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 from statistical tables => z2= -0.525 z1= 0.525 z1=(x-500)/100=0.525 => x=500+52.5=552.5 z2=(x-500)/100=-0.525 => x=500-52.5=447.5 therefore the middle 40% is between 447.5 and 552.5. Now my question. Is that the correct method and approach, it is a bonus if I got it right. I only want to check the method really. Thanks.