Question

How Laude? Many educational institutions award three levels of Latin honors often based on GPA. These are cumlaude (with high praise), magnacum laude (with great praise), and summa cumlaude (with highest praise). Requirements vary from school to school. Suppose the GPAs at State College are normally distributed with a mean of 2.75 and standard deviation of 0.47.

(a) Suppose State College awards the top 2% of students (based on GPA) with the summa cumlaude honor. What GPA gets you this honor? Round your answer to 2 decimal places.
????? GPA or higher

(b) Suppose State College awards the top 10% of students (based on GPA) with the magna cumlaude honor. What GPA gets you this honor? Round your answer to 2 decimal places.
???? GPA or higher

(c) Suppose State College awards the top 20% of students (based on GPA) with the cumlaude honor. What GPA gets you this honor? Round your answer to 2 decimal places.
???? GPA or higher

Answer 1

Solution:-

Given that,

mean = = 2.75

standard deviation =  = 0.47

(a)Using standard normal table,

P(Z > z) = 2%

= 1 - P(Z < z) = 0.02  

= P(Z < z) = 1 - 0.02

= P(Z < z ) = 0.98

= P(Z < 2.054 ) = 0.98

z =2.054

Using z-score formula,

x = z * +

x =2.054 * 0.47+2.75

x = 3.72

(b)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 < 1.282 ) = 0.90

z =1.282

Using z-score formula,

x = z * +

x =1.282 * 0.47+2.75

x = 3.35

(c)Using standard normal table,

P(Z > z) = 20%

= 1 - P(Z < z) = 0.20  

= P(Z < z) = 1 - 0.20

= P(Z < z ) = 0.80

= P(Z < 0.842 ) = 0.80

z =0.842

Using z-score formula,

x = z * +

x =0.842 * 0.47+2.75

x = 3.15