Problem 2. Gaussian Integrals The famous Bell shaped curve (also called a normal distribution or simply a Gaussian) is defined by the function P(x) (1) Try to memorize it. A graph of the normal distribution is shown below. It is often applied to student grades. In this case represents the grade, and x-μ represents the deviation from the mean grade μ. The probability of finding a grade between x and x + dx is supposedly given by P(x) dx. Here is the standard deviation. Approximately 67% of grades are supposed to lie within plus or minus "one sigma" of the mean, which is shown by the band in the figure below. In my experience with student grades the bell curve is almost never realized. We will set the mean to zero μ = 0 below, and it will often be zero in this course. P(x) 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 -4 -3 And so, he found cleverly Consider the integral of the Gaussian: [² 2= [" dre-2²³² [ dx for n odd. -2 1 V2πσ2 I = 1 - L² dre ∞ (c) Consider integrals of the form Gauss invented a rather clever trick for evaluating this integral by squaring it: [ da fody e-9(²³²+1²). ∞ -1 0 1 (x-μ) / 0 e-(x-μ)²/20² dy e-By² Then integral can be evaluated simply by changing to Polar coordinates, dx dy = rdrdo: 2π 1² = 1²ht de fon dx e-Bx² I = I2n = √²/3. 3 = πT rdre-Br² FR (a) By a simple change of variables use the result for I to show that *P(x) dx = 1. 2 This means that the probability density P(x) is correctly normalized. (b) Explain why = **P(x) x" dr = 0, 3 4 -Bx² - L Use the generating function trick of last week to find I2 and 14. e x²n dx. (4) G (6) (7) X