Question

The undergraduate grade point averages​ (UGPA) of students taking an admissions test in a recent year can be approximated by a normal​ distribution, as shown in the figure. ​(a) What is the minimum UGPA that would still place a student in the top 10​% of​ UGPAs? ​(b) Between what two values does the middle 50​% of the UGPAs​ lie? u=3.32, o= 0.16

Answer 1

a)

µ=   3.32                  
σ =    0.16  

P(X>x) = 0.10
P(X≤x) =   0.9                  
                      
Z value at    0.9   =   1.2816   (excel formula =NORMSINV(   0.9   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.282   *   0.16   +   3.32  
X   =   3.53   (answer)          

b)

µ =    3.32                          
σ =    0.16                          
proportion=   0.50   
proportion left    0.50 is equally distributed both left and right side of normal curve                       
z value at   0.25   = ±   0.674   (excel formula =NORMSINV(   0.50   / 2 ) )      
                              
z = ( x - µ ) / σ                              
so, X = z σ + µ =                              
X1 =   -0.674   *   0.16   +   3.32   =   3.21  
X2 =   0.674   *   0.16   +   3.32   =   3.43