251distrex4 12/2/04 (Open this document in 'Page Layout' view!)
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 the 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 , a standard deviation of 100 we write x
and if x is Normally distributed with a mean of 50 and
~ N

50 , 100

.
4) Any probability for a Normally distributed variable can be found using the Standardized
Normal Distribution by using the transformation z
 x



.
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 76 th 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 76 th percentile of x ~ N 2 9 .
Make a diagram: Show a Normal curve with a mean at zero. Show z
.
24
, the 76 th 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 the interior of the Standardized Normal table the closest we can come to .2600 is P

0
 z



.
2600
0 .
71


. From
.
2611 .
We thus decide that

2

0 .
71
 

2
 z
.
24
6 .
39


0 .
71
8 .
39 .
, so the value of x can then be written x
   z
.
24

To check this write P
 x

8 .
39


P

 z

8 .
39
9

2



.
7611

.
76

P
 z

0 .
71


P
 z

0
 
0
 z

71


.
5

.
2611

251distrex4 12/2/04 b. Find the 24 th 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
P
P

 z
0
 z
.
76


 z
 z
.
24
,2600 is P

0
.
2400


 z
 z
.
76
. We can also see that
.
2600
0 .
71 x can then be written

P


 z z
.
76
.
24
(It’s
 z

0
 z
.
76
because 76% is above it.) so that

.
2600 . Because z
.
76
  z
.
24
,
. From the interior of the Standardized Normal table the closest we can come to

.
2611 x
  
. We thus decide that z
.
76
 
2

0 .
71
 

2 z
.
24


6 .
39
0 .
71

or

4 .
39 z
.
76
.
  z
.
24
 
0 .
71 , so the value of
To check this write

P
 z

0
 

0 .
71
P
 x
 z


0


4 .
39

.
5



.
2611
P

 z


.
2389

4 .
39
9

2



.
24

P
 z
 
0 .
71
 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 above it and 38% between it and zero. Show z
.
12
as a point above zero with 12%
 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,
The upper point, z
.12
will have P

0 z
.12
and
 z
 z
.
12

 z
 .88
76
, so that
%
2


.
3800
.
88 z z
.
12


.
7600 .
, and by symmetry z
.
88
  z
.
12
. (We split 76% in halves around zero.) From the interior of the Normal table the closest we can come to .3800 is
P

0
 
.


.
3790 or P

0
 
.


.
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.
We are best off using z
.
12

The interval for x can then be written
P


8 .
575
 x

12 .
575


.
76 .
, and making our interval for is -1.175 to 1.175. x
   z
.
12
  
.
 
 
.
or -8.575 to 12.575 or
To check this P


8 .
575
 x

12 .
575


P



8 .
575
9

2
 z

12 .
575
9

2



P


1 .
175
 z

1 .
175

2 P

0
 z

1 .
175


2 (.
3800 )

.
7600 .