Chapter 6 Normal Probability Distributions

1
Chapter 6Normal Probability Distributions

• Overview
• The Standard Normal Distribution
• Normal Distributions Finding Probabilities
• Normal Distributions Finding Values
• The Central Limit Theorem

2
Overview

• Continuous random variable
• Normal distribution

x - µ 2
( )
1
?
2
y
e
? 2 p
3
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.
4
Because the total area under the density curve is
equal to 1, there is a correspondence between
area and probability.
5
Heights of Adult Men and Women

• Women
• µ 63.6
• ? 2.5

Men µ 69.0 ? 2.8
63.6
69.0
Height (inches)
6
Definition

• Standard Normal Deviation
• a normal probability distribution that has a
• mean of 0 and a standard deviation of 1

7
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
13.5
13.5
x - 3s
x - 2s
x - s
x
x 2s
x 3s
x s
8
Probability of Half of a Distribution
0.5
0
9
Notation

• P(a lt z lt b)
• denotes the probability that the z score is
  between a and b
• P(z gt a)
• denotes the probability that the z score is
  greater than a
• P (z lt a)
• denotes the probability that the z score is
  less than a

10
Finding the Area to the Right of z 1.27
This area is 0.1020
z 1.27
0
11
To findz 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

12
Table E Standard Normal Distribution

• µ 0 ? 1

0 x
z
13
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
z 1.58
0
Score (z )
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.
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
0 degrees and 1.58 degrees.
Area 0.4429
P ( 0 lt x lt 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.

16
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 lt x lt 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.

17
Finding the Area Between z 1.20 and z 2.30
Area A is 0.1044
A
z 1.20
z 2.30
0
18
Other Normal Distributions

• If ? ? 0 or ?? ? 1 (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.

19
Converting to Standard Normal Distribution

P
?
x
(a)
20
Converting to Standard Normal Distribution

P
P
?
x
z
0
(a)
(b)
21
Probability of Weight between 143 pounds and 201
pounds
201 - 143
z
2.00
29
x 143
s ????29
Weight
143 201
z
0 2.00
22
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
Weight
143 201
z
0 2.00
23
Finding a probability when given a z-score using
the TI83

• 2nd
• Distributions
• 2normalcdf
• (lower bound, upper bound, mean, s.d.)

24
Finding a z - score when given a
probabilityUsing 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.

25
Finding z Scores when Given Probabilities
95
5
5 or 0.05
0.45
0.50
1.645
0
(z score will be positive)
Finding the 95th Percentile
26
Finding z Scores when Given Probabilities
90
10
Bottom 10
0.40
0.10
-1.28
0
(z score will be negative)
Finding the 10th Percentile
27
Finding z Scores when Given ProbabilitiesUsing
the TI83

• 2nd
• Distributions
• 3Invnorm
• (,mean,s.d)

28
Cautions to keep in mind

• 1. Dont 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.

29
Finding z Scores when Given Probabilities
95
5
5 or 0.05
0.45
0.50
1.645
0
(z score will be positive)
Finding the 95th Percentile
30
Finding z Scores when Given Probabilities
90
10
Bottom 10
0.40
0.10
-1.28
0
(z score will be negative)
Finding the 10th Percentile
31
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
  x value(s) being sought.
• 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.

32
Finding P10 for Weights of Women
90
10
40
50
Weight
143
x ?
33
Finding P10 for Weights of Women
The weight of 106 lb (rounded) separates the
lowest 10 from the highest 90.
0.10
0.40
0.50
Weight
143
x 106
0
-1.28
34
Forgot to make z score negative???
UNREASONABLE ANSWER!
0.10
0.40
0.50
Weight
143
x 180
0
1.28
35
REMEMBER!
Make the z score negative if the value is located
to the left (below) the mean. Otherwise, the z
score will be positive.
36
Definition
Sampling Distribution of the mean the
probability distribution of sample means, with
all samples      having the same sample size n.
37
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.

38
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 will approach
??????????????
n

39
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).

40
Notation

• the mean of the sample means
• the standard deviation of sample mean
• ???
• (often called standard error of the mean)

µx µ
?
?x
n
41
Distribution of 200 digits from Social Security
Numbers (Last 4 digits from 50 students)
20
Frequency
10

• 0

0 1 2 3 4 5 6 7
8 9
Distribution of 200 digits
42
SSN digits x
43
Distribution of 50 Sample Means for 50 Students
15
Frequency
10
5

• 0

0 1 2 3 4 5 6 7
8 9
44

• As the sample size increases, the sampling
  distribution of sample means approaches a normal
  distribution.

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, 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.

46
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
150
??? 29
0
0.24
47
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
150
?x? 4.83333
0
1.45
48
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 gt 150) 0.4052
• b.) if 36 different women are randomly selected,
  their mean    weight is greater than 150 lb.
  P(x gt 150) 0.0735It is much easier for an
  individual to deviate from the mean than it is
  for a group of 36 to deviate from the mean.

49
Finding a z - score for a group Using TI83

• 2nd
• Distributions
• 2normalcdf
• (lower bound, upper bound, mean, s.e)
• Replace s.d. with standard error