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This of course was to be expected from our previous experience with the sample mean and the central limit theorem Both the median and the mean could be expected, without surprise, to be estimators of the population mean, or a measure of central tendency rather than estimators of the maximum of the population, which is what we seek It is intuitively clear that if the race cars we saw were numbered 6, 17 and 45, then there are at least 45 cars in the race But how many more

Installing your own Syn c h ro n i z at ionCont ext is tricky and should only be done when you own the thread As noted, there is no compositional mechanism to support multiple contexts on the same thread, so there is an inherent race anytime multiple components want to install their own You can chain contexts together, but this only works in some limited circum stances, such as when you just need to wrap calls to add some kind of logic such as tracing For example, here's a general purpose implementation that passes through all C reateCopy , Post , S e n d , OperationSt a rted, and Ope rationCompl eted method calls It overrides Wa it, however, to enable wrapping waits in arbi trary pre and post delegates

Payroll Best Practices / 71

u s i n g System; u s i ng System Thread i n g ; de legate o b j e c t PreWaitNot i f i c ation ( I n t Pt r [ ] waitHand l e s , bool Wa itAl l ,

int m i l l i seconds Timeout )j d e l egate void PostWa itNot i f i c a t i o n ( I n t Pt r [ ] waitHa n d le s , boo I WaitAl l , i n t m i l l i s e condsT imeout , int ret , E x c e ption e x , o b j e c t state )j c l a s s Bloc k i ngNot ifySy n c h ro n i z at ionContext : Syn c h ro n i z a t ionContext

Example 15 considers a binomial event where we toss a coin ve times In the rst sample space, S, we wrote out all the possible orders in which the tosses could occur This is of course impossible if we tossed the coin, say, 10, 000 times! In the second sample space, S1 ,we simply looked at the number of heads that occurred The difference is that the sample points are not equally likely In the rst sample space, where we enumerated the result of each toss, using the fact that the tosses are independent, and assuming that the coin is loaded, where P(H) = p and P(T ) = 1 p = q, we nd, to use two examples, that P(TTTTT ) = q5 and P(HTTHH) = p3 q2 Now we can relate the two sample spaces In S1 , P(0) = P(0 heads) = P(TTTTT ) = q5 Now P(1 head) is more complex since the single head can occur in one of the ve possible places Since these sample points are mutually disjoint, P(1 head) = 5 p q4 There are 10 points in S where two heads appear Each of these points has probability p2 q3 so P(2 heads) = 10 p2 q3 We nd, similarly, that P(3 heads) = 10 p3 q2 , P(4 heads) = 5 p4 q, and, nally, P(5 heads) = p5 So the sample points in S1 are far from being equally likely If we add all these probabilities, we nd q5 + 5 p q4 + 10 p2 q3 + 10 p3 q2 + 5 p4 q + p5 which we recognize as the binomial expansion (q + p)5 that is 1 since q = 1 p