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 5​% of​ UGPAs? ​(b) Between what two values does the middle 50​% of the UGPAs​ lie? standard deviation 0.19 and mean of 3.32

Answer 1

Solution :

Given that,  

mean = = 3.32

standard deviation = = 0.19

(a)

Using standard normal table ,

P(Z > z) = 5%

1 - P(Z < z) = 0.05

P(Z < z) = 1 - 0.05

P(Z < 1.65) = 0.95

z = Using z-score formula,

x = z * +

x = 1.65 * 0.19 + 3.32 = 3.63

minimum UGPA that is 3.63

(b)

Middle 50% as the to z values are -0.674 and 0.674

Using z-score formula,

x = z * +

x = -0.674 * 0.19 + 3.32 = 3.19

and

x = 0.674 * 0.19 + 3.32 = 3.45

Two value are 3.19 and 3.45