Chapter 5
Normal Probability Distribution
Lesson 5-1/5-2
The Standard Normal Distribution
Review

Random Variable


Probability Distribution


Describes the probability for each value of the random
variable.
Discrete Random Variable


A variable having a single numerical value, determine by
chance, for each outcome of some procedure.
Has either a finite number of values or a countable number
values (0, 1, 2, ..)
Continuous Random Variable

Has infinitely many values, and those values are often
associated with measurements on a continuous scale with
no gaps or interruptions.
Overview

Continuous Random Variable has distribution
with a graph that is symmetric and bell-shape
and it can by describe the given equation, we
say that it has a normal distribution.
f (x) 
e
1 x   


2  
 2
2
Density Curve

Density Curve (or probability density
function) is 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.
Density Curve
How do we use the density curve to find probabilities?


The area under the graph of a density
function over some interval represents the
probability of observing a value of random
variable in the interval.
The total area under the curve is equal to 1,
there is a correspondence between the area
and probability.
Example – Density Curve
Percent Histogram
IQ
(Intervals of size 20)
40
Percent
30
20
10
0
55
75
95
IQ
115
135
Example – Density Curve
(Area of rectangle = probability)
IQ
(Intervals of size 20)
Density
0.02
0.01
0.00
55
75
95
IQ
115
135
Uniform Distribution

Uniform Distribution is a probability
distribution in which the continuous
random variable values are spread
evenly over the range of possibilities;
the graph results in a rectangle shape.
Example – Uniform Distribution
Using Area to Find Probability
Suppose that your friend is always late. Let x represent the time from
when you are suppose to meet your friend until he shows up. Further
suppose that your friend could be on time or up to 30 minutes late.
What is the probability that your friend will be between 10 to 20 minutes
late the next time you meet him?
P(x)
1
30
w  30 ht 
1
30
A  (20  10) 
x
10
20
30
Time Late (minutes)
A  30 
1 1

30 3
P (10  x  20) 
1
3
1
1
30
Example – Page 237, #2
Refer to the continuous uniform distribution depicted in Figure
5-2, assume that a class length between 50.0 min and 52 min
is randomly selected, and find the probability that the given
time is selected.
Greater than 51.0 min
1
ht

w 2
2
50
51
52
P ( x  51)  w  ht
1
 (52  51) 
2
 0.50
The most important continuous
probability distribution is the normal
distribution
Bell-shaped curve
0.08
Mean = 70 SD = 5
0.07
0.06
0.05
0.04
Mean = 70 SD = 10
0.03
0.02
0.01
0.00
40
50
60
70
Grades
80
90
100
Random variables with a
normal distribution have
a bell-shaped probability
distribution (histogram).
Characteristics of
normal distribution





Symmetric, bell-shaped curve.
Shape of curve depends on population
mean  and standard deviation .
Center of distribution is .
Spread is determined by .
Most values fall around the mean, but
some values are smaller and some are
larger.
Example – Density Curve
Decrease interval size...
IQ
(Intervals of size 10)
Density
0.02
0.01
0.00
55
65
75
85
95
IQ
105
115
125
135
Example – Density Curve
Decrease interval size more….
IQ
(Intervals of size 5)
0.03
Density
0.02
0.01
0.00
50
60
70
80
90
100
IQ
110 120
130 140
Examples of normal random variables



testosterone level of male students
head circumference of adult females
length of middle finger of Honors Stats
students
Probability above 75?
Probability student scores higher than 75?
0.08
0.07
Density
0.06
0.05
P(X > 75)
0.04
0.03
0.02
0.01
0.00
55
60
65
70
Grades
75
80
85
Probability = Area under curve





Calculus?! You’re kidding, right?
But somebody did all the hard work for us!
We just need a table of probabilities for every
possible normal distribution.
But there are an infinite number of normal
distributions (one for each  and )!!
Solution is to “standardize.”
Standardizing



Z score is called the standard normal.
 Mean (μ) is 0
 Standard Deviation (σ ) is 1
Find the z – score
x μ
z
σ
Find the probability
 Use the Z tables
 Use the TI (normalcdf or normalpdf)
Standard Normal Distribution



