[tex]\mathbb P(0.13<X<7.39)=\mathbb P\left(-3<\dfrac{X-3.76}{1.21}<3\right)=\mathbb P\left(|X-\mu|<3\sigma)[/tex]
where [tex]\mu[/tex] is the mean duration and [tex]\sigma[/tex] is the standard deviation. By Chebychev's theorem, this probability is bounded below by
[tex]\mathbb P(|X-\mu|<3\sigma)\ge1-\dfrac1{3^2}=\dfrac89[/tex]
