Download File

Chapter 6
Normal Probability Distributions
Overview
The Standard Normal Distribution
Normal Distributions: Finding
Probabilities
Normal Distributions: Finding Values
The Central Limit Theorem
1
Overview
 Continuous random variable
 Normal distribution
Curve is bell shaped
and symmetric
µ
Score
y=
e
1
2

x-µ
2
(  )
2p
2
Definitions
 Density Curve (or probability
density function) the graph of a
continuous probability distribution
1. The total area under the curve must
equal 1.
2. Every point on the curve must have a
vertical height that is 0 or greater.
3
Because the total area under
the density curve is equal to 1,
there is a correspondence
between area and probability.
4
Heights of Adult Men and Women
Women:
µ = 63.6
 = 2.5
Men:
µ = 69.0
 = 2.8
63.6
69.0
Height (inches)
5
Definition
Standard Normal Deviation
a normal probability distribution that has a
mean of 0 and a standard deviation of 1
6
The Empirical Rule
Standard Normal Distribution: µ = 0 and  = 1
99.7% of data are within 3 standard deviations of the mean
95% within
2 standard deviations
68% within
1 standard deviation
34%
34%
2.4%
2.4%
0.1%
0.1%
13.5%
x - 3s
x - 2s
13.5%
x-s
x
x+s
x + 2s
x + 3s
7
Probability of Half of a Distribution
0.5
0
8
Notation
P(a < z < b)
denotes the probability that the z score is
between a and b
P(z > a)
denotes the probability that the z score is
greater than a
P (z < a)
denotes the probability that the z score is
less than a
9
Finding the Area to the Right of z = 1.27
This area is
0.1020
0
z = 1.27
10
To find:
z Score
the distance along horizontal scale of the
standard normal distribution; refer to the
leftmost column and top row of theTable
Area
the region under the curve; refer to the
values in the body of the Table
11
Table E
Standard Normal Distribution
=1
µ=0
0
x
z
12
Definition
Standard Normal Deviation
a normal probability distribution that has a
mean of 0 and a standard deviation of 1
Area found in
Table A2
Area = 0.3413
0.4429
-3
-2
-1
0
1
2
3
0
z = 1.58
Score (z )
13
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
14
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing water
between 0 degrees and 1.58 degrees.
Area = 0.4429
P ( 0 < x < 1.58 ) = 0.4429
0
1.58
The probability that the chosen
thermometer will measure freezing water
between 0 and 1.58 degrees is 0.4429.
15
Example:
If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of 1 degree
for freezing water, and if one thermometer is randomly
selected, find the probability that it reads freezing water
between -2.43 degrees and 0 degrees.
Area = 0.4925
P ( -2.43 < x < 0 ) = 0.4925
-2.43
0
The probability that the chosen thermometer
will measure freezing water between -2.43
and 0 degrees is 0.4925.
16
Finding the Area Between z = 1.20 and z = 2.30
Area A is
0.1044
A
0
z = 1.20 z = 2.30
17
Other Normal Distributions
0
  1
If
or
(or both), we will convert
values to standard scores using Formula 5-2,
then procedures for working with all normal
distributions are the same as those for the
standard normal distribution.
z=
x-µ

18
Converting to Standard Normal
Distribution
P
(a)

x
19
Converting to Standard Normal
Distribution
x-
z=

P
P
(a)

x
(b)
0
z
20
Probability of Weight between 143
pounds and 201 pounds
z=
x = 143
s = 29
143
201
201 - 143
29
= 2.00
Weight
z
0
2.00
21
Probability of Weight between 143
pounds and 201 pounds
OR - 47.72% of women have
weights between 143 lb and
201 lb.
x = 143
s = 29
143
201
Weight
z
0
2.00
22
Finding a probability when given a
z-score using the TI83
•
•
•
•
2nd
Distributions
2:normalcdf
(lower bound, upper bound, mean, s.d.)
23
Finding a z - score when given a probability
Using the Table
1. Draw a bell-shaped curve, draw the centerline, and
identify the region under the curve that corresponds to
the given probability. If that region is not bounded by
the centerline, work with a known region that is
bounded by the centerline.
2. Using the probability representing the area bounded by
the centerline, locate the closest probability in the body
of Table E and identify the corresponding z score.
3. If the z score is positioned to the left of the centerline,
make it a negative.
24
Finding z Scores when Given Probabilities
95%
5%
5% or 0.05
0.45
0.50
0
1.645
(z score will be positive)
Finding the 95th Percentile
25
Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
0.10
0.40
-1.28
0
(z score will be negative)
Finding the 10th Percentile
26
Finding z Scores when Given Probabilities
Using the TI83
•
•
•
•
2nd
Distributions
3:Invnorm
(%,mean,s.d)
27
Cautions to keep in mind
1. Don’t confuse z scores and areas.
Z scores are distances along the horizontal scale,
but areas are regions under the normal curve.
Table A-2 lists z scores in the left column and
across the top row, but areas are found in the
body of the table.
2. Choose the correct (right/left) side of the graph.
3. A z score must be negative whenever it is
located to the left of the centerline of 0.
28
Finding z Scores when Given Probabilities
95%
5%
5% or 0.05
0.45
0.50
0
1.645
(z score will be positive)
Finding the 95th Percentile
29
Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
0.10
0.40
-1.28
0
(z score will be negative)
Finding the 10th Percentile
30
Procedure for Finding Values
Using the Table and By Formula
1.
Sketch a normal distribution curve, enter the given probability or
percentage in the appropriate region of the graph, and identify the
value(s) being sought.
x
2.
Use Table E to find the z score corresponding to the region bounded
by x and the centerline of 0. Cautions:

