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: 6,524 Thanks: 588 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

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