Recent content by fobster

Got a question I need a little bit of help. Assume the scores on an aptitude are normally distributed with mean=500 and standard deviation=100 What is the middle 40%? My workings p(x1≤x≤x2)=p(z1≤z≤z2) => p(z1≤z≤z2)=p(z≤z2)-p(z≥z1)=p(z≤z2)-[1-p(z≤z1)] p(z≤z2)=0.7 p(z≤z1)=0.3...

So is this right? p(a manager and in office A)=0.3*0.1=0.03 p(a manager)=0.095 0.03/0.095=0.31579

An insurance company runs three offices, A, B and C. The company's employess are distirbuted as follows; 30% work in office A, 20% in Off. B and 50% in Off. C. In office A 10% are managers, in office B 20% are managers and in office C 5% are managers a. What is the total proportion of...