六由。 。。 FIG.7:Two histogram with different o. 2.Erample of continuous variables It is simple enough to generalize to continuous distribution.Technically we need"infini- tesimal intervals".Thus Probability that an individual (chosen p(x)da at random)lies between r and (+dr) p(r)is the probability of getting r,or probability density.The probability that lies between a and b(a finite interval)is given by the integral of p() and the rules we deduced for discrete distributions translate in the obvious way: 1 = xp(r)dr n-ronoyke o2=(A)〉=(x2)-(2 Example:Suppose I drop a rock off a cliff of height h.As it falls,I snap a million photographs,at random intervals.On each picture I measure the distance the rock has fallen.Question:What is the average of all these distance?That is to say,what is the time average of the distance traveled? Solution:The rock starts out at rest,and picks up speed as it falls;it spends more time near the top,so the average distance must be less than h/2.Ignoring air resistance,the distancer at timet is () 14 FIG. 7: Two histogram with different σ. 2. Example of continuous variables It is simple enough to generalize to continuous distribution. Technically we need ”infini￾tesimal intervals”. Thus    Probability that an individual (chosen at random) lies between x and (x + dx)    = ρ(x)dx ρ(x) is the probability of getting x, or probability density. The probability that x lies between a and b (a finite interval) is given by the integral of ρ(x) Pab = Z b a ρ(x)dx and the rules we deduced for discrete distributions translate in the obvious way: 1 = Z +∞ −∞ ρ(x)dx hxi = Z +∞ −∞ xρ(x)dx hf(x)i = Z +∞ −∞ f(x)ρ(x)dx σ 2 ≡ ­ (∆x) 2 ® = ­ x 2 ® − hxi 2 Example: Suppose I drop a rock off a cliff of height h. As it falls, I snap a million photographs, at random intervals. On each picture I measure the distance the rock has fallen. Question: What is the average of all these distance? That is to say, what is the time average of the distance traveled? Solution: The rock starts out at rest, and picks up speed as it falls; it spends more time near the top, so the average distance must be less than h/2. Ignoring air resistance, the distance x at time t is x(t) = 1 2 gt2 14