Need help understanding how an answer was achieved: (my worked out answer is inside)
Particle physicists use particle track detectors to determine the lifetime of short-lived particles. A muon has a mean lifetime of 2.2 micro/seconds and makes a track 9.5 cm long before decaying into an electron and two neutrinos. What was the speed of the muon?

Question

Need help understanding how an answer was achieved:  (my worked out answer is inside)
Particle physicists use particle track detectors to determine the lifetime of short-lived particles. A muon has a mean lifetime of 2.2 micro/seconds and makes a track 9.5 cm long before decaying into an electron and two neutrinos. What was the speed of the muon?

errrwut · Answer

$$T_0 = 2.2 µs$$

$$L = 9.5cm$$

Using,

$$\beta = \frac{L*\sqrt{1-\beta^2}}{c*T_0}$$

Solving for 
$$\beta$$

$$\beta=\frac{L}{\sqrt{L^2+c^2*T^2}}$$

You can neglect c for the mean time? 

I get 

$$\beta = 0.974218$$

However, the solution is supposedly 
$$\beta = 1.4*10^{-4}$$

errrwut · Answer

I also posted more information here: http://openstudy.com/updates/5807fa6ee4b05a233fe7f33c