4. find the value of z such that .80 of the area under the curve lies 
between -z and z

Start with this normal curve,



we are trying to find where the middle 80% of the area starts and
stops, as indicated by the area between these two line segments:



Therefore there is 20% of the area which is not between those two
values. Due to symmetry, 10% will be left of the lower value and
the other 10% will be right of the upper value.  So now we have:

 
  
Now I don't know what kind of normal table you have in your book,
as there are two kinds currently in use, or whether you're not 
using a table at all but using a TI-83 or TI-84 or similar 
graphical calculator.

In one kind of table, there are negative values given for z.  If 
you have that kind of table, you'll just look through the body of 
the table and find the value closest to .1000.  The closest you 
find is .1003, where z = -1.28.  So the answer to the question 
is 1.28, and the middle 80% of the area will extend all the way 
from -1.28 to +1.28.

If you have the other kind of table, where negative values for z 
are not given, then you have to look through the body of your 
table to find the closest z-value for half the area, or .4.  The 
closest you can find to .4 is .3997, which corresponds to a 
z-value of 1.28. It's the area between the y-axis and the upper 
bound of the area, which is half of the total area. 

If you're using a TI-83 or TI-84 then

From the cleared main screen

PRESS 2nd VARS ---    for DISTR
PRESS 3        ---    on the screen you see InvNorm(
Type .1)       ---    on the screen you see InvNorm(.1)
Press ENTER    ---    on the screen you see -1.281551567

That's the lower value so the upper value in 1.281551567.
Rounded off to two decimal places, z = 1.28, same as with
tables.

Edwin