Mean of 0
Standard deviation of 1
Total area under its density curve is 1
Example – Page 238, #10
Using the Standard Distribution. Assume readings on the thermometers
are normally distributed with a mean of 0° and a standard deviation of
1.00° C. Draw a sketch and find the probabilities of each reading in
degrees.
Less than -2.75
Use Table A-2
P ( z  2.75) 
-2.75
0
z
Example – Page 238, #10
Use Table A-2
P ( z  2.75)  0.0030
Use TI
-2.75
0
z
Example – Page 238, #10
2nd Vars
μ = 0, σ = 1
P ( z  2.75)  0.0030
normalcdf ( E 99, 2.75,0,1)
-2.75
0
z
Example – Page 238, #14
Greater than 1.96
P ( z  1.96) 
Use Table A-2
0
1.96
z
Example – Page 238, #14
Use Table A-2
1 – 0.9750 = 0.0250
P ( z  1.96)  1  0.9750
 0.0250
Use TI
μ = 0, σ = 1
normalcdf (1.96, E 99,0,1)
 0.02499
Area =0.9750
0
1.96
z
Example – Page 238, #22
Between -1.18 and 2.15
P ( 1.18  z  2.15)  0.8652
Use Table A-2
P ( z  2.15)  0.9842
P ( z  1.18)  0.1190
-1.18
0
2.15
0.9842 – 0.1190=0.8652
z
Example – Page 238, #22
Between -1.18 and 2.15
P ( 1.18  z  2.15) 
Use TI
μ = 0, σ = 1
normalcdf ( 1.18,2.15,0,1)
 0.86522
-1.18
0
2.15
z
The Empirical Rule
-3
-2
-1
0
1
2
3
Example – Page 238, #30
Find the indicated area under the curve of the standard
normal distribution, then convert it to a percentage and
fill in the blank.
68.27 % of the area between z = -1 and z = 1
About ______
(or within 1 standard deviation of the mean).
P (1  z  1)  normcdf (1,1,0,1)  68.27%
Notation

P(a < z < b)


P(z > a)


Denotes the probability that the z score is
between a and b.
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.
Finding z scores when given a
Probability


Given a probability (area) under the
curve
Percentile


P75 is the called the 75th percentile.
Find the z score using


The body of Table A-2
invNorm(area, μ, σ)
Example – Page 239, #38
Assume that the readings on the thermometers are normally
distributed with mean 0° and standard deviation 1.00° C
Find the P20, the 20th percentile.
0.84   0.2000
P  z  ______
.2000
z=?
Example – Page 239, #38
0.84   0.2000
P  z  ______
.2000
μ = 0, σ = 1
invNorm (0.2000,0,1)  0.84
z=?
Example – Page 239, #40
Assume that the readings on the thermometers are normally
distributed with mean 0° and standard deviation 1.00° C
If 3% of the thermometers are
rejected because they have
readings that are high and
another 3% are rejected
because they have readings that
are too low, find the two
readings that are cutoff values
separating the rejected
thermometers from others.
.03
z=?
.03
z=?
1 – 0.03 = 0.97
Example – Page 239, #40
Use TI
μ = 0, σ = 1
1.88  0.03
P (z  _______)
.03
.03
invNorm (0.03,0,1)  1.88
z=?
1.88
P (z  _______)
 0.03
invNorm (1  0.03,0,1)  1.88
z=?
1 – 0.03 = 0.97
Lesson 5-3
Applications of Normal Distribution
Nonstandard Normal
Distribution

If μ ≠ 0 or σ ≠ 1 (or both), we will convert
values to standard scores using the formula:
z 

x μ
σ
Then the procedures for working with all
normal distributions are the same as those
for the standard normal distribution.
Converting to Standard
Normal Distribution
z=
x–

Example – Page 246, #2
Assume adults have IQ scores that are normally distributed with a mean
of 100 and a standard deviation of 15.
Find the probability that a randomly selected adult has an IQ greater than
131.5 (the requirement for membership in the Mensa organization).
z 
x μ
σ

