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
