Page 208 - Applied statistics and probability for engineers

186 Chapter 5/Joint Probability Distributions

The particular linear function that represents the average of random variables with identical
means and variances is used quite often in subsequent chapters. We highlight the results for
this special case.
Mean and Variance
of an Average If X = ( X + X + … + X p) p with E(X ) = μ for i =1, 2,… , p,
2
1
i
E X ( ) = μ (5-28a)
2
If X , X , …, X are also independent with V(X ) = σ for i = 1, 2, …, p,
1 2 p i 2
V X ( ) = σ (5-28b)
p
The conclusion for V X ( ) is obtained as follows. Using Equation 5-27 with c i = 1/ p and V(X )
i
2
= σ yields
2
2
=
2
1
(
V X) ( / p) σ +…+ (1/ ) σ 2 = σ 2 p /
p

pterms
Another useful result concerning linear functions of random variables is a reproductive
property that holds for independent, normal random variables.
Reproductive
Property of the If X , X , …, X are independent, normal random variables with E(X ) = μ and
2
1
p
Normal Distribution V X i ( ) = σ , for i = 1, 2, …, p, i i
2
i
Y = c X + c X + …+ c X p
p
1
2
1
2
is a normal random variable with
E Y ( ) = μ + c μ + …+ c p μ p
2
2
c 1 1
and
2
2
2
2
2
V Y ( ) = c σ + c σ +…+ c p σ 2 p (5-29)
2
2
1
1
The mean and variance of Y follow from Equations 5-25 and 5-27. The fact that Y has a normal
distribution can be obtained from moment generating functions in a later section of this chapter.
Example 5-32 Linear Function of Independent Normal Random Variables Let the random variables X
1
and X denote the length and width, respectively, of a manufactured part. Assume that X is normal
2 1
with E(X ) = 2 cm and standard deviation 0.1 cm and that X is normal with E(X ) = 5 cm and standard deviation 0.2 cm.
1 2 2
Also assume that X and X are independent. Determine the probability that the perimeter exceeds 14.5 cm.
1 2
Then Y = 2X + 2X is a normal random variable that represents the perimeter of the part. We obtain E(Y) = 14 cm
1 2
and the variance of Y is
4
V Y ( ) = × .4 0 1 2 + × .2 2 = .2
0
0
Now
(
.
−
.
P Y >14 5) = P ⎛ ⎜ ⎝ Y −μ Y σ Y > 14 5 14 ⎞ ⎟
0 2 ⎠
.
= ( . 0 13
P Z >1 12) = .
Example 5-33 Beverage Volume An automated illing machine ills soft-drink cans. The mean ill volume is

12.1 luid ounces, and the standard deviation is 0.1 oz. Assume that the ill 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 oz?