Chebyshev's inequality states that
[tex]\mathbb P(|X-\mu|\ge k\sigma)\le\dfrac1{k^2}\implies\mathbb P(|X-\mu|\le k\sigma)\ge1-\dfrac1{k^2}[/tex]
We assume [tex]\mu=\bar x=28.19[/tex] and [tex]\sigma=s=4.06[/tex], so that the inequality is
[tex]\mathbb P(|X-28.19|\le4.06k)\ge1-\dfrac1{k^2}[/tex]
and since we want the probability to be at least 0.75, we have
[tex]1-\dfrac1{k^2}=\dfrac34\implies k=2[/tex]
which means we get from Chebyshev's inequality that
[tex]\mathbb P(|X-28.19|\le8.12)=\mathbb P(20.07\le X\le36.31)\ge0.75[/tex]
