Normal approximation of the binomial distribution with
"\\mu=np=80*0.48=38.4,\\;\\;\\sigma=\\sqrt{np(1-p)}=\\sqrt{80*0.48(1-0.48)}\\approx4.47"
can be used.
So, the probability that they will break their previous export record "80*0.4=32" orders is
"P(X>32)=P(Z>\\frac{32-38.4}{4.47})=P(Z>-1.43)=0.9236."
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