Lesson Objective

Learn the mechanics of using
the table for the Normal Distribution.

Given a region for a variable that
follows the Normal Distribution,
find the probability that a randomly
selected item will fall in this region.

Given a probability, find the region for a
normally distributed variable that
corresponds to this probability.
 Department of ISM, University of Alabama, 1995-2003
M11-Normal Distribution 1
1
The Normal Distributions
a.k.a., “The Bell Shaped Curve”
Describes the shape for some
quantitative, continuous random variables.
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
2
Normal Population Distribution
has two parameters:


= mean determines the location.
= standard deviation determines
spread, variation, scatter.
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
3
Notation:
X ~ N(  = 66,  = 9) or N(66, 9)
Z = the number of standard deviations
that an X - value is from the mean.
Z =


X-
Z ~ N(  = 0,  = 1 )
or N(0,1)
Z follows the “Standard Normal Distribution”
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
4
Empirical Rule of the Normal Distribution
______, ±3
______, ±2
____, ±1
-4
-3
-2
-1
0
1
 Department of ISM, University of Alabama, 1995-2003
2
3
4
M12-Normal Distribution 2
5
Recall
The “area” under the curve within a range of
X values is equal to proportion
of the population within that range of
X values.
Question: How do we compute “areas”?
• Geometry formulas
• Calculus (integration)
• Tables
• Excel
• Minitab
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
6
Reading the Standard Normal Table
(finding areas under the normal curve)
Step 1 for all problems:
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
7
Standard
Normal
Table
2nd decimal place
Table gives
P(0 < z < ?) =
Find
P(0 < z < 1.72) =
.4573
Up to the 1st decimal place
-4.0
 Department of ISM, University of Alabama, 1995-2003
-3.0
-2.0
-1.0
0
0.0
1.72
1.0
M12-Normal Distribution 2
2.0
8
3.0
What proportion of Z values
are between –1.23 and +2.05?
P(-1.23 < Z < 2.05) = ?
=
=
-4.0
-3.0
What proportion of Z values
are between +1.23 and +2.05?
P(1.23 < Z < 2.05) = ?
=
=
-4.0
-3.0
 Department of ISM, University of Alabama, 1995-2003
-1.23
-2.0
-1.0
0
0.0
1.0
2.05
2.0
3.0
Z
4.0
3.0
Z
4.0
?
-2.0
-1.0
0 1.23 2.05
0.0
1.0
2.0
M12 Normal Distribution 2
9
Weights of packages are normally
distributed with mean of 10 lbs.
and standard deviation of 4.0 lbs.
Find the proportion of packages
that weigh between 10 and 15.72
lbs.
X = weight of packages.
X ~ N( = 10,  = 4.0)
P( 10.0 < X < 15.72) = ?
P( 0 < Z < 1.43)
=
-4.0
-3.0
-2.0
-1.0
10 15.72
0.0
1.0
2.0
3.0
X
Z
4.0
15.72 – 10.0
Z=
=
4.0
10.0 – 10.0
Z=
=
4.0
 Department of ISM, University of Alabama, 1995-2003
M12 Normal Distribution 2
10
Same situation.
What proportion of packages
weigh more than 15.72 lbs?
X = weight of packages.
X ~ N( = 10,  = 4.0)
P( X > 15.72) = ?
P( Z > 1.43) =
=
?
-4.0
-3.0
-2.0
-1.0
10
0
0.0
15.72
1.43
1.0
2.0
3.0
X
Z
4.0
15.72 – 10.0
Z=
= 1.43
4.0
 Department of ISM, University of Alabama, 1995-2003
M12 Normal Distribution 2
11
.5000
Same situation.
What proportion of packages
weigh less than 14.2 lbs?
X = weight of packages.
X ~ N( = 10,  = 4.0)
P( X < 14.2) = ?
P( Z < 1.05) = .5 + .3531
= .8531
.3531
-4.0
-3.0
-2.0
-1.0
10 14.2
0 1.05
0.0
1.0
2.0
3.0
X
Z
4.0
14.2 – 10.0
Z=
= 1.05
4.0
 Department of ISM, University of Alabama, 1995-2003
M12 Normal Distribution 2
12
Same situation.
What proportion of packages
weigh between 5.08 and 18.2 lbs?
X = weight of packages.
X ~ N( = 10,  = 4.0)
-4.0
-3.0
 Department of ISM, University of Alabama, 1995-2003
-2.0
-1.0
0.0
1.0
2.0
3.0
M12-Normal Distribution 2
4.0
13
Same situation.
What proportion of packages
weigh either less than 2.4 lbs
or greater than 11.0 lbs?
Homework
X = weight of packages.
X ~ N( = 10,  = 4.0)
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
14
Same situation.
Find the weight such that
10% of all packages weigh
less than this weight.
.40
.10
This is a backwards problem!
We are given the probability;
we need to find the boundary.
X = weight of packages.
X ~ N( = 10,  = 4.0)
P( X < ?) = .10
P( Z < -1.28 ) = .10
-4.0
-3.0
?
-1.28
-2.0
-1.0
10
0
0.0
1.0
2.0
3.0
X
Z
4.0
? – 10
–1.28 =
4
? = 10 – 1.28 • 4
= 10 – 5.12
= 4.88 pounds
10% weigh less than 4.88 pounds;
90% weigh moreM12
than
4.88
pounds.
Normal
Distribution
2
 Department of ISM, University of Alabama, 1995-2003
15
Standard
Normal
Table
Table gives
P(0 < z < ?) =
Find
P( __ < z < 0) =.40
Find the Z value to cut off the top
.4000
10%.
-4.0
 Department of ISM, University of Alabama, 1995-2003
-3.0
-1.28
-2.0
-1.0
0
0.0
1.0
M12 Normal Distribution 2
2.0
16
3.0
Standard
Normal
Table
Table gives
P(0 < z < ?) =
.25 .25
Find the Z values that define the middle 50%.
-4.0
 Department of ISM, University of Alabama, 1995-2003
-3.0
-2.0
?
-1.0
0
0.0
1.0
M12 Normal Distribution 2
?
2.0
17
3.0
Standard
Normal
Table
Table gives
P(0 < z < ?) =
Find the Z values that define the middle 95%.
-4.0
 Department of ISM, University of Alabama, 1995-2003
-3.0
-2.0
-1.0
0.0
1.0
M12 Normal Distribution 2
2.0
18
3.0
Normal Functions in Excel
• NORMDIST –
Used to compute areas under any
normal curve. Can also compute
height of curve (not useful except
for drawing normal curves).
• NORMSDIST - Used to compute
areas under a standard normal
( N(0,1) or Z curve ).
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
19
Normal Functions in Excel
• NORMINV - Used to
find the X value corresponding to
a given cumulative probability for
any normal distribution.
• NORMSINV - Used to find the Z
value corresponding to a given
cumulative probability for a
standard normal distribution.
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
20
Practice problems.
You MUST know
1. P( Z < –1.92) =
how to work ALL
2. P( Z < 2.56) =
of these problems
and the following
3. P( Z > 0.80) =
practice problems
4. P( Z = 1.42) =
to pass this course.
5. P( .32 < Z < 2.48) =
6. P( -1.75 < Z < 1.75) =
7. P( Z < 4.25) =
8. P( Z > 4.25) =
9. P(-.05 < Z < .05) =
10. Find Z such that only 12% are smaller.
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
21
Practice problem answers
1. .0274
2. .9948
3. 1.0 – .7881 = .2119
4. .0
5. .9934 – .6255 = .0689
6. .9599 – .0401 = .9198
7. 1.0000
8. .0000
9. .0398
10. -1.175
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
22
Question:
What do we do when we have a normal
population distribution, but the mean is
not “0” and/or the standard deviation is
not “1”?
Use the Universal Translator
Z=
X–

