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 15​% of​ UGPAs? ​(b) Between what two values does the middle 50​% of the UGPAs​ lie? mean=3.24 SD=0.21

Answer 1

Given that,

mean = = 3.24

standard deviation = =0.21

Using standard normal table,

P(Z > z) = 10%

= 1 - P(Z < z) = 0.10

= P(Z < z ) = 1 - 0.10

= P(Z < z ) = 0.90

= P(Z < z <1.28 ) = 0.90  

z = 1.28 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x= 1.28*0.21+3.24

x= 3.5088

x=4

(B)

middle 50% of score is

P(-z < Z < z) = 0.50

P(Z < z) - P(Z < -z) = 0.50

2 P(Z < z) - 1 = 0.50

2 P(Z < z) = 1 + 0.50= 1.50

P(Z < z) =1.50 / 2 = 0.75

P(Z < 0.67) = 0.75

z  ± 0.67 using z table

Using z-score formula  

x= z * +

x= -0.67 *0.21+3.24

x= 3.0993

z = 067

Using z-score formula  

x= z * +

x= 0.67 *0.21+3.24

x= 3.3807

answer=3.0993 to 3.3807