a normal distribution has a man of 80 and a standard deviation of 14. Determine the value above which 80 percent of the value will occur Solution from a table: left-tail | P(z<-0.842) = 0.2 z score = obs-exp / sd -0.842=(x-80)/14 x=68.2.
a normal distribution has a man of 80 and a standard deviation of 14. Determine the value above which 80 percent of the value will occur Solution from a table: left-tail | P(z<-0.842) = 0.2 z score = obs-exp / sd -0.842=(x-80)/14 x=68.2.