Let X be the volume of milk in a randomly chosen carton.
X∼N(4.82,σ2)
P(X<4.64)=0.05P(Z<4.64−4.82σ)=0.054.64−4.82σ=−1.64494.64−4.82=−1.6449σσ=0.10943=0.109∎ (to 3 s.f.)
[apply standardization and inverse norm to find mu/sigma]