Question

5.3.29 Find the​ z-scores for which 15​% of the​ distribution's area lies between minusz and z.

Answer 1

proportion=   0.15                          
proportion left 1-0.15 = 0.85   is equally distributed both left and right side of normal curve                       
z score at   0.85/2 = ±   0.189   (excel formula =NORMSINV(   0.85   / 2 ) )      
                              
so,

P (-0.189<Z< 0.189 ) = 15%