131.5  100
 2.10
15
0.0179
P (x  131.5)  0.0179
 normalcdf (131.5, E 99,100,15)
100
131.5
x
0
2.10
z
Example – Page 246, #4
Assume adults have IQ scores that are normally distributed with a mean
of 100 and a standard deviation of 15.
Find the probability that a randomly selected adult has an IQ between 110
and 120 (referred to as bright normal)
0.1613
120
x
1.33
110
0.67
0
P (110  x  120)  0.1613
 normalcdf (110,120,100,15)
100
x = 110 – 120 , μ = 100, σ = 15
z
Example – Page 246, #6
Assume adults have IQ scores that are normally distributed with a mean
of 100 and a standard deviation of 15.
Find P80, which is the IQ separating the bottom 80% from the top 20%.
area = .80, μ = 100, σ = 15
112.6  0.8000
P (x  ______)
 invNorm (0.8,100,15)
0.80
100
?
x
0.84
z
Example – Page 246, #8
Assume adults have IQ scores that are normally distributed with a mean
of 100 and a standard deviation of 15.
Find the IQ separating the top 55% from the others.
0.55
area = 1 – 0.55, μ = 100, σ = 15
98.1  0.55
P (x  ______)
 invNorm (1  0.55,100,15)
?
– 0.13
100
x
0
z
Example – Page 248, #14
Replacement times for CD players are normally distributed with a mean of
7.1 years and standard deviation of 1.4 years.
A). Find the probability that a randomly selected CD player will have
replacement time less than 8.0 years.
x = 8, μ = 100, σ = 15
0.7398
7.1 8
P (x  8)  0.7398
normalcdf (E 99,8,7.1,1.4)
0
0.64
x
z
Example – Page 248, #14
Replacement times for CD players are normally distributed with a mean of
7.1 years and standard deviation of 1.4 years.
B). If you want to provide a warranty so that only 2% of the CD players
will be replaced before the warranty expires, what is the time length
of the warranty?
area = 0.02, μ = 100, σ = 15 0.02
P (x  ______)
4.2  0.02
InvNorm (0.02,7.1,1.4)
?
– 2.07
7.1
x
0
z
Lesson 5-4
Sampling Distributions and
Estimators
Concept of a
Sampling Distributions
1.
2.
3.
4.
A research question is posed.
Information (data) is collected so that
the question posed can be answered.
The information is organized and
summarized.
Draw conclusions from the data.
Sampling Distribution of the Mean
 Sampling Distribution of the mean is the probability
distribution of sample means, with all samples having the
same size n.
 Obtaining the sampling distribution of the means is as
follows:
1. Obtain a simple sample size n.
2. Compute the sample mean
3. Assuming we are sampling from a finite population
repeat 1 and 2 until all simple random samples of size
n have been obtained.
Sampling Variability

The value of a statistic, such as the
sample mean x , depends on the
particular values included in the sample,
and generally varies from sample to
sample. This variability of a statistics is
called sampling variability.
The Sampling Distribution of the
Proportion

The Sampling Distribution of the Proportion
is the probability distribution of sample
proportions, with all samples having the
same sample size n.
Example
A population consists of values 1, 2, and 5. We randomly
select sample size 2 with replacement. There are 9 possible
samples.
A). Find the mean and standard deviation of the population.
μ  2.7
σ  1.7
Example
A population consists of values 1, 2, and 5. We randomly
select sample size 2 with replacement. There are 9 possible
samples.
B). For each sample, find the mean, median, range,
variance and standard deviation.
C). For each statistic, find the mean from part (b)
Example
Example
D) Find the sampling distribution of the mean
Sample Mean
1.0
1.5
2.0
3.0
3.5
5.0
Frequency
1
2
1
2
2
1
Probability
1/9
2/9
1/9
2/9
2/9
1/9
Interpreting of
Sampling Distribution


