A
ALEX_AK
Member
Hello one question here.
The solution uses 'z bar' which approximates z(1) and z(2). Why is this so? m is less than 20 here and is not large enough to use this approximation.
The solution uses 'z bar' which approximates z(1) and z(2). Why is this so? m is less than 20 here and is not large enough to use this approximation.