My Math Forum What is the probability that the average velocity is ≥ 10^(−9 ) cm/sec?

 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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top