5-51. If X and Y have a bivariate normal distribution with 0, show that X and Y are independent.
5-52. Show that the probability density function fXY(x, y;
X, Y, X, Y, ) of a bivariate normal distribution integrates to one. [Hint: Complete the square in the exponent and use the fact that the integral of a normal probability density function for a single variable is 1.]
5-53. If X and Y have a bivariate normal distribution with joint probability density fXY(x, y;X, Y, X,Y, ), show that the marginal probability distribution of X is normal with mean Xand standard deviation X. [Hint: Complete the square in the exponent and use the fact that the integral of a normal probability density function for a single variable is 1.]
5-4 LINEAR FUNCTIONS OF RANDOM VARIABLES
A random variable is sometimes defined as a function of one or more random variables. In this section, results for linear functions are highlighted because of their importance in the remainder of the book. For example, if the random variables X1and X2denote the length and width, respectively, of a manufactured part, Y 2X12X2is a random variable that represents the perimeter of the part. As another example, recall that the negative binomial random variable was represented as the sum of several geometric random variables.
In this section, we develop results for random variables that are linear combinations of random variables.
Given random variables X1, X2, , Xpand constants c1, c2, , cp,
(5-24) is a linear combinationof X1, X2,p, Xp.
Yc1X1c2X2p cpXp
p p
Linear Combination
Now, E(Y ) can be found from the joint probability distribution of X1, X2, , Xpas follows.
Assume X1, X2, , Xpare continuous random variables. An analogous calculation can be used for discrete random variables.
cp 冮
冮
p 冮
xpfX1X2pXp1x1, x2,p, xp2 dx1 dx2pdxp
c2 冮
冮
p 冮
x2fX1X2pXp1x1, x2,p, xp2 dx1 dx2pdxp,p, c1 冮
冮
p 冮
x1fX1X2pXp1x1, x2,p, xp2 dx1 dx2pdxp
E1Y2 冮
冮
p 冮
1c1x1c2x2 p cpxp2fX1X2pXp1x1, x2,p, xp2 dx1 dx2pdxp
p
p
182 CHAPTER 5 JOINT PROBABILITY DISTRIBUTIONS
By using Equation 5-10 for each of the terms in this expression, we obtain the following.
If
(5-25) E1Y2c1E1X12c2E1X22p cpE1Xp2
Yc1X1c2X2p cpXp, Mean of a
Linear Function
Furthermore, it is left as an exercise to show the following.
If X1, X2, , Xpare random variables, and then in general,
(5-26) If X1, X2, , Xpare independent,
(5-27) V1Y2c21V1X12c22V1X22p c2pV1Xp2
p
V1Y2c21V1X12c22V1X22p c2pV1Xp22 a
ij a cicjcov1Xi, Xj2 Yc1X1c2X2 pcpXp, p
Variance of a Linear Function
Note that the result for the variance in Equation 5-27 requires the random variables to be independent. To see why the independence is important, consider the following simple exam- ple. Let X1denote any random variable and define X2 X1. Clearly, X1and X2are not inde- pendent. In fact, XY 1. Now, YX1X2is 0 with probability 1. Therefore, V(Y )0, regardless of the variances of X1and X2.
EXAMPLE 5-30 Negative Binomial Distribution In Chapter 3, we found that if Y is a negative binomial random variable with parameters p and r,
where each Xiis a geometric random variable with parameter YX1X2pXr,
p, and they are independent. Therefore, and . From Equation 5-25, , and from Equation 5-27, V1Y2r11p2p2. E1Y2rp
V1Xi211p2p2 E1Xi21p
An approach similar to the one applied in the above example can be used to verify the formulas for the mean and variance of an Erlang random variable in Chapter 4. An important use of Equation 5-27 is in error propagation and this is presented in the following example.
EXAMPLE 5-31 Error Propagation
A semiconductor product consists of three layers. If the vari- ances in thickness of the first, second, and third layers are 25, 40, and 30 nanometers squared, what is the variance of the thickness of the final product?
Let X1, X2, X3, and X be random variables that denote the thickness of the respective layers, and the final product. Then,
XX1X2X3
The variance of X is obtained from Equaion 5-27:
Consequently, the standard deviation of thickness of the final product is 951/2= 9.75 nm and this shows how the variation in each layer is propagated to the final product.
