Note that $3.56 is 2 standard deviations below the mean, and $3.92 is 2 standard deviations above the mean.
Chebyshev's theorem says that
[tex]P(|X-\mu|\le k\sigma)\ge1-\dfrac1{k^2}[/tex]
In this case we take [tex]k=2[/tex], and we're given [tex]\mu=\$3.74[/tex] and [tex]\sigma=\$0.09[/tex], so
[tex]P(\$3.56<X<\$3.92)\ge1-\dfrac14=0.75[/tex]
making the minimum percentage 75%.
