ≤ ) = (     X <  x 2) = (  X <  x 2)  (4-2)
                                                                                   P x 1 <
                                                  x 2
                                                                x 2
               Example 4-1     Electric Current  Let the continuous random variable X denote the current measured in a thin
                               copper wire in milliamperes. Assume that the range of X is [4.9, 5.1] mA, and assume that the
               probability density function of X is  f x ( ) = 5 for 4 9. ≤ ≤x  5 1. What is the probability that a current measurement is
                                                               .
               less than 5 milliamperes?
                 The probability density function is shown in Fig. 4-4. It is assumed that  f x ( ) = 0 wherever it is not specii cally
               deined. The shaded area in Fig. 4-4 indicates the probability.

                                                 (
                                                            (
                                               P X < 5) =  ∫ 5  f x dx )  =  5 ∫  5  dx = 0 5
                                                                           .
                                                         4 9 .     4 9
                                                                   .
               As another example,                             5 1 .
                                                (
                                                                   ( )
                                               P 4.95 <  X < 5.1) =  ∫  f x dx = 0 75.
                                                               4 95
                                                                .
               Example 4-2     Hole Diameter  Let the continuous random variable X denote the diameter of a hole drilled in
                               a sheet metal component. The target diameter is 12.5 millimeters. Most random disturbances to
               the process result in larger diameters. Historical data show that the distribution of X can be modeled by a probability
                                           . )
                                     − (
               density function  f x ( ) = 20 e  20  x −12 5  , for x ≥ 12 .5.
                 If a part with a diameter greater than 12.60 mm is scrapped, what proportion of parts is scrapped? The density func-
               tion and the requested probability are shown in Fig. 4-5. A part is scrapped if X > 12.60. Now,
                                            (
                                                            (
                                           P X > 12.60) =  ∞ ∫  f x dx )  =  ∞ ∫  20 e − ( x − 12 5.  )  dx
                                                                        20
                                                                   .
                                                         .
                                                        12 6      12 6
                                                                  ∞
                                                      = − e −20( ( x − 12 5. )  12 6 .  = 0 135
                                                                       .
                 What proportion of parts is between 12.5 and 12.6 millimeters? Now
                                                         .
                                        (
                                                           ( )
                                                                              .
                                      P 12.5 <  X < 12.6) =  12 6  f x dx = −  e − ( x − 12 5.  )  12 6  = .865
                                                                      20
                                                        ∫
                                                                                 0
                                                                              .
                                                         .
                                                        12 5                 12 5
                                                                 (
                                                                                   P X 12 6) =
                                                                                                −
               Because the total area under f x ( ) equals 1, we can also calculate P 12 5. <  X <12.6) = − (  >  .  1 0 135 =  0 865.
                                                                                1
                                                                                                         .
                                                                                                   .
                 Practical Interpretation: Because 0.135 is the proportion of parts with diameters greater than 12.60 mm, a large
               proportion of parts is scrapped. Process improvements are needed to increase the proportion of parts with dimensions
               near 12.50 mm.
                                    f(x)                                   f(x)
                                     5
                                                4.9    5.1           x               12.5 12.6              x
                                   FIGURE 4-4  Probability density         FIGURE 4-5  Probability density function
                                   function for Example 4-1.               for Example 4-2.
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