25403095 nm2 V1X2V1X12V1X22V1X32
The particular linear function that represents the average of p random variables, with identical means and variances, is used quite often in subsequent chapters. We highlight the results for this special case.
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5-4 LINEAR FUNCTIONS OF RANDOM VARIABLES 183
The conclusion for is obtained as follows. Using Equation 5-27, with and V(Xi) 2, yields
p terms
Another useful result concerning linear functions of random variables is a reproductive propertythat holds for independent, normal random variables.
V1X2 11p222 p 11p222 2p
ci1p
V1X2
If with E(Xi) for i1, 2, , p,
(5-28a) If X1, X2, , Xpare also independent with V(Xi) 2for i1, 2, , p,
(5-28b) V1X2 2
p
p p
E1X2 X1X1X2p Xp2p p
Mean and Variance of
an Average
If X1, X2, , Xp are independent, normal random variables with E(Xi) i and , for i1, 2, , p,
is a normal random variable with
and
(5-29) V1Y2c2121c2222 pc2p2p
E1Y2c11c22 p cpp
Yc1X1c2X2p cpXp
V1Xi2 2i p p Reproductive
Property of the Normal Distribution
The mean and variance of Y follow from Equations 5-25 and 5-27. The fact that Y has a nor- mal distribution can be obtained from supplemental material on moment-generating functions on the Web site for the book.
EXAMPLE 5-32 Linear Function of Independent Normal Random Variables Let the random variables X1 and X2 denote the length and
width, respectively, of a manufactured part. Assume that X1is normal with E(X1)2 centimeters and standard deviation 0.1 centimeter, and that X2is normal with E(X2)5 centime- ters and standard deviation 0.2 centimeter. Also, assume that X1and X2are independent. Determine the probability that the perimeter exceeds 14.5 centimeters.
Then, Y2X12X2is a normal random variable that represents the perimeter of the part. We obtain E(Y )14
centimeters and the variance of Y is
Now,
P1Z1.1220.13
P1Y14.52P3 1Y Y2Y114.514210.24
V1Y240.1240.220.2
μ
184 CHAPTER 5 JOINT PROBABILITY DISTRIBUTIONS
EXERCISES FOR SECTION 5-4 EXAMPLE 5-33 Beverage Volume
Soft-drink cans are filled by an automated filling machine. The mean fill volume is 12.1 fluid ounces, and the standard devia- tion is 0.1 fluid ounce. Assume that the fill volumes of the cans are independent, normal random variables. What is the proba- bility that the average volume of 10 cans selected from this process is less than 12 fluid ounces?
Let X1, X2, , X10denote the fill volumes of the 10 cans.
The average fill volume (denoted as ) is a normal random variable with
X p
Consequently,
P1Z 3.1620.00079 P1X122PcX X
X 1212.1 10.001 d E1X212.1 and V1X20.12
10 0.001
5-54. X and Y are independent, normal random variables with E(X )0, V(X )4, E(Y )10, and V(Y )9.
Determine the following:
(a) (b)
(c) (d)
5-55. X and Y are independent, normal random variables with
Determine the following:
(a) (b)
(c) (d)
5-56. Suppose that the random variable X represents the length of a punched part in centimeters. Let Y be the length of the part in millimeters. If E(X )5 and V(X )0.25, what are the mean and variance of Y ?
5-57. A plastic casing for a magnetic disk is composed of two halves. The thickness of each half is normally distributed with a mean of 2 millimeters and a standard deviation of 0.1 millimeter and the halves are independent.
(a) Determine the mean and standard deviation of the total thickness of the two halves.
(b) What is the probability that the total thickness exceeds 4.3 millimeters?
5-58. Making handcrafted pottery generally takes two major steps: wheel throwing and firing. The time of wheel throwing and the time of firing are normally distributed random variables with means of 40 min and 60 min and stan- dard deviations of 2 min and 3 min, respectively.
(a) What is the probability that a piece of pottery will be fin- ished within 95 min?
(b) What is the probability that it will take longer than 110 min?
