Chapter 5
Normal Probability Distributions
5-1
Continuous Random Variables
5-2
Standard Normal Distribution
5-3
Applications of Normal Distributions:
• Finding Probabilities
• Finding Values
5-4
Determining Normality
5-5
The Central Limit Theorem
5-6
Normal Distribution as Approximation to
Binomial Distribution
1
5-1
Continuous Random
Variables
Uniform (rectangular) distribution
Normal distribution
Standard Normal distribution
Note:
– Continuous random variables are typically represented by
curves
– Discrete random variables are typically represented by
histograms
2
Definitions
 Density Curve (or probability density
function) is the graph of a continuous
probability distribution
Properties
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
Two Important Points:
1. Because the total area under the
density curve is equal to 1, there is
a correspondence between area
and probability.
2. To find probability in a continuous
distribution just find the area under
the curve (Calculus)
4
Uniform Distribution
 Rectangular Distribution
• Simplest type of continuous probability
distribution
• random variable values are spread evenly over
the range of possibilities;
• graph results in a rectangular shape.
• Finding probability is the same as finding the
area of a rectangle
5
Using Area to
Find Probability of Class Length
Find P(51.5 < x < 52)
6
Definitions
 Uniform Distribution
• Formal definition of density function
y=
1
for a < x < b
b-a
From previous example: a = 50 and b = 52
1
1
y=
=
2
52 - 50
7
Examples:
Let X be a uniform random variable, find
the density function for each and graph
1. If X is between 0 and 5
Find a. p(x > 4) b. p(1 < x < 5)
2. If X is between 6 and 12
Find a. Find p(x > 6) b. p(x < 5) c. p(x > 12)
Go to excel
8
Normal Distribution
 Continuous random variable
 Normal distribution
Curve is bell shaped
and symmetric
µ
Score
Formula
y=
e
1
2

x-µ
2
(  )
2p
9
5-2
The Standard Normal
Distribution
10
Definition
Standard Normal Distribution:
a normal probability distribution that has a
mean of 0 and a standard deviation of 1, and the
total area under its density curve is equal to 1.
y=
e
1
2
Area found in
z - table or
calculator
z2
2p
0.4429
0
z = 1.58
Note: Typically ‘z’ denotes a standard normal random variable
11
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
12
Z - Table
 Uses calculus to find the area
under density function curve
associated with a particular zvalue
 Table available on website
 Calculator is easier
13
POSITIVE Z SCORES
14
NEGATIVE Z SCORES
15
To find:
z Score
the distance along horizontal scale of the
standard normal distribution; refer to the
leftmost column and top row of Table
Area
the region under the curve; refer to the
values in the body of Table
16
TI-83 Calculator
Finding Areas of Standard Normal Distribution
1. Press 2nd Distr
2. Choose normcdf
3. Enter normcdf(low z value, high z value)
4. Press enter
Note: if low z value is negative infinity use –E99
if high z value is positive infinity use E99
17
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, at the freezing point of
water, the reading is less than 1.58 degrees.
P (z < 1.58) = 0.9429
The probability that the chosen thermometer will measure
freezing water less than 1.58 degrees is 0.9429.
18
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, at the freezing point of
water, the reading is less than 1.58 degrees.
P (z < 1.58) = 0.9429
94.29% of the thermometers have readings less than
1.58 degrees.
19
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 (at the freezing
point of water) above –1.23 degrees.
P (z > –1.23) = 0.8907
The probability that the chosen thermometer with a reading
above –1.23 degrees is 0.8907.
20
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 (at the freezing
point of water) above –1.23 degrees.
P (z > –1.23) = 0.8907
89.07% of the thermometers have readings above –1.23
degrees.
21
Example:
A thermometer is randomly selected.
Find the probability that it reads (at the freezing point of
water) between –2.00 and 1.50 degrees.
P (z < –2.00) = 0.0228
P (z < 1.50) = 0.9332
P (–2.00 < z < 1.50) =
0.9332 – 0.0228 = 0.9104
The probability that the chosen thermometer has a
reading between – 2.00 and 1.50 degrees is 0.9104.
22
Example:
A thermometer is randomly selected.
Find the probability that it reads (at the freezing point of
water) between –2.00 and 1.50 degrees.
P (–2.00 < z < 1.50) = 0.9104
If many thermometers are selected and tested at the
freezing point of water, then 91.04% of them will read
between –2.00 and 1.50 degrees.
23
Let’s try several examples examples (sketch a
graph)
1)
P(Z < 1.45)
8)
P(-1.35 < Z < 1.45)
2)
P(Z < -1.45)
9)
P(.56 < Z < 1.45)
3)
P(Z > 1.45)
4)
P(Z > -1.45)
5)
P(0 < Z < 1.45)
6)
P(-1.45 < Z < 0)
7)
P(-1.45 < Z < 1.45)
Test Problems
24
Frequently used Z – Scores
1)
Z = 1.28
2)
Z = 1.645
3)
Z = 1.96
4)
Z = 2.33
5)
Z = 2.58
To illustrate why, find
1)
P(Z > 1.28)
2)
P(Z > 1.645)
3)
P(Z > 1.96)
4)
P(Z > 2.33)
5)
P(Z > 2.58)
25
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
26
Finding a z - score when given a
probability (Area)
1. Draw a bell-shaped curve, draw the centerline,
and identify the region under the curve that
corresponds to the given probability.
2.Use the invNorm calculator function to
determine the corresponding z score.
27
Finding z Scores when Given Probabilities
Finding the 95th Percentile
95%
5%
5% or 0.05
0.45
0.50
0
invNorm(.95)
1.645
(z score will be positive)
28
Finding z Scores when Given Probabilities
Finding the 10th Percentile
90%
10%
Bottom 10%
0.10
0.40
z
0
(z score will be negative)
29
Finding z Scores when Given Probabilities
Finding the 10th Percentile
90%
10%
Bottom 10%
0.10
0.40
-1.28
(z score will be negative)
0
invNorm(.10)
30
Examples:
Find the Z score that corresponds the following percentiles
1)
50th
2)
25th
3)
75th
Find the Z score(s) for the following areas from the mean
4)
0.4495
5)
0.3186
31
Find the Z score when given the probability
(sketch each)
1) P(0 < z < a) = .34
2) P(-b < z < b) = .8442
3) P(z > c) = .056
4) P(z > d) = .8844
5) P(z < e) = .3400
These are similar to HW #4 (Extra Credit)
32
5-3 Applications of Normal
Distributions
If   0 or  1 (or both), we will convert
values to standard scores using the z-score
formula we learned in Chapter 2, then procedures
for working with all normal distributions are the
same as those for the standard normal
distribution.
z=
x-µ

