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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