5-59. In the manufacture of electroluminescent lamps, sev- eral different layers of ink are deposited onto a plastic sub- strate. The thickness of these layers is critical if specifications regarding the final color and intensity of light are to be met.
Let X and Y denote the thickness of two different layers of ink.
It is known that X is normally distributed with a mean of 0.1 millimeter and a standard deviation of 0.00031 millimeter and Y is also normally distributed with a mean of 0.23 millimeter and a standard deviation of 0.00017 millimeter. Assume that these variables are independent.
P13X2Y282 P13X2Y182 V13X2Y2 E13X2Y2
E1X22, V1X25, E1Y26, and V1Y28.
P12 X3Y402 P12 X3Y302 V12X3Y2 E12 X3Y2
(a) If a particular lamp is made up of these two inks only, what is the probability that the total ink thickness is less than 0.2337 millimeter?
(b) A lamp with a total ink thickness exceeding 0.2405 mil- limeter lacks the uniformity of color demanded by the customer. Find the probability that a randomly selected lamp fails to meet customer specifications.
5-60. The width of a casing for a door is normally distrib- uted with a mean of 24 inches and a standard deviation of 1兾8 inch. The width of a door is normally distributed with a mean of 23-7兾8 inches and a standard deviation of 1兾16 inch.
Assume independence.
(a) Determine the mean and standard deviation of the differ- ence between the width of the casing and the width of the door.
(b) What is the probability that the width of the casing minus the width of the door exceeds 1兾4 inch?
(c) What is the probability that the door does not fit in the casing?
5-61. An article in Knee Surgery Sports Traumatology, Arthroscopy [“Effect of Provider Volume on Resource Utilization for Surgical Procedures” (2005, Vol. 13, pp.
273–279)] showed a mean time of 129 minutes and a standard deviation of 14 minutes for ACL reconstruction surgery for high-volume hospitals (with more than 300 such surgeries per year). If a high-volume hospital needs to schedule 10 surger- ies, what are the mean and variance of the total time to com- plete these surgeries? Assume the times of the surgeries are in- dependent and normally distributed.
5-62. Soft-drink cans are filled by an automated filling machine and the standard deviation is 0.5 fluid ounce. Assume that the fill volumes of the cans are independent, normal random variables.
(a) What is the standard deviation of the average fill volume of 100 cans?
(b) If the mean fill volume is 12.1 ounces, what is the proba- bility that the average fill volume of the 100 cans is below 12 fluid ounces?
(c) What should the mean fill volume equal so that the proba- bility that the average of 100 cans is below 12 fluid ounces is 0.005?
JWCL232_c05_152-190.qxd 1/7/10 2:33 PM Page 184
(d) If the mean fill volume is 12.1 fluid ounces, what should the standard deviation of fill volume equal so that the probability that the average of 100 cans is below 12 fluid ounces is 0.005?
(e) Determine the number of cans that need to be measured such that the probability that the average fill volume is less than 12 fluid ounces is 0.01.
5-63. The photoresist thickness in semiconductor manufac- turing has a mean of 10 micrometers and a standard deviation of 1 micrometer. Assume that the thickness is normally distributed and that the thicknesses of different wafers are independent.
(a) Determine the probability that the average thickness of 10 wafers is either greater than 11 or less than 9 micrometers.
(b) Determine the number of wafers that need to be measured such that the probability that the average thickness ex- ceeds 11 micrometers is 0.01.
(c) If the mean thickness is 10 micrometers, what should the standard deviation of thickness equal so that the probability that the average of 10 wafers is either greater than 11 or less than 9 micrometers is 0.001?
5-64. Assume that the weights of individuals are indepen- dent and normally distributed with a mean of 160 pounds and a standard deviation of 30 pounds. Suppose that 25 people squeeze into an elevator that is designed to hold 4300 pounds.
(a) What is the probability that the load (total weight) exceeds the design limit?
(b) What design limit is exceeded by 25 occupants with prob- ability 0.0001?
5-65. Weights of parts are normally distributed with variance . Measurement error is normally distributed with mean zero and variance 0.5 , independent of the part weights, and adds to the part weight. Upper and lower specifications are centered at 3 about the process mean.
2 2