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180 CHAPTER 5 JOINT PROBABILITY DISTRIBUTIONS

An important use of the bivariate normal distribution is to calculate probabilities involving
two correlated normal random variables.

EXAMPLE 5-34 Suppose that the X and Y dimensions of an injection-molded part have a bivariate normal
distribution with X 0.04, Y 0.08. X 3.00. Y 7.70, and 0.8. Then, the prob-
ability that a part satisfies both specifications is

P12.95 X 3.05, 7.60 Y 7.802

This probability can be obtained by integrating f XY (x, y; X , Y , X Y , ) over the region
2.95 x 3.05 and 7.60 y 7.80, as shown in Fig. 5-7. Unfortunately, there is often no
closed-form solution to probabilities involving bivariate normal distributions. In this case, the
integration must be done numerically.

EXERCISES FOR SECTION 5-6

5-79. Let X and Y represent concentration and viscosity of a 5-82. Suppose that X and Y have a bivariate normal distri-
chemical product. Suppose X and Y have a bivariate normal bution with joint probability density function f XY (x, y; X , Y ,
distribution with X 4, Y 1, X 2, and Y 1. Draw X , Y , ).
a rough contour plot of the joint probability density function (a) Show that the conditional distribution of Y, given that
for each of the following values for : X x is normal.
(a) 0 (b) 0.8 (b) Determine E1Y 0 X x2 .
(c) 0.8 (c) Determine V1Y 0 X x2 .
5-80. Let X and Y represent two dimensions of an injec- 5-83. If X and Y have a bivariate normal distribution with
tion molded part. Suppose X and Y have a bivariate normal 0, show that X and Y are independent.
distribution with X 0.04, Y 0.08, X 3.00, 5-84. Show that the probability density function f XY (x, y;
Y 7.70, and Y 0. Determine P(2.95 X 3.05, X , Y , X , Y , ) of a bivariate normal distribution integrates
7.60 Y 7.80). to one. [Hint: Complete the square in the exponent and use the
5-81. In the manufacture of electroluminescent lamps, fact that the integral of a normal probability density function
several different layers of ink are deposited onto a plastic for a single variable is 1.]
substrate. The thickness of these layers is critical if specifi- 5-85. If X and Y have a bivariate normal distribution with
cations regarding the final color and intensity of light of joint probability density f XY (x, y; X , Y , X , Y , ), show
the lamp are to be met. Let X and Y denote the thickness that the marginal probability distribution of X is normal
of two different layers of ink. It is known that X is nor- with mean X and standard deviation X . [Hint: Complete
mally distributed with a mean of 0.1 millimeter and a the square in the exponent and use the fact that the integral
standard deviation of 0.00031 millimeter, and Y is also of a normal probability density function for a single variable
normally distributed with a mean of 0.23 millimeter and a is 1.]
standard deviation of 0.00017 millimeter. The value of for
5-86. If X and Y have a bivariate normal distribution with
these variables is equal to zero. Specifications call for a
joint probability density f XY (x, y; X , Y , X , Y , ), show that
lamp to have a thickness of the ink corresponding to X in
the correlation between X and Y is . [Hint: Complete the
the range of 0.099535 to 0.100465 millimeters and Y in
square in the exponent].
the range of 0.22966 to 0.23034 millimeters. What is the
probability that a randomly selected lamp will conform to
specifications?

5-7 LINEAR COMBINATIONS OF RANDOM VARIABLES

A random variable is sometimes defined as a function of one or more random variables.
The CD material presents methods to determine the distributions of general functions of
random variables. Furthermore, moment-generating functions are introduced on the CD