Note: z scores are real numbers that have no units
33
Converting to Standard
Normal Distribution
z=
x–

34
Probability of Sitting Heights
Less Than 38.8 Inches
 = 36
 = 1.4
38.8 – 36.0
z =
= 2.00
1.4
35
Probability of Sitting Heights
Less Than 38.8 Inches
 = 36
 = 1.4
P ( x < 38.8 in.) = P(z < 2)
normalcdf (-E99,2) = 0.9772
36
Probability of Weight between 150
pounds and 200 pounds
µ = 150
σ = 25
x
z=
150
200
200 - 150
25
= 2.00
Weight
z
0
2.00
37
Probability of Weight between 150
pounds and 200 pounds
normalcdf (0,2) = .4772
µ = 150
σ = 25
x
150
200
Weight
z
0
2.00
38
Probability of Weight between 150
pounds and 200 pounds
There is a 0.4772 probability
of randomly selecting
someone with a weight
between 150 and 200 lbs.
µ = 150
σ = 25
x
150
200
Weight
z
0
2.00
39
Probability of Weight between 150
pounds and 200 pounds
OR - 47.72% have weights
between 150 lb and 200 lb.
µ = 150
σ = 25
x
150
200
Weight
z
0
2.00
40
Normal Distribution:
Finding Values
From Known Areas
41
Cautions to keep in mind
1. Don’t confuse z scores and areas.
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.
42
Procedure for Finding Values
Using TI-83 Calculator
1.
2.
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
Find the z score

invNorm(area)

Make the z score negative if it is located to the left of the centerline.
3.
Use x = µ + (z • ) to find the value for x or invNorm(area,µ,σ)
4.
Refer to the sketch of the curve to verify that the solution makes
sense.
43
Finding P10 for Weights
x =  + (z ● )
µ = 150
σ = 25
x = 150 + (-1.28 • 25) = 118.75
0.10
0.40
0.50
x
z
x=?
150
-1.28
0
Weight
44
Finding P10 for Weights
The weight of 119 lb (rounded) separates
the lowest 10% from the highest 90%.
0.10
0.40
x
z
x = 119
-1.28
0.50
150
Weight
0
45
In Class Example
A tire company finds the lifespan for one brand of
its tires is normally distributed with a mean of
48,500 miles and a standard deviation of 5000
miles.
a) What is the probability that a randomly selected
tire will be replaced in 40,000 miles or less?
b) If the manufacturer is willing to replace no more
than 10% of the tires, what should be the
approximate number of miles for a warranty?
Similar to hw #9 – Test problem
46
REMEMBER!
Make the z score negative if the
value is located to the left (below)
the mean. Otherwise, the z score
will be positive.
47
Extra Credit
A teacher informs her statistics class that a test if
very difficult, but the grades will be curved. Scores
are normally distributed with a mean of 25 and
standard deviation of 5. The grades will be curved
according to the following scheme.
A’s: Top 10%
B’s: Between the top 10% and top 30%.
C’s: Between the bottom 30% and top 30%
D’s: Between the bottom 10% and bottom 30%
F’s: Bottom 10%
Find the numerical limits for each grade. Include
sketch.
48
5-4
Determining Normality
49
Determining Normality
Use Descriptive Methods
1. Histogram
2. Dot Plot
3. Stem-leaf
50
Determining Normality
Use Normal Probability Plots
• Plot the observed data versus expected z-score of the
data.
• Expected z-scores are based on the position of the
data in ascending order similar to percentiles.
• If the data are from a normal distribution then the
Normal Probability Plot will be approximately
linear. Note: X = µ + Zσ (linear relationship between X and Z)
• Due to the tedious nature of this process, use
technology.
51
TI-83 Calculator
Normal Probability Plot
1. Enter data in L1
2. Choose statplot (2nd Y=)
3. Press Enter
4. Plot on should be ON
5. Cursor to Type and choose the last plot type
6. Press Zoom
7. Choose 9
52
5-5
The Central Limit
Theorem
(sampling distribution of the mean)
53
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
54
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
55
Frequency
Distribution of 50 Sample Means
for 50 Students
15
10
5
0
0
1
2
3
4
5
6
7
8
9
56
Definition
Sampling Distribution of the mean
the probability distribution of
sample means, with all samples
having the same sample size n.
See Excel
57
Sampling Distributions
http://onlinestatbook.com/stat_sim/sampling_dist/index.html
58
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.
59
Central Limit Theorem
Conclusions:
1. The distribution of sample x will, as the
sample size increases, approach a normal
distribution regardless of the original
population 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
60
Notation
the mean of the sample means
µx = µ
the standard deviation of sample mean