Refer to the BODY of Table E to find the closest area, then
identify the corresponding z score.

Make the z score negative if it is located to the left of the
centerline.
3.
Using the Formula, enter the values for µ, , and the z score found
in step 2, then solve for x.
x = µ + (z • )
4.
Refer to the sketch of the curve to verify that the solution makes
sense in the context of the graph and the context of the problem.
31
Finding P10 for Weights of Women
10%
90%
40%
x=?
143
50%
Weight
32
Finding P10 for Weights of Women
The weight of 106 lb (rounded) separates
the lowest 10% from the highest 90%.
0.10
0.40
x = 106
-1.28
0.50
143
Weight
0
33
Forgot to make z score negative???
UNREASONABLE ANSWER!
0.10
0.40
x = 180
1.28
0.50
143
Weight
0
34
REMEMBER!
Make the z score negative if the
value is located to the left (below)
the mean. Otherwise, the z score
will be positive.
35
Definition
Sampling Distribution of the mean
the probability distribution of
sample means, with all samples
having the same sample size n.
36
Central Limit Theorem
Given:
1. The random variable x has a distribution (which
may or may not be normal) with mean µ and
standard deviation .
2. Samples all of the same size n are randomly
selected from the population of x values.
37
Central Limit Theorem
Conclusions:
1. The distribution of sample x will, as the
sample size increases, approach a normal
distribution.
2. The mean of the sample means will be the
population mean µ.
3. The standard deviation of the sample means
n
will approach 
38
Practical Rules Commonly Used:
1. For samples of size n larger than 30, the distribution of
the sample means can be approximated reasonably well
by a normal distribution. The approximation gets better
as the sample size n becomes larger.
2. If the original population is itself normally distributed,
then the sample means will be normally distributed for
any sample size n (not just the values of n larger than 30).
39
Notation
the mean of the sample means
µx = µ
the standard deviation of sample mean

x = n
(often called standard error of the mean)
40
Distribution of 200 digits from
Social Security Numbers
Frequency
(Last 4 digits from 50 students)
20
10
0
0
1
2
3
4
5
6
7
8
9
Distribution of 200 digits
41
x
SSN digits
1
5
9
5
9
4
7
9
5
7
2
6
2
2
5
0
2
7
8
5
8
3
8
1
3
2
7
1
3
3
7
7
3
4
4
4
5
1
3
6
6
3
8
2
3
6
1
5
3
4
6
7
3
7
3
3
8
3
7
6
4
6
8
5
5
2
6
4
9
4.75
4.25
8.25
3.25
5.00
3.50
5.25
4.75
5.00
2
6
1
9
5
7
8
6
4
0
7
4.00
5.25
4.25
4.50
4.75
3.75
5.25
3.75
4.50
6.00
42
Frequency
Distribution of 50 Sample Means
for 50 Students
15
10
5
0
0
1
2
3
4
5
6
7
8
9
43
As the sample size increases,
the sampling distribution of
sample means approaches a
normal distribution.
44
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the probability
that her weight is greater than 150 lb.
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150 lb.
45
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, the probability that
her weight is greater than 150 lb. is 0.4052.
0.5 - 0.0948 = 0.4052
0.0948
 = 143
= 29
0
150
0.24
46
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
b.) if 36 different women are randomly selected, the
probability that their mean weight is greater than 150 lb is
0.0735.
z = 150-143 = 1.45
29
36
0.5 - 0.4265 = 0.0735
0.4265
x = 143
x = 4.83333
0
150
1.45
47
Example: Given the population of women has normally
distributed weights with a mean of 143 lb and a standard
deviation of 29 lb,
a.) if one woman is randomly selected, find the
probability that her weight is greater than 150 lb.
P(x > 150) = 0.4052
b.) if 36 different women are randomly selected, their
mean weight is greater than 150 lb.
P(x > 150) = 0.0735
It is much easier for an individual to deviate from the
mean than it is for a group of 36 to deviate from the mean.
48
Finding a z - score for a group
Using TI83
•
•
•
•
2nd
Distributions
2:normalcdf
(lower bound, upper bound, mean, s.e)
– Replace s.d. with standard error
49