251distrex4 1/14/08 (Open this document in 'Page Layout' view!)
Percentiles and intervals about the mean for the Normal Distribution.
We already know the four rules below from the previous section (251disrtex2)
1) The Standardized Distribution is symmetrical about zero.
2) The area under the Normal curve, like the area under the curve for all continuous
distributions, is 1 and since the zero is the halfway point (median) under the curve, this implies
that Pz  0  .5 and Pz  0  .5 .
3) The standard notation to say that a given distribution is Normal with a certain mean and
standard deviation is ~ N  ,   . This means that, if z represents a variable with the Standardized
Normal distribution, we write z ~ N 0, 1 and if x is Normally distributed with a mean of 50 and
a standard deviation of 100 we write x ~ N 50, 100  .
4) Any probability for a Normally distributed variable can be found using the Standardized
x
Normal Distribution by using the transformation z 
.

If we solve this for x , we find x    z
Since there is no particular reason to repeat the problems from class, here are some that I had available that
seem to cover all the bases.
Assume that x ~ N (2,9) . Do the following:
a. Find x.24 (the 76th percentile.)
b. Find the x.76 (What percentile is this?).
c. Find a symmetrical interval about the mean with 76% probability.
. Solution: Material in italics below is a description of the diagrams you were asked to make or a general
explanation, and the written description will not be part of your solution.
In general, for the following problems 1) the values of z you need here must came from the z-table (Table
of the Standardized Normal Distribution), not the t-table because they aren't on the t-table. You can find
values like z .25 or z .40 on the t-table, but I didn't ask for them. 2) Numbers like z .24 are values of z not
probabilities, you can't find them by taking 0.24 on the z part of the table and then reading a probability
and claiming that it is a value of z .
a. . Find the 76th percentile of x ~ N (2,9) .
Make a diagram: Show a Normal curve with a mean at zero. Show z .24 , the 76th percentile, as a point
above zero with 24% above it and 26% between it and zero. The area below zero is 50%, so that the entire
area below it is 76%..
From the diagram, we want one point z .24 so that Pz  z.24   .7600 or P0  z  z.24   .2600 . From
the interior of the Standardized Normal table the closest we can come to .2600 is P0  z  0.71  .2611 .
We thus decide that z.24  0.71 , so the value of x can then be written x    z.24
 2  0.719  2  6.39  8.39 . norm
8.39  2 

To check this write Px  8.39   P  z 
 Pz  0.71  Pz  0  P0  z  71  .5  .2611
9 

 .7611  .76
251distrex4 1/14/08 (Open this document in 'Page Layout' view!)
251distrex4 12/2/04
b. Find the 24th percentile. of x ~ N (2,9) .
Make a diagram: Show a Normal curve with a mean at zero. Show  z .24 as a point below zero with 24%
below it and 26% between it and zero. The area above zero is 50%..
From the diagram, we want one point z.76   z.24 (It’s z .76 because 76% is above it.) so that
Pz  z.76   .2400 . We can also see that Pz.76  z  0  .2600 . Because z.76   z.24 ,
P0  z  z.24   .2600 . From the interior of the Standardized Normal table the closest we can come to
,2600 is P0  z  0.71  .2611 . We thus decide that z.24  0.71 or z .76   z .24  0.71 , so the value of
x can then be written x    z.76  2  0.719  2  6.39  4.39 . norm
 4.39  2 

To check this write Px  4.39   P  z 
  Pz  0.71
9


norm
 Pz  0  P0.71  z  0  .5  .2611  .2389  .24
c. A symmetrical interval about the mean with 76% probability for x ~ N (2,9) .
Make a diagram: Show a Normal curve with a mean at zero. Show z .12 as a point above zero with 12%
above it and 38% between it and zero. Show  z .12 as a point below zero with 12% below it and 38%
between it and zero The area between  z .12 and z .12 is thus twice 38% or 76%.
From the diagram, we want two points, z .12 and z .88 , so that P z.88  z  z.12  .7600 .

The upper point, z .12 will have P0  z  z .12   76 %

 .3800 , and by symmetry z.88   z.12 . (We
2
split 76% in halves around zero.) From the interior of the Normal table the closest we can come to .3800 is
P 0  z  117
.  .3790 or P 0  z  118
.  .3810 . So the point we want is roughly half way between 1.17
and 1.18. Either 1.17 or 1.18 would be acceptable in an assignment unless you can find a more exact value
in the last line of the t table.
norm ttable
We are best off using z.12  1175
, and making our interval for is -1.175 to 1.175.
.
The interval for x can then be written x    z.12  2  1175
or -8.575 to 12.575 or
.  9  2  10575
.
P8.575  x  12.575   .76 .
12 .575  2 
  8.575  2
z
To check this P 8.575  x  12 .575   P 
  P1.175  z  1.175 
9
9


2P0  z  1.175   2(.3800 )  .7600 .