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
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
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
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
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 ],...)
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)
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)=0.3173 Prob(t > 1.5)=0.1336 Prob(t > 2.0)=0.0455 Prob(t > 3.0)=0.0027... Difficult to calculate (not on most calculators), but results holds true for all Gaussians. See Appendix A in Taylor for Tables.
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. 0.32 D. 0.16 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%
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
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
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
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
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 www.stat.sc.edu/~west/javahtml/clt.html Will not try to prove C.L.T. (Laplace), but will try to give you an intuitive feel for (after CQs)
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=0.0156. The second set has a probability of 20/64=0.3125
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