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               184     CHAPTER 5 JOINT PROBABILITY DISTRIBUTIONS


                                    Then, Y   2X 1   2X 2 is a normal random variable that represents the perimeter of the
                                 part. We obtain, E(Y)   14 centimeters and the variance of Y is

                                                                   2
                                                                             2
                                                      V1Y2   4 
 0.1   4 
 0.2   0.0416
                                 Now,

                                               P1Y   14.52   P31Y    2	    114.5   142	 10.04164
                                                                       Y
                                                                    Y
                                                            P1Z   1.122   0.13
               EXAMPLE 5-38      Soft-drink cans are filled by an automated filling machine. The mean fill volume is 12.1 fluid
                                 ounces, and the standard deviation is 0.1 fluid ounce. Assume that the fill volumes of the cans
                                 are independent, normal random variables. What is the probability that the average volume of
                                 10 cans selected from this process is less than 12 fluid ounces?
                                    Let  X 1 ,  X , p  ,  X 10  denote the fill volumes of the 10 cans. The average fill volume
                                            2
                                           X
                                 (denoted as  ) is a normal random variable with
                                                                             0.1 2
                                                    E1X 2   12.1  and  V1X 2        0.001
                                                                              10

                                 Consequently,

                                                                    X    X   12   12.1
                                                      P1X   122   P  c                d
                                                                        X     10.001
                                                                 P1Z   3.162   0.00079



               EXERCISES FOR SECTION 5-7
               5-87.  If X and Y are independent, normal random variables  to be met. Let X and Y denote the thickness of two different
               with  E(X)   0,  V(X)   4,  E(Y)   10, and  V(Y)   9.  layers of ink. It is known that X is normally distributed with a
               Determine the following:                        mean of 0.1 millimeter and a standard deviation of 0.00031
               (a) E12X   3Y 2   (b) V12X   3Y 2               millimeter and Y is also normally distributed with a mean of
               (c) P12X   3Y   302  (d) P12 X   3Y   402       0.23 millimeter and a standard deviation of 0.00017 millime-
               5-88.  Suppose that the random variable  X represents the  ter. Assume that these variables are independent.
               length of a punched part in centimeters. Let Y be the length  (a) If a particular lamp is made up of these two inks only,
               of the part in millimeters. If E(X)   5 and V(X)   0.25, what  what is the probability that the total ink thickness is less
               are the mean and variance of Y?                    than 0.2337 millimeter?
                                                               (b) A lamp with a total ink thickness exceeding 0.2405 mil-
               5-89.  A plastic casing for a magnetic disk is composed of
                                                                  limeters lacks the uniformity of color demanded by the
               two halves. The thickness of each half is normally distributed
                                                                  customer. Find the probability that a randomly selected
               with a mean of 2 millimeters and a standard deviation of
                                                                  lamp fails to meet customer specifications.
               0.1 millimeter and the halves are independent.
               (a) Determine the mean and standard deviation of the total  5-91.  The width of a casing for a door is normally distrib-
                  thickness of the two halves.                 uted with a mean of 24 inches and a standard deviation of
               (b) What is the probability that the total thickness exceeds  1 8 inch. The width of a door is normally distributed with a
                  4.3 millimeters?                             mean of 23 and 7 8 inches and a standard deviation of 1 16
                                                               inch. Assume independence.
               5-90.  In the manufacture of electroluminescent lamps, sev-
                                                               (a) Determine the mean and standard deviation of the differ-
               eral different layers of ink are deposited onto a plastic sub-
                                                                  ence between the width of the casing and the width of the
               strate. The thickness of these layers is critical if specifications
                                                                  door.
               regarding the final color and intensity of light of the lamp are
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