Statistics that target population parameters are
(unbiased estimators):
 Mean
 Variance
 Proportion
Statistics that do not target population parameters
are:
 Median
 Range
 Standard Deviation
Example – Page 256, #2
Data set 14 in Appendix B includes a sample measure reading
levels for 12 pages randomly selected from J. K. Rowling’s
Harry Potter and the Sorcerer’s Stone. The mean of the 12
Flesch-Kincaid grade level values is 5.08. The value of 5.08
is one value that is part of a sampling distribution. Describe the
sampling distribution.
Is the probability distribution for all such sample means. It
consists of all values of sample means, along with their
corresponding probabilities.
Example – Page 256, #4
The Gallup organization conducted a poll of 1015 randomly
selected students in grades K through 12 and found that
10% attended private or parochial school.
A). Is 10% (or 0.10) result a statistic or a parameter?
Statistics because it is calculated from a sample, not
a population.
Example – Page 256, #4
The Gallup organization conducted a poll of 1015 randomly
selected students in grades K through 12 and found that
10% attended private or parochial school.
B). What is the sampling distribution suggested by the
given data?
It is the probability distribution for all sample proportions
found for all possible sample of size 1015. It consists
of all sample proportions along with their corresponding
probabilities.
Example – Page 256, #4
The Gallup organization conducted a poll of 1015 randomly
selected students in grades K through 12 and found that
10% attended private or parochial school.
C). Would you feel more confident in the results if the
sample size were 2000 instead of 1015?
Yes, because as the sample size increases, the sample
statistics tend to vary less and they tend to be closer
to the population parameter (measured characteristics
of the population).
Example – Page 257, #6
Here are the numbers of sales per day that were made
by Kim Ryan, a courteous telemarketer who worked four
days before being fired: 1, 11, 9, 3. Assume sample size
2 are randomly selected with replacement from this
population of four values.
A). List the 16 different possible samples and find the mean
of each of them.
Population mean μ = 6
Example – Page 257, #6
Sample
1, 1
x
1
Sample
1, 11
6
1, 9
11, 1
5
6
1, 3
11, 11
2
11
11, 9
9, 1
9, 9
3, 1
3, 9
10
5
9
2
6
11, 3
9, 11
9, 3
3, 11
3, 3
7
10
6
7
3
x
Example – Page 257, #6
B). Identify the probability of each sample, and then describe
the sampling distribution of the sample means.
Example – Page 257, #6
Sampling Distribution
x
P (x )
Frequency
1
1
1/16
2
2
2/16
3
1
1/16
5
2
2/16
6
4
4/16
7
2
2/16
9
1
1/16
10
2
2/16
11
1
1/16
16 /16
x  P (x )
1/16
4/16
3/16
10/16
24/16
14/16
9/16
20/16
11/16
 96 /16
Example – Page 257, #6
C). Find the mean of the sampling distribution
96
μx 
6
16
96
μx   x  P ( x ) 
6
16
D) Is the mean of the sampling distribution equal to the
the mean of the population?
Yes, the sample mean is unbiased estimator of the
population.
Example – Page 258, #10
After constructing a new manufacturing machine, 5
prototype car headlights are produced and it is found
that 2 are defective (D) and 3 are acceptable (A).
Assume that two headlights are randomly selected with
replacement from the population.
A). After identifying the 25 different possible samples,
find the proportion of defects in each of them.
Example – Page 258, #10
Population (proportion of defects)
p
2
 0.40
5
binomial distribution
x = number of defects per sample
9
P (x  0)  binompdf (2,0.40,0)  0.36 
25
12
P (x  1)  binompdf (2,0.40,1)  0.48 
25
4
P (x  2)  binompdf (2,0.40,2)  0.16 
25
Example – Page 258, #10
B). Describe the sampling distribution of proportions of
defects.
9
P (x  0) 
25
12
P (x  1) 
25
4
P (x  2) 
25
9 with 0 defects
0
pˆ   0
2
12 with 1 defects
1
pˆ   .50
2
4 with 2 defects
pˆ 
2
1
2
Example – Page 258, #10
Sampling Distributions of Proportions of defects
p̂
P  p̂ 
pˆ  P (pˆ)
0
9/25
0/25
0.50
12/25
6/25
1
4/25
4/25
 25 / 25  1
10 / 25  0.40
Example – Page 258, #10
C). Find the mean of the sampling distribution
10
μpˆ   pˆ  P (pˆ) 
 .40
25
D). Is the mean of the sampling distribution equal to
population proportion of defects?
Yes, the sample proportion is an unbiased estimator
of the population.
Lesson 5-5
The Central Limit Theorem
Central Limit
Given
1). Random variable x has a distribution
(which may or may not be normal) with
mean μ and standard deviation σ.
2). Samples all of the same size n are
randomly selected from the population of
x values
Central Limit Theorem
Conclusion
1). The distribution of sample means x will, as
the sample size increases, approach a
normal distribution.
2). The mean of the sample means will be the
the population mean μ .
3). The standard deviation of the sample means
will approach σ /√n .
Practical Rules Commonly Used


For the samples of size n larger than 30, the
distribution of the samples means can be
approximated reasonable well by a normal
distribution.
 The approximation gets better as the sample
size n becomes larger
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.
Notations

The mean of the sample means
x  

The standard deviation of sample means
x 


n
Often called the standard error of the mean
Applying the
Central Limit Theorem


Probability of an individual from a normally
distributed population:
 Use normalcdf (LB ,UB ,μ,σ)
Probability of a group from a normally
distributed population:

