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 55​% of​ UGPAs?​(b) Between what two values does the middle 5050​% of the UGPAs​ lie?

3.3842.76Grade point average mu equals 3.38μ=3.38 sigma equals 0.19σ=0.19 x

Answer 1

Solution:
Given in the question
The undergraduate grade point averages​ (UGPA) of students taking an admissions test in a recent year can be approximated by a normal​ distribution with
Mean () = 3.38
Standard deviation () = 0.19
Solution(a)
we need to calculate minimum UGPA that would still place a student in the top 55​%
So p-value = 0.55, from Z table we found Z-score = 0.1257
So UGPA Can be calculated as
X = + Z-score * = 3.38 + 0.1257*0.19 = 3.40
So Minimum UGPA = 3.4 that would still place a student in the top 55%
Solution(b)
Here we need to calculate Upper and lower limit for middle 50%
so alpha = 1-0.5 = 0.5
alpha/2 = 0.5/2 = 0.25
Upper alpha/2 = 0.75 so from Z table we found Z-score = 0.6745
Lower alpha/2 = 0.25 so from Z table we found Z-score = -0.6745
So UGPA's score can be calculated as
Upper limit = + Z-score * = 3.38 + 0.6745*0.19 = 3.51
Lower limit = + Z-score * = 3.38 - 0.6745*0.19 = 3.25
So middle 50% of the UGPA's are in between 3.25 to 3.51