Find Standard Deviation Using Chebyshev's Theorem I need some help finding the standard deviation using Chebyshev's theorem.  Here's the problem:

You have concluded that at least $77.66\%$ of the $3,075$ runners took between $60.5$ and $87.5$ minutes to complete the $10$ km race.  What was the standard deviation of these $3,075$ runners?

I set up the formula as follows:
$$.7766 = 1 - \frac{1}{k^2}$$
I got $k = 2.115721092$, which makes some sense because I know that a standard deviation of $2$ yields $75\%$, so I expected a slightly higher percentage $(77.66)$ to yield a slightly higher standard deviation.
Thanks for any hints.
 A: Given a random variable $X$ of finite expectation $\mu$ and standard deviation $\sigma$, Chebyshev's theorem states that $$P(X\not\in (\mu-k\sigma,\mu+k\sigma))\leq \frac{1}{k^2}$$

The probability of $X$ lying at least $k$ standard deviations away
  from the mean is less than or equal to $\frac{1}{k^2}$.

Given the stated conclusion, it must be that $\mu=\frac{60.5+87.5}{2}=74$ and $k\sigma=87.5-74=13.5.$
As for the value of $k$, your equation is correct: $$77.66\%=1-\frac 1{k^2}$$
$$\implies k\simeq2.11572109187$$
Therefore $$\sigma=\frac{13.5}{k}\simeq 6.38$$
A: There is no upper or lower bound on the standard deviation (apart from $0$) given that information.  Perhaps


*

*everybody finished in exactly $61$ minutes to give a standard deviation of $0$

*$2389$ people finished in $61$ minutes and $686$ finished in $100000$ minutes (almost $10$ weeks) to give a standard deviation of over $41600$


But if you knew that no more than $77.66\%$ finished the race in that time and you knew that the average time was $74$ minutes (halfway between $60.5$ and $87.5$ then you could state a lower bound for the standard deviation of about $\frac{87.5-74}{2.115721092} \approx 6.38$.  To give an example close to this lower bound, consider


*

*$2387$ people finished in $74$ minutes, $344$ finished in $60.4$ minutes and $344$ finished in $87.6$ minutes to give a standard deviation of just over $6.43$ 

