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 October 26th, 2017, 08:56 AM #1 Member   Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 What is the probability that the average velocity is ≥ 10^(−9 ) cm/sec? Physicists say that particles in a long tube are constantly moving back and forth along the tube, each with a velocity Vk (in cm/sec) at any given moment that is normally distributed, with mean µ = 0 and variance σ ^(2) = 1. Suppose there are 10^(20) particles in the tube. What is the probability that the average velocity is ≥ 10^(−9 ) cm/sec? For this I tried to find the mean and got E(X)=o, and V(X)=10^(20), the I tried to calculate the normal cdf to find the area, but couldnt solve it October 26th, 2017, 01:10 PM #2 Global Moderator   Joined: May 2007 Posts: 6,834 Thanks: 733 You got the mean and variance for X, which is the sum. You want the mean (=0) and variance for the average, which is $\displaystyle 10^{-20}$, so the standard deviation is $\displaystyle 10^{-10}$. Now do your calculation. October 26th, 2017, 06:34 PM #3 Newbie   Joined: Oct 2017 From: US Posts: 13 Thanks: 1 That is like 10 sigma (10^-9 / 10^-10 =10) with a p value of 7.62e-24, which is pretty close to 0. Tags 10−9, average, ≥, cm or sec, probability, velocity Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ashlegen Algebra 4 February 17th, 2014 01:10 AM gorgekiter Calculus 3 October 20th, 2013 05:54 PM sayzteff Physics 0 November 24th, 2012 07:13 AM Krijnve Advanced Statistics 3 April 17th, 2010 07:38 AM mathman2 Calculus 10 October 19th, 2009 04:15 PM

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