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 1010​% 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 0.18σ=0.18 x A normal curve labeled mu = 3.38 and sigma = 0.18 is over a horizontal x-axis labeled Grade point average from 2.76 to 4 in increments of 0.31 and is centered on 3.38.

​(a) The minimum UGPA that would still place a student in the top

1010​%

of UGPAs is

nothing.

​(Round to two decimal places as​ needed.)

​(b) The middle

5050​%

of UGPAs lies between

nothing

on the low end and

nothing

on the high end.

​(Round to two decimal places as​ needed.)

Answer 1

(A) Given that

mean = 3.38

standard deviation (sigma) = 0.18

Area = 0.90 (for top 10%, the area must be above bottom 90%)

So, required minimum GPA = invNorm(area, mean, sigma)

= invNorm(0.90,3.38,0.18)

= 3.61

(B) Middle 50% means 25% in the bottom and 25% above 50%, i.e. 75%

Using invNorm(area, mean, sigma)

mean = 3.38

standard deviation (sigma) = 0.18

Area = 0.25

Low end = invNorm(0.25,3.38,0.18)

= 3.26

and

Using invNorm(area, mean, sigma)

mean = 3.38

standard deviation (sigma) = 0.18

Area = 0.75

High end = invNorm(0.75,3.38,0.18)

= 3.50