Let \(X\sim\operatorname{Bin}(160,0.7)\), with mean \(112\) and variance \(33.6\). Since \(np=112\) and \(n(1-p)=48\) are large, use a normal approximation. More than 120 means \(X\ge121\); the corrected boundary is 120.5: \[\operatorname{P}(X>120)\approx1-\Phi\!\left(\frac{120.5-112}{\sqrt{33.6}}\right)=0.0713\quad\text{(3 s.f.)}.\]