A typical large dog runs with a mean speed of $58.667$ feet/sec and a standard deviation of $2 $ feet per second. My reaction time is $0.7$ seconds, with a standard deviation of $0.1$ seconds. (Chebyshev problem) a) My dog is sitting at my feet with a $70$ foot lead on and we see a squirrel at the same time. He rans the full length of the lead before I could step on it. What is the maximum probability that my reaction time was $> 1$ sec? b) Given my reaction time was 1 second, what is the probability of having a large dog move that fast.
I am stuck on it and this is not a homework problem.
I'll try. Define: $\tau$ = "reaction time", $v$ = "speed of the dog.
The distance run by your dog in time $\tau$ is: $\Delta x= \tau v$. This must be greater than the lead length: $$\tau v > 70$$ This defines a domain on the 2-d plane:
Let's start from problem b):
Here we work only "on the orange dashed line". If you knew the form of the distribution $\rho_v(v)$ you could compute the "probability that your dog runs fast enough to escape, given a reaction time $=1s$" as:
$$P(v>\frac{70}{\tau} | \tau = 1 ) = \int_{70}^\infty dv \rho_v(v)$$
Since we don't know $\rho_v(v)$, we can bound the solution via Tchebychev's theorem. You are looking for speeds that are at least $\frac{70-58.667}{2}$ standard deviations away from the mean, so:
$$P(v>\frac{70}{\tau} | \tau = 1 ) \leq \frac{1}{(\frac{70-58.667}{2})^2} = 0.176$$
About problem a):
Again, if you knew the two distributions $\rho_v(v)$ and $\rho_\tau (\tau)$ you could compute the "probability that your reaction time was >1, given that the dog escaped" as:
$$P(\tau>1 | \tau v > 70) =\frac{ \int_1^\infty d\tau \int_{70/\tau}^\infty dv \rho_v(v) \rho_\tau(\tau)}{\int_0^\infty d\tau \int_{70/\tau}^\infty dv \rho_v(v) \rho_\tau(\tau)}= \frac{P(yellow)}{P(yellow + blue)}$$
Since we don't know them, we can bound the solution via Tchebychev's theorem. First we apply it to $\rho_v$:
$$P(\tau>1 | \tau v > 70) \leq \frac{ \int_1^\infty d\tau \rho_\tau(\tau) \frac{1}{(\frac{70/\tau - 58.667}{2})^2} }{ \int_0^\infty d\tau \rho_\tau(\tau) \frac{1}{(\frac{70/\tau - 58.667}{2})^2} } $$
I don't see now a clear way to enforce a Tchebychev-like bound also for $\rho_\tau(\tau)$! Any ideas?
edit: it could be possible that the request in your problem is to give a Tchebichev-like bound for $P(\tau>1)$ irrespectively of the fact that the dog actually escaped. In that case the problem is identical to case b) and we can write:
$$P(\tau>1) \leq \frac{1}{(\frac{1-0.7}{0.1})^2} = \frac{1}{9}$$