Use normalcdf (LB ,UB ,μx ,σ x )
Example – Page 267, #2
Assume the men’s weights are normally distributed with
mean given by μ = 172 lb and a standard deviation given
by σ = 29 lb
A). If 1 man is randomly selected, find the probability
that his weight is greater than 180 lb.
x = men’s weight, μ = 172, σ = 29
P (x  180)  normalcdf (180, E 99,172,29)  0.3913
Example – Page 267, #2
Assume the men’s weights are normally distributed with
mean given by μ = 172 lb and a standard deviation given
by σ = 29 lb
B). If 100 men are randomly selected, find the probability
that they have mean weight greater than 180 lb.
P (x  180)  normalcdf (180, E 99,172,2.9)  0.0029
μx  μ  172
σ
29
σx 

 2.9
n
100
Example – Page 268, #8
Engineers must consider the breadths of male heads when
designing motorcycle helmets. Men have breadths that are
normally distributed with mean of 6.0 in and a standard
deviation of 1.0 in.
A). If one male is randomly selected, find the probability
that his head breadth is less than 6.2 in.
x = men’s breadths, μ = 6, σ = 1
Example – Page 268, #8
A). If one male is randomly selected, find the probability
that his head breadth is less than 6.2 in.
μ6
P (x  6.2)  0.5793
σ 1
normalcdf (E 99,6.2,6,1) 
58% probability that an
individual man’s head breadth
is less than 6.2 in
0.5793
6
6.2
x
0
0.2
z
Example – Page 268, #8
B). The safeguard Helmet company plans an initial production
run of 100 helmets. Find the probability that 100 randomly
selected men have a mean head breadth less 6.2.
P (x  6.2)  0.9772
σ
1
σX 

 0.10
n
100
μ6
σ 1
normalcdf (E 99,6.2,6,.1) 
98% probability that an 100 men’s head breadth is less
than 6.2 in
Example – Page 268, #8
C). The production managers sees the result from part (b)
and reasons that all helmets should be made for men
with head breadths less than 6.2 in., because they would
fit all but a few men. What is wrong with this reasoning?
The mean head breadth of 100 men is very likely to be
less than 6.2 in., there could be many individual men
that could not use the helmets because they have head
breadth greater than 6.2 in. Based on the results from
part (a), these helmets would not fit about
(1 – 0.5793 = 0.4207) 42% of men.
Example – Page 268, #12
Membership in Mensa requires an IQ score above 131.5.
Nine candidates take IQ tests, and their summary results
indicated that their mean IQ score is 133. (IQ scores are
normally distributed with a mean of 100 and a standard
deviation of 15.)
Example – Page 268, #12
A). If 1 person is randomly selected, find the probability
of getting someone with an IQ of least 133.
μ  100
x = IQ, μ = 100, σ = 15
P (x  133)  0.01390
normalcdf (133, E 99,100,15)
σ  15
Example – Page 268, #12
B). If 9 people are randomly selected, find the probability
that their mean IQ score is at least 133.
μ  100
11
P (x  133)  2.06  10  0.0001
σ  15
σ
15
σX 

5
n
9
normalcdf (133, E 99,100,5)
Example – Page 268, #12
C). Although the summary results are available, the
individual IQ test scores have been lost. Can it be
concluded that all 9 candidates have IQ scores above
133 so that they are eligible for Mensa membership?
No, if the mean is 133, all we know is that the scores
clustered around 133. Some scores could be considerably
higher and others considerable lower.
Example – Page 268, #16
Coaching for the SAT Test: Scores for men on the verbal portion
Of the SAT I test are normally distributed with a mean of 509
and a standard deviation of 112 (based on the data from the
College Board). Randomly selected men are given the
Columbia Review Course before taking the SAT test. Assume
that the course has no effect.
A) If 1 of the men is randomly selected, find the probability
that his score is at least 590.
x = men’s SAT Score, μ = 509, σ = 112
Example – Page 268, #16
A) If 1 of the men is randomly selected, find the probability
that his score is at least 590.
  509
P(x  590)  0.2348
 normalcdf (590, E 99,509,112)
  112
Example – Page 268, #16
B) If 16 of the men are randomly selected, find the probability
that their mean score is at least 590.
  509
P(x  590)  0.0019
σ
112
σX 