Example: Suppose X ~ N(120, 10).
11. Find P ( X > 150 ).
12. Find the quartiles of this distribution.
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
23
What proportion of Z values
are between 0 and 1.43?
ori
.4236
P( 0 < Z < 1.43) = ?
= .4236
-4.0
-3.0
-2.0
What proportion of Z values
are between -1.43 and 0?
-1.0
0
0.0
1.43
1.0
2.0
3.0
Z
4.0
0
1.0
2.0
3.0
Z
4.0
.4236
P(-1.43 < Z < 0) = ?
= .4236
-4.0
-3.0
 Department of ISM, University of Alabama, 1995-2003
-1.43
-2.0
-1.0
0.0
M12-Normal Distribution 2
24
What proportion of Z values
are greater than 1.43?
P(Z > 1.43) = ?
= .5 - .4236
= .0764
-4.0
.5000
.4236
-3.0
-2.0
-1.0
0
0.0
1.43
1.0
.0764
2.0
3.0
Z
4.0
2.0
3.0
Z
4.0
.5000
.4236
What proportion of Z values
are less than 1.43?
P(Z < 1.43) = ?
= .5 + .4236
= .9236
-4.0
-3.0
 Department of ISM, University of Alabama, 1995-2003
-2.0
-1.0
0
0.0
1.43
1.0
M12-Normal Distribution 2
25
What proportion of Z values
are between –1.23 and 2.05?
P(-1.23 < Z < 2.05) = ?
= .4798 + .3907
= .8705
-4.0
.4798
.3907
-3.0
-1.23
-2.0
-1.0
0
0.0
-4.0
2.05
2.0
3.0
Z
4.0
3.0
Z
4.0
.4798
.4236
What proportion of Z values
are between 1.23 and 2.05?
P(1.23 < Z < 2.05) = ?
= .4798 - .4236
= .0562
1.0
.0562
-3.0
 Department of ISM, University of Alabama, 1995-2003
-2.0
-1.0
0 1.23 2.05
0.0
1.0
2.0
M12-Normal Distribution 2
26
Same situation.
Find the proportion of packages
that weigh between 4.28 and 10.0
lbs.
X = weight of packages.
X ~ N( = 10,  = 4.0)
P( 4.28 < X < 10.0) = ?
= P( -1.43 < Z < 0)
= .4236
-4.0
.4236
-3.0
4.28 10
-1.43 0
-2.0
-1.0
0.0
1.0
2.0
3.0
X
Z
4.0
4.28 – 10.0
Z=
= -1.43
4.0
10.0 – 10.0
Z=
=0
4.0
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
27
.4750
.2734
?
Same situation.
What proportion of the packages
weigh between 13.0 and 17.84 lbs?
X = weight of packages.
X ~ N( = 10,  = 4.0)
P( 13.0 < X < 17.84) = ?
= P( .75 < Z < 1.96)
= .4750 - .2734
= .2016
-4.0
-3.0
-2.0
-1.0
10 13 17.84
0 .75 1.96
0.0
1.0
2.0
3.0
X
Z
4.0
17.84 – 10.0
Z=
= 1.96
4.0
13.0 – 10.0
Z=
= .75
4.0
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
28
Same situation.
Find the weight such that
a. 16% weigh more
less than this value.
b. You have the boundaries
of the middle 80%.
c. The top 25% weigh more.
d. You have the quartiles.
-4.0
-3.0
-2.0
-1.0
0.0
1.0
2.0
3.0
X
Z
4.0
X = weight of packages.
X ~ N( = 10,  = 4.0)
 Department of ISM, University of Alabama, 1995-2003
M12-Normal Distribution 2
29