Fundamentals of Probability and Statistics for Engineers

96 Fundamentals of Probability and Statistics for EngineersEx ample 4. 11. Problem: an inspection is made of a group of n televisionpicture tubes. If each passes the inspection with probability p and fails withprobability q (p ‡ q ˆ 1), calculate the average number of tubes in n tubes thatpass the inspection.Answer: this problem may be easily solved if we introduce a random variableX j to represent the outcome of the jth inspection and defineX j ˆ 1;0;Then random variable Y, defined byif the jth tube passes inspection;if the jth tube does not pass inspection.has the desired property that its value is the total number of tubes passing theinspection. The mean of X j isTherefore, as seen from Equation (4.38), the desired average number is given byWe can also calculate the variance of Y. If X 1 ,...,X n are assumed to beindependent, the variance of X j is given byEquation (4.41) then givesY ˆ X 1 ‡ X 2 ‡‡X n ;EfX j gˆ0…q†‡1…p† ˆp:m Y ˆ EfX 1 g‡‡EfX n gˆnp: 2 j ˆ Ef…X j p† 2 gˆ…0 p† 2 …q†‡…1 p† 2 p ˆ pq: 2 Y ˆ 2 1 ‡‡2 n ˆ npq:Ex ample 4. 12. Problem: let X 1 ,...,X n be a set of mutually independentrandom variables with a common distribution, each having mean m. Showthat, for every ">0, and as n !1,PY nm " ! 0; where Y ˆ X 1 ‡‡X n : …4:44†Note: this is a statement of the law of large numbers. The random variable Y / ncan be interpreted as an average of n independently observed random variablesfrom the same distribution. Equation (4.44) then states that the probability thatthis average will differ from the mean by greater than an arbitrarily prescribedTLFeBOOK