Suppose that you have an exponential random variable X with mean 7. Using properties of exponential distributions, you also know that the standard deviation is 7. Also, you should note that for an exponential distribution the random variable represents time and thus can never be smaller than 0. It follows then that
\begin{equation*}
P( \mu - 1.8 \sigma \le X \le \mu + 1.8 \sigma) = P( 7 - 1.8 \cdot 7 \le X \le 7 + 1.8 \cdot 7) \\ = P( 0 \le X \le 19.6) = F(19.6) \approx 0.939.
\end{equation*}
since the exponential distribution has a known distribution function.
However, using the Chebyshev’s Theorem,
\begin{equation*}
P( \mu - 1.8 \sigma \le X \le \mu + 1.8 \sigma) = P( \big | X - \mu \big | \lt 1.8 \cdot \sigma ) \gt 1 - \frac{1}{{1.8}^2} \approx 0.691.
\end{equation*}
The difference in these two results is not a problem since the first is designed to give you a precise answer with the knowledge that X itself has a known probability function whereas in the second case you only presume that X has the desired mean and standard deviation. With less information, you get a less precise lower bound but since the lower bound \(= 0.691 < 0.939 = \) exact value, then there is no conflict.