Physicists say that particles in a long tube are constantly moving back and forth along the tube, each with a velocity \(V_k\) (in cm/sec) at any given moment that is normally distributed, with mean \(\mu= 0\) and variance \(\sigma^2 = 1\). Suppose there are \(10^{20}\) particles in the tube.
Find the mean and variance of the average velocity of the particles.
What is the probability that the average velocity is \(\geq 10^{-9}\) cm/sec?
We will have to assume the particles in the long tube are independent and identically distributed. This may be a reasonable assumption as long as most particles don’t collide. The average velocity \(A_N\) is the average of \(N = 10^{20}\) random variables each with mean zero and variance 1.
Thus, we write \[A_N = \frac{1}{N} \sum_{k=1}^{N} V_{k}\]
Clearly \[E[ A_N ] = \frac{1}{N} \sum_{k=1}^{N} E[V_{k}] = 0\].
The variance \(Var[ A_N]\) requires the independence assumption to be tractable.
\[Var[ A_N ] = Var\left( \frac{1}{N} \sum_{k=1}^{N} V_{k} \right) = \frac{1}{N^2}\sum_{k=1}^{N} Var(V_{k}) = \frac{1}{N^2}\sum_{k=1}^{N} 1 = \frac{1}{N}\]
Thus, the variance of the average velocity is \(10^{-20} cm^2/sec^2\).
We know that \(N=10^{20}\) so \(\sqrt{N} = 10^{10}\) and \(1/sqrt{N}=10^{-10}\).
By the Central Limit Theorem, we know
\[\begin{align} P\left[ A_N > 10^{-9} \right] & = & P\left[ A_N > 10^{-9} \right] \\ & = & P\left[ \frac{ A_N }{1/\sqrt{N}} > \frac{10^{-9}}{10^{-10}} \right] \\ & = & P\left[ \frac{ A_N }{1/\sqrt{N}} > 10 \right] \\ & = & 1 - \Phi(10) \\ & \approx & 0 \\ \end{align} \]
To show that a 10 standard deviation event has probability nearly equal to zero, we use R to calculate the value:
(prob_speed = 1 - pnorm(10))## [1] 0We conclude the probability that the average velocity is greater than \(10^{-9}\) cm/sec is nearly zero.