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October 26th, 2017, 09:56 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, 02:10 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,416 Thanks: 557 
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, 07:34 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 7.62e24, which is pretty close to 0.


Tags 
10−9, average, ≥, cm or sec, probability, velocity 
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