Download br7ch06 - La Sierra University

Understandable Statistics
Seventh Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Chapter Six
Normal Distributions
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
1
The Normal Distribution
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
2
Properties of The Normal
Distribution

The curve is bell-shaped with the
highest point over the mean, .
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
3
Properties of The Normal
Distribution

The curve is symmetrical about a
vertical line through .
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
4
Properties of The Normal
Distribution

The curve approaches the horizontal
axis but never touches or crosses it.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
5
Properties of The Normal
Distribution
–


The transition points between cupping
upward and downward occur
above  +  and  –  .
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
6
The Normal Density Function
( x   )2
P( x) 
2 2
e
 2
This formula generates the density curve which
gives the shape of the normal distribution.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
7
The Empirical Rule
Approximately 68% of the data values lie is
within one standard deviation of the mean.
68%

One standard deviation from the mean.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
8
The Empirical Rule
Approximately 95% of the data values lie within
two standard deviations of the mean.
95%
x
Two standard deviations from the mean.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
9
The Empirical Rule
Almost all (approximately 99.7%) of the data
values will be within three standard deviations of
the mean.
99.7%
x
Three standard deviations from the mean.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
10
Application of the Empirical
Rule
The life of a particular type of light bulb
is normally distributed with a mean of
1100 hours and a standard deviation of
100 hours.
What is the probability that a light bulb of
this type will last between 1000 and 1200
hours?
Approximately 68%
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
11
Control Chart
a statistical tool to track data over a
period of equally spaced time
intervals or in some sequential
order
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
12
Statistical Control
A random variable is in statistical
control if it can be described by the
same probability distribution when
it is observed at successive points in
time.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
13
To Construct a Control Chart
• Draw a center horizontal line at .
• Draw dashed lines (control limits) at
    and   .
• The values of  and  may be target
values or may be computed from past
data when the process was in control.
• Plot the variable being measured using
time on the horizontal axis.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
14
Control Chart





1
2
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
3
4
5
6
7
15
Control Chart





1
2
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
3
4
5
6
7
16
Out-Of-Control Warning
Signals
I
One point beyond the 3 level
II
A run of nine consecutive points on
one side of the center line at target 
III
At least two of three consecutive
points beyond the 2 level on the same
side of the center line.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
17
Probability of a False Alarm
Warning Signal
Probability of false
alarm
I Point beyond 3
0.003
II Nine conscecutive
points on same side of

III At least 2/3 points
beyond 2
0.004
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
0.004
18
Is the Process in Control?





1
2
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
3
4
5
6
7
19
Is the Process in Control?





1 2 3 4 5 6 7 8 9 10 11 12 13
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
20
Is the Process in Control?





1
2
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
3
4
5
6
7
21
Is the Process in Control?





1
2
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
3
4
5
6
7
22
Z Score
• The z value or z score tells the number of
standard deviations the original
measurement is from the mean.
• The z value is in standard units.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
23
Formula for z score
x
z

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
24
Calculating z-scores
The amount of time it takes for a pizza
delivery is approximately normally
distributed with a mean of 25 minutes
and a standard deviation of 2 minutes.
Convert 21 minutes to a z score.
x   21  25
z

 2.00

2
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
25
Calculating z-scores
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Convert 29.7 minutes to a z score.
x   29.7  25
z

 2.35

2
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
26
Interpreting z-scores
Mean delivery time = 25 minutes
Standard deviation = 2 minutes
Interpret a z score of 1.6.
x  z    1.6( 2 )  25  28 .2
The delivery time is 28.2 minutes.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
27
Standard Normal Distribution:

