FIG.7:Two histogram with different. 4.What is the average(or mean)age?Answer: 国++300+2@+2e+5四-警-21 14 In general the average value ofj(which we shall write thus:())is 6)=2N0-2PU 1=0 a nechanics it is called the expectation value.Nevertheless the yalue is not sgay9器seew 的=2PU) In general,the average value of some function of j is given by U6》-立f6PU j= Beware(2〉≠)2 o三/(a)2〉=VG)-02 △j=j- For the two distributions in the above figure,we have 1=V25.2-25-V0.2 2=V31.0-25=v6 Erample of riables It is simple enough to generalize to continuous distribution.Technically we need "infinitesimal intervals".Thus =p(r)dr 9 FIG. 7: Two histogram with different σ. 4. What is the average (or mean) age? Answer: (14) + (15) + 3 (16) + 2 (22) + 2 (24) + 5 (25) 14 = 294 14 = 21 In general the average value of j (which we shall write thus: ⟨j⟩) is ⟨j⟩ = ∑jN(j) N = ∑∞ j=0 jP(j) In quantum mechanics it is called the expectation value. Nevertheless the value is not necessarily the one you can expect if you made a single measurement. In the above example, you will never get 21. 5. What is the average of the squares of the ages? Answer: You could get 142 = 196, with probability 1/14, or 152 = 225, with probability 1/14, or 162 = 256, with probability 3/14, and so on. The average is ⟨ j 2 ⟩ = ∑∞ j=0 j 2P(j) In general, the average value of some function of j is given by ⟨f(j)⟩ = ∑∞ j=0 f(j)P(j) Beware ⟨ j 2 ⟩ ̸= ⟨j⟩ 2 . Now, there is a conspicuous difference between the following two histograms, even though they have the same median, the same average, the same most probable value, and the same number of elements: The first is sharply peaked about the average value, whereas the second is broad and flat. (The first might represent the age profile for students in a big-city classroom, and the second, perhaps, a rural one-room schoolhouse.) We need a numerical measure of the amount of ”spread” in a distribution, with respect to the average. The most effective way to do this is compute a quantity known as the standard deviation of the distribution σ ≡ √⟨ (∆j) 2 ⟩ = √ ⟨j 2⟩ − ⟨j⟩ 2 ∆j = j − ⟨j⟩ For the two distributions in the above figure, we have σ1 = √ 25.2 − 25 = √ 0.2 σ2 = √ 31.0 − 25 = √ 6 2. Example of continuous variables It is simple enough to generalize to continuous distribution. Technically we need ”infinitesimal intervals”. Thus { Probability that an individual (chosen at random) lies between x and (x + dx) } = ρ(x)dx