Page 113 - Fundamentals of Probability and Statistics for Engineers

96 Fundamentals of Probability and Statistics for Engineers

Example 4.11. Problem: an inspection is made of a group of n television

picture tubes. If each passes the inspection with probability p and fails with
probability q (p q ˆ 1), calculate the average number of tubes in n tubes that
‡
pass the inspection.
Answer: this problem may be easily solved if we introduce a random variable
X j to represent the outcome of the jth inspection and define
1; if the jth tube passes inspection;

X j ˆ
0; if the jth tube does not pass inspection.
Then random variable Y , defined by
Y ˆ X 1 ‡ X 2 ‡ ‡ X n ;

has the desired property that its value is the total number of tubes passing the
inspection. The mean of X j is

EfX j gˆ 0 q†‡ 1 p†ˆ p:

Therefore, as seen from Equation (4.38), the desired average number is given by
m Y ˆ EfX 1 g ‡ ‡ EfX n gˆ np:

We can also calculate the variance of Y . If X 1 ,. . ., X n are assumed to be
independent, the variance of X j is given by

2
2
2
2
ˆ Ef X j p† gˆ 0 p† q†‡ 1 p† p ˆ pq:
j
Equation (4.41) then gives
2
2
2
ˆ ‡ ‡ ˆ npq:
Y
1
n

Example 4.12. Problem: let X 1 ,..., X n be a set of mutually independent

random variables with a common distribution, each having mean m. Show
that, for every "> 0, and as n !1,
Y

P m " ! 0; where Y ˆ X 1 ‡ ‡ X n : 4:44†

n

Note: this is a statement of the law of large numbers. The random variable Y n /
can be interpreted as an average of n independently observed random variables
from the same distribution. Equation (4.44) then states that the probability that
this average will differ from the mean by greater than an arbitrarily prescribed

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