=0

=1
-1
0
1
Values are converted to z

scores wherexz =

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
28
Importance of the Standard
Normal Distribution:
Standard
Normal
Distribution:
Any Normal
Distribution:
0
1
Areas will be equal.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
1
29
Use of the Normal Probability
Table
(Table 5) - Appendix II
Entries give the probability that a
standard normally distributed
random variable will assume a
value to the left of a given negative
z-score.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
30
Use of the Normal Probability
Table
(Table 5a) - Appendix II
Entries give the probability that a
standard normally distributed
random variable will assume a
value to the left of a given positive z
value.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
31
To find the area to the left of
z = 1.34
_____________________________________
z … 0.03
0.04
0.05 ..…
_____________________________________
.
.
1.2 … .8907
.8925
.8944 ….
1.3 … .9082
.9099
.9115 ….
1.4 … .9236
.9251
.9265 ….
.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
32
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area to the left of a given
negative z :
Use Table 5 (Appendix II) directly.
z
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
0
33
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area to the left of a given
positive z :
Use Table 5 a (Appendix II) directly.
0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
z
34
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area between z values on either
side of zero:
Subtract area to left of z1 from area to left
of z2 .
z1
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
0
z2
35
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area between z values on the
same side of zero:
Subtract area to left of z1 from area to left
of z2 .
0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
z1
z2
36
Patterns for Finding Areas
Under the Standard Normal
Curve
To find the area to the right of a positive z
value or to the right of a negative z value:
Subtract from 1.0000 the area to the left of the
given z.
Area under
entire curve
= 1.000.
0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
z
37
Use of the Normal Probability
Table
a.
.8925
P(z < 1.24) = ______
b.
.4452
P(0 < z < 1.60) = _______
c.
.4911
P( - 2.37 < z < 0) = ______
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
38
Normal Probability
d.
.9974
P( - 3 < z < 3 ) = ________
e.
.9322
P( - 2.34 < z < 1.57 ) = _____
f.
.0774
P( 1.24 < z < 1.88 ) = _______
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
39
Normal Probability
g.
.2254
P( - 2.44 < z < - 0.73 ) = _______
h.
.9495
P( z < 1.64 ) = __________
i.
.0084
P( z > 2.39 ) = _________
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
40
Normal Probability
j.
.9236
P ( z > - 1.43 ) = __________
k.
.0034
P( z < - 2.71 ) = __________
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
41
Application of the Normal
Curve
The amount of time it takes for a pizza delivery is
approximately normally distributed with a mean of 25
minutes and a standard deviation of 2 minutes. If you order
a pizza, find the probability that the delivery time will be:
a.
between 25 and 27 minutes.
.3413
a. ___________
b.
less than 30 minutes.
.9938
b. __________
c.
less than 22.7 minutes.
.1251
c. __________
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
42
Inverse Normal Distribution
Finding z scores when probabilities
(areas) are given
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
43
Find the indicated z score:
Find the indicated z score:
.8907
0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
z=
1.23
44
Find the indicated z score:
.6331
.3669
z = – 0.34
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
45
Find the indicated z score:
.3560
.8560
0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
z=
1.06
46
Find the indicated z score:
.4792
.0208
z = – 2.04
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
0
47
Find the indicated z score:
.4900
0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
z=
2.33
48
Find the indicated z score:
.005
z = – 2.575
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
0
49
Find the indicated z score:
A
= .005
–z
B
0
z
 2.575 or  2.58
If area A + area B = .01, z = __________
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
50
Application of Determining z
Scores
The Verbal SAT test has a mean score of
500 and a standard deviation of 100.
Scores are normally distributed. A major
university determines that it will accept
only students whose Verbal SAT scores
are in the top 4%. What is the minimum
score that a student must earn to be
accepted?
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
51
...students whose Verbal SAT
scores are in the top 4%.
Mean = 500, standard deviation = 100
.9600
= .04
z = 1.75
The cut-off score is 1.75 standard deviations
above the mean.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
52
Application of Determining z
Scores
Mean = 500, standard deviation = 100
.9600
= .04
z = 1.75
The cut-off score is 500 + 1.75(100) = 675.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
53
Normal Approximation Of The
Binomial Distribution:
Under certain conditions, a
binomial random variable has a
distribution that is approximately
normal.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
54
Using the normal distribution
to approximate the binomial
distribution
If n, p, and q are such that:
np and nq
are both greater than 5.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
55
Mean and Standard Deviation:
Binomial Distribution
  np and   npq
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
56
Experiment: tossing a coin 20
times
Problem: Find the probability of getting
exactly 10 heads.
Distribution of the number of heads appearing should look like:
0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
10
20
57
Using the Binomial
Probability Formula
n=
20
x=
10
P(10) = 0.176197052
p=
0.5
q = 1  p = 0.5
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
58
Normal Approximation of the
Binomial Distribution
First calculate the mean
and standard deviation:
 = np = 20 (.5) = 10
  np(1  p ) 
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
20(.5)(.5)  5  2.24
59
The Continuity Correction
Continuity Correction: to compute the
probability of getting exactly 10 heads, find
the probability of getting between 9.5
and 10.5 heads.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
60
The Continuity Correction
Continuity Correction is needed because
we are approximating a discrete
probability distribution with a
continuous distribution.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
61
The Continuity Correction
We are using the area under the
curve to approximate the area of the
rectangle.
9.5 - 10.5
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
62
Using the Normal Distribution
P(9.5 < x < 10.5 ) = ?
for x = 9.5: z =  0.22
P(z <  0.22 ) = .4129
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
63
Using the Normal Distribution
for x = 10.5: z = = 0.22
P( z < .22) = .5871
P(9.5 < x < 10.5 ) .5871
= - .4129 = .1742
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
64
Application of Normal
Distribution
If 22% of all patients with high blood pressure
have side effects from a certain medication,
and 100 patients are treated, find the
probability that at least 30 of them will
have side effects.
Using the Binomial Probability Formula we would need
to compute:
P(30) + P(31) + ... + P(100) or 1  P( x < 29)
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
65
Using the Normal
Approximation to the
Binomial Distribution
Is it appropriate to use the
normal distribution?
Check: n p =
nq=
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
66
Using the Normal
Approximation to the
Binomial Distribution
n p = 22
n q = 78
Both are greater than five.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
67
Find the mean and standard
deviation
 = 100(.22) = 22
and  = 100(.22)(.78) 
17.16  4.14
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
68
Applying the Normal
Distribution
To find the probability that at least 30 of
them will have side effects, find P( x  29.5)
Find this area
22
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
29.5
69
Applying the Normal
Distribution
The probability that
at least 30 of the
patients will have
side effects is 0.0351.
z = 29.5 – 22 = 1.81
4.14
Find P( z  1.81)
.9649
= .0351
0
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
1.81
70
Reminders:
• Use the normal distribution to
approximate the binomial only if both np
and nq are greater than 5.
• Always use the continuity correction
when approximating the binomial
distribution.
Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .
71