x = n
(often called standard error of the mean)
61
Central Limit Theorem
Conditions for Application:
1. Original population is not normal or
unknown
• Sample must be sufficiently large (n > 30)
2. The original population is normal.
• Sample may be of any size
Test Question
62
Describing the Sampling
Distribution of the Mean
As the sample size increases,
the sampling distribution of
sample means approaches a
normal distribution with
µx = µ
Test Question

x = n
63
Example: The mean age of students at a community is
25 with a standard deviation of 8. A sample of 64 students
is drawn. Describe the sampling distribution of the mean.
Solution:
X~N(25,1)
Test Question
64
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.
65
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. (same approach
as section 5.3.
z = 150-143 = 0.24
29
P (X > 150) = P(z > .24)
Normalcdf (.24,E99)= 0.4052
0.5948
 = 143
= 29
0
x
150
z
0.24
66
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.
P (X > 150) = P(z > .24)
Normalcdf (.24,E99)= 0.4052
0.5948
 = 143
= 29
0
x
150
z
0.24
67
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, find the
probability that their mean weight is greater than 150 lb.
P (x > 150)
x = 143
150
x = 29 = 4.83333
36
x
68
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
P (x > 150) = P(z > 1.45)
Normalcdf (1.45,E99)= 0.0735
0.9265
x = 143
x = 4.83333
0
150
x
z
1.45
69
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 lbs.
P(x > 150) = 0.4052
b.) if 36 different women are randomly selected, find the
probability that their mean weight is greater than 150
lbs.
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.
70
5-6
Normal Distribution as
Approximation to
Binomial Distribution
71
Approximate a Binomial Distribution
with a Normal Distribution if:
np  5
nq  5
Basically if ‘n’ is large your ok.
Typically this approximation is only used
when ‘n’ is large anyway.
72
Approximate a Binomial Distribution
with a Normal Distribution if:
np  5
nq  5
then µ = np and  = npq
and the random variable has
a
distribution.
(normal)
73
Approximate a Binomial Distribution
with a Normal Distribution:
http://bcs.whfreeman.com/ips4e/cat_010/applets/CLT-Binomial.html
74
Definition:
Continuity Correction:
• Needed when we use the normal distribution
(continuous) as an approximation to the
binomial distribution (discrete)
• Made to a discrete whole number x in the
binomial distribution by representing the
single value x by the interval from
x - 0.5 to x + 0.5.
75
Procedure for Using a Normal Distribution to
Approximate a Binomial Distribution
1. Verify np  5 and nq  5.
2. Calculate µ = np and  = npq.
3. Identify the continuity correction x - 0.5 to x + 0.5.
4. Draw a normal curve and enter the values of µ
4. Change x by replacing it with x - 0.5 or x + 0.5, as
appropriate.
5. Find the area corresponding to the desired probability.
76
Finding the Probability of “At Least” 520 Men
Among 1000 Accepted Applicants
77
P(x  520)
520, 521, 522, ….
Use P(x  519.5)
.
520
519.5
P(x > 520)
521, 522, 523, . . .
Use P(x > 520.5)
P(x  520)
0, 1, . . . 518, 519, 520
Use P(x  520.5)
521
520.5
520
520.5
P(x < 520)
0, 1, . . . 518, 519
Use P(x < 519.5)
519
519.5
78
P(x=520)
520
519.5 520.5
Interval represents discrete number 520
79