Physics 1140 Lecture 6: Gaussian Distributions

Transcription

1 Physics 1140 Lecture 6: Gaussian Distributions February 21/22, 2008 Homework #3 due Monday, 5 PM Should have taken data for Lab 3 this week - due Tues. Mar. 4, 5:30 PM Final (end of lectures) is next week in your correct (Thursday or Friday) lecture section

2 Exam Next Week (Feb 28 or 29) Come to your regular Thursday or Friday lecture section Bring a scientific calculator and one (1) sheet of paper on which you can hand-write anything about the course (no xeroxes or printouts) The material covered is Chapter 1-5 of Taylor and what I presented in the lectures Most of what I covered in lecture is in Chapter 1-5 of Taylor (except for some pretty basic probability and correlated errors) No correlated errors on exam Can t promise there won t be some multiple choice on it, but the vast majority will not be multiple choice (show your work for partial credit!!) My office hours (Wed 2-4) next week will be in 1140 lab area (not office) answering any questions you may have about exam material

3 Review of last week: Standard Deviation x A series of measurements of a quantity is taken with apparatus A. The distribution of A results is shown. Apparatus A is replaced with apparatus B, and a new series of measurements of the same quantity is taken. The B results are shown (same x min and x max ) CQ1: Which series of measurements has the smaller standard deviation σ? A. A B. B C. Both A and B have the same standard deviation D. Impossible to tell from the information given Answer: B. The B distribution has a width about ½ that of A

4 Review of last week: Error on Mean x The standard deviation of distribution A is twice that of B. However, there are 4 times as many measurements in A as in B CQ2: Which series of measurements has the smaller error (or uncertainty) on the mean σ μ? A. A B. B C. Both A and B have the same error on the mean D. Impossible to tell from the information given Answer: C. σ μ A = σ N-1 A / N A = 2σ N-1 B / 4N B = σ N-1 B / N B = σ μ B

5 μ TRUE, σ TRUE Do Not Change (Though Our Estimates of Them May) We assume that our measurements are drawn from a distribution with properties μ TRUE, σ TRUE (often called the parent distribution) Errors are random, and if we only draw a few measurements, μ MEAS and σ N-1 will most likely not give the values μ TRUE, σ TRUE But if parent distribution a Gaussian, can make the same probabilistic predictions true to all Gaussian distributions (68% probability μ TRUE between [μ MEAS -σ μ, μ MEAS +σ μ ],...)

6 Functional Form of Gaussian Distribution G(x) = [( 1/(σ 2π )] e -(x-μ) 2 /(2σ 2 ) Equation symmetric around μ (mean) -> G(μ - Δ) = G(μ + Δ) Factor of 1/(σ 2π) insures that - + G(x)dx = 1 Often use f(x) = A e -(x-μ) 2 /2σ 2 -> f(μ) = A - + f(x)dx = Aσ 2π (area of unnormalized Gaussian)

7 Calculating Probabilities with Gaussians Two weeks ago introduced concept of t = x predicted -x measured / σ If x has an uncertainty associated with it, or it is another predicted measurement we expect x measured to be identical to, remember to add errors in quadrature if uncorrelated ( σ = σ 2 predicted +σ2 measured ), otherwise σ = σ measured Probability(t > 1.0)= Prob(t > 1.5)= Prob(t > 2.0)= Prob(t > 3.0)= Difficult to calculate (not on most calculators), but results holds true for all Gaussians. See Appendix A in Taylor for Tables.

