P(z > c) = 0.2464

Since this is less than .5 and
z is greater than c, we know it
is a "right tail" of the normal
curve beginning on the right side
of the normal curve.  That is, c
will be a positive number. So the area 
between z=0 and z=c, which is what 
we can read in the table, wiil be
0.5-0.2464 = 0.2536

So we look in the body of the z-table
for 0.2536.  We don't find this exactly,
as we can only find that 0.2536 is between 
the table values 0.2517 and 0.2549.  Since
0.2536 is closer to 0.2549 than it is to
0.2517, we will choose the z-value that
corresponds to 0.2549, which is z=0.69.
Thus c=0.69.
----------------------------------------
[More accuracy can be gotten by using a
TI-84 calculator, although most likely
you teacher will not allow it:

press CLEAR
press 2ND
press VARS
press 3   

Here you see invNorm(
Then type 1-0.2464)
Now you should see  invNorm(1-0.2464)

press ENTER

Read .6858622141)

Most teachers will not allow calculators,
but I do not understand why, since you still
have to understand the normal curve to know
to subtract from 1.)
-------------------------------------
P(z < c) = 0.9798

Since this is greater than .5 and
z is less than c, we know it
is a "left tail" of the normal
curve beginning on the right side
of the normal curve.  In other words,
we know that c is positive. So the area 
between z=0 and z=c, which is what 
we can read in the table, wiil be 
0.9798-0.5 = 0.4798

So we look in the body of the table
for 0.4798.  This time we do find this
value exactly, and the z-value that
corresponds to it is z=2.05.
Thus c=2.05.

(On the TI-84,

press CLEAR
press 2ND
press VARS
press 3

Here you see invNorm(
Then type .9798)
Now you should see  invNorm(.9798)

press ENTER

Read 2.049635638.  Yes, I know, your
teacher will not allow you to use
a calculator, but I still don't know
why).

Edwin