Question

Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.57 and a standard deviation of 0.4. Describe where the highest and lowest 32% of grade point averages lie.

Answer 1

Solution:

Given, the Normal distribution with,

   = 2.57

= 0.4

1) Highest 32%

For highest 32% data ,

P(X > x) = 0.32

In terms of z

P(Z > z) = 0.32

P(Z < z) = 1 - P(Z < z) = 1 - 0.32 = 0.68

But from z table ., P(Z < 0.468 ) = 0.68

So , z = 0.468

Using z score formula ,

x = + z* = 2.57 + (0.468 * 0.4) = 2.7572

The highest 32% grade point is at 2.7572

2)

Lowest 32%

For Lowest 32% data ,

P(X < x) = 0.32

In terms of z

P(Z < z) = 0.32

But from z table ., P(Z < -0.468 ) = 0.68

So , z = -0.468

Using z score formula ,

x = + z* = 2.57 + (-0.468 * 0.4) = 2.3828

The lowest 32% grade point is at 2.3828