Is this possible?
Say we have a normal distribution of scores from range $~0 ~-~ 100~$, with $~(2/3)^{rd}~$ of score falling in between $~40 ~-~ 80~$.
Is the mean just $~50~$?
or do we calculate that somehow via $~Z~$ score given that:
We have $~P(40<X<80) = ($whatever the $~Z~$ value of $0.66$ is$)$.
I would have two equations to solve for mean & s.d. if there were two probabilities.
Feel like I am missing something.
Any help would be greatly appreciated!
Edit: When the distribution of the population is normal, then the distribution of the sample mean is also normal (http://www.stat.yale.edu/Courses/1997-98/101/sampmn.htm) - Would this make is possible to infer the sd since 0 = 50 - 3*SD?