Solution:
Given:
Recall that the z-value is expressed as
![\begin{gathered} z=(x-\mu)/(\sigma) \\ \text{where} \\ \mu\Rightarrow\operatorname{mean}\text{ value} \\ \sigma\Rightarrow s\tan dard\text{ deviation} \end{gathered}]()
Thus,

A) maximum score that can be in the bottom 10% of scores
using the table of z-values,
for the bottom 10% scores, we have

To evaluate x, substitute the value of z into equation 1.
Thus,

Thus, the maximum score that can be in the bottom 10% of scores is 2.5
B) Two values for which the middle 80% of scores lie.
From the z score values shown below:
The z scores of the value are

Thus,

Thus, the two values for which the middle 80% of scores lie are 2.5 and 4.86.