Let \(X\sim B(200,0.15)\). Since \(np=30\) and \(n(1-p)=170\) are large, use \(Y\sim N(30,25.5)\), with 25.5 denoting the variance. More than 40 means at least 41, so the continuity correction is 40.5: \[P(X>40)\approx P(Y>40.5)=1-\Phi\!\left(\frac{40.5-30}{\sqrt{25.5}}\right).\] The standardised boundary is \(2.07931\ldots\), giving a probability of about \(0.0188\).