8 Which σ to Use? σ μ = σ N-1 / N σ tells us how widely individual measurements (like 1 race) are N-1 distributed σ μ tells us how widely measurements of the mean are distributed CQ3: A quantity x is measured 100 times and the mean μ and the standard deviation σ are determined. What is the probability that the 101st measurement of x will give a value x > μ + σ? A. 0 B. 1 C D Answer: D Fraction of area of Gaussian between [μ - σ, μ + σ] is 68%, fraction above μ + σ is 16%, fraction below μ - σ is 16%

9 Full Width at Half Maximum (of Gaussian) G(x) = [( 1/(σ 2π )] e -(x-μ) 2 /(2σ 2 ) (normalized version) At peak G(μ) = 1/(σ 2π) At what positions (x ) is the value of the function half this ±½ [ G(x ± ) = 1/(2σ 2π) ]? ½ 1/(2σ 2π) = 1/(σ 2π)e -(x-μ) 2 /(2σ 2 ) -> 1/2 = e -(x-μ) 2 /(2σ 2 ) -(x-μ) 2 /(2σ 2 ) = ln(1/2) = -ln(2) -> (x-μ) 2 = 2σ 2 ln(2) x ±½ = μ ± 2σ2 ln(2) FWHM = x +½ - x = 2σ 2ln(2) = 2.35σ -½ So you can estimate the σ of a Gaussian curve by looking for the 2 points where the curve is 1/2 its max, and dividing the (x) difference by 2.35 Just like we can estimate μ from the x value of the peak

10 Standard Deviation CQ Could also estimate σ from G(x) peak value [ σ = 1/(G(μ) 2π ) ] but this requires the Gaussian is properly normalized (which it often is not) and that you know that (which you often don t) FWHM does not required normalized Gaussian CQ4: What approximately is the standard deviation (or σ) of this Gaussian? A. 1 B. 50 C. 120 D. 250 Answer: B FWHM ~ 120 σ = FWHM/2.35 ~ 50

11 Where Do Gaussians Come From? Probably nothing less Gaussian looking than the distribution of outcomes of many flips of a fair coin Assign 1 to a HEADS, 0 to a TAILS - plot is expected outcome Your exact results might vary, but we can predict the statistical behavior Break experiment down into 1600 sets of 4 contiguous coin flips and add up the score in each set Only 1 way to get a score of 0 -> TTTT Same for a score of 4 -> HHHH

12 Some Scores Come From >1 Configuration 4 ways to get a score of 1 -> HTTT, THTT, TTHT, TTTH Also 4 ways to get a score of 3 -> THHH, HTHH, HHTH, HHHT 6 ways to get a score of 2 -> HHTT, HTHT, HTTH, THHT, THTH, TTHH Can calculate frequencies for various scores (even for unfair coins) from the Binomial Theorem Plot the expected outcome of the 1600 sets -> these fit to a pretty good Gaussian with μ=2, σ~1 Only real problem is there are no > 2σ tails Entries/Score Score

13 Central Limit Theorem This is an example of the Central Limit Theorem at work Better tails with bigger sets, or something that partitions finer (dice) -> see Will not try to prove C.L.T. (Laplace), but will try to give you an intuitive feel for (after CQs)

14 Simple Probability CQs We flip a coin 6 times each, for two sets. The first set comes out HHHHHH. The second set comes out HTHTHT CQ4: Which has the more probable outcome? A. First set B. Second set C. Both sets are equally probable Answer: C (each 1/64 likely). Each coin flip is 50:50 and unaffected by previous flips We flip a coin 6 times each, for two sets. The first set comes out 6 Heads. The second set comes out 3 Heads, 3 Tails CQ5: Which has the more probable outcome? A. First set B. Second set C. Both sets are equally probable Answer: B. The first set has a probability of 1/64= The second set has a probability of 20/64=0.3125

15 How Does This Affect My Measurements? Imagine 14 different factors, each of which can knock your measurement +ε/2 or -ε/2 (equally probable) These 14 factors are uncorrelated with each other Chance they all line up +ε/2 is 1/16384 Chance they all line up -ε/2 is also 1/16384 Most likely (3432/16384) they cancel each other out But frequently (3003/16384) they ll be 2 more +ε/2 than -ε/2 (& vice versa) Expect 52 that are >2.95σ from mean for Gaussian; see 30 (out to 3.75σ) Entries/Score Score