 28
n
16
 normalcdf (590, E 99,509,28)
  112
Example – Page 268, #16
C) In finding the probability for part (b), why can the central
limit theorem be used even though the sample size does
not exceed 30?.
When the original distribution is normal, the sampling
distribution of the x is normal regardless of the sample
size
Example – Page 268, #16
D) If the random sample of 16 men does result in a mean
score of 590, is there strong evidence to support the
claim that the course is actually effective? Why or why
not?
Yes, since 0.0019 < 0.05 there is strong evidence to
support the claim that the course is effective – assuming
as stated, that the 16 participants were truly a random
sample (and not self-selected or otherwise chosen in
some biased manner).
Lesson 5-6
Normal as Approximation to
Binomial
Review
Binomial Probability Distribution
1). The procedure must have fixed number of trials.
2). The trials must be independent.
3). Each trial must have two outcomes.
4). The probabilities must remain constant for each trial.
5). Solve by using:
binompdf (n , p , x )
binomcdf (n , p ,  x )
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)
Continuity Correction
When we use the normal distribution (which
is continuous) as an approximation to the
binomial distribution (which is discrete), a
continuity correction is 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.
Correction for Continuity
P(X = 18) ≈ P(17.5 < X < 18.5)
Continuity Correction
P (x  120)
P (x  119.5)
x = at least 120
= 120, 121, 122, . . .
P (x  120)
x = more than 120
= 121, 122, 123, . . .
P (x  120.5)
Continuity Correction
P (x  120)
P (x  120.5)
x = at most 120
= 0, 1, . . . 118, 119, 120
P (x  120)
x = fewer than 120
= 0, 1, . . . 118, 119
P (x  119.5)
Continuity Corrections
x = exactly 120
P (x  120)
P (119.5  x  120.5)
Interval represents discrete number 120
Procedure for Using a Normal Distribution
to Approximate a Binomial Distribution
1) Establish that the normal distribution is suitable
approximation to the binomial distribution by verifying
np ≥ 5 and nq ≥ 5.
2) Find the values of the parameters μ and σ by calculating
μ = np and σ = √npq .
3) Identify the discrete values of x (the number of
successes). Change discrete value x by replacing it with
the interval x – 0.50 to x + 0.50.
4) Draw a normal curve and enter the value of μ, σ and
either x – 0.50 or x + 0.50
5) Find the area corresponding to the desired probability.
Example – Page 279, #2
The given values are discrete. Use the continuity and describe the
region of the normal distribution that corresponds to the indicated
probability.
Probability of at least 24 students understanding
continuity correction
P ( x  24)
the area to the right of 23.5
P ( x  23.5)
Example – Page 279, #4
The given values are discrete. Use the continuity and describe the
region of the normal distribution that corresponds to the indicated
probability.
Probability that the number of working vending machines
in the United States is exactly 27.
P ( x  27)
the area between 26.5 and 27.5
P (26.5  x  27.5)
Example – Page 279, #6
The given values are discrete. Use the continuity and describe the
region of the normal distribution that corresponds to the indicated
probability.
Probability that the number of defective Wayne Newton
CDs is between 15 and 20 inclusive
P ( x  15) and P ( x  20)
the area between 14.5 and 20.5
P (14.5  x  20.5)
Example – Page 279, #10
A) Find the indicated binomial probability. B) If np ≥ 5 and nq ≥ 5,
also estimate the indicated probability by using the normal
distribution as an approximation to the binomial distribution: if
np < 5 or nq < 5, then state that the normal approximation is
not suitable.
With n = 12 and p = 0.8, find P(7)
A) P ( x  7)  0.053
binompdf (12,0.8,7)
Example – Page 279, #10
A) Find the indicated binomial probability. B) If np ≥ 5 and nq ≥ 5,
also estimate the indicated probability by using the normal
distribution as an approximation to the binomial distribution: if
np < 5 or nq < 5, then state that the normal approximation is
not suitable.
With n = 12 and p = 0.8, find P(7)
B)
np  5
nq  5
12(0.8)  5
12(0.2)  5
9.6  5
2.4  5
Normal approximation
is not suitable
Example – Page 279, #14
Estimate the probability of getting at least 65 girls in 100
births. Assume that boys and girls are equally likely. Is
it unusual to get at least 65 girls in 100 births.
Let x = # of girls born
binomial:
Check for normal:
np  5
nq  5
100(0.50)  5
100(1  .50)  5
50  5
50  5
n  100
p  0.50
P ( x  65)
Example – Page 279, #14
n  100
P ( x  65)
μ  np
 100(.50)
 50
p  0.50
σ  npq
 100(0.50)(0.50)
5
0.0019
Pc ( x  64.5)  0.0019
normalcdf (64.5, E 99,50,5)
Since 0.0019 < 0.05 it is
unusual to get at least 65 girls
in 100 births
50
64.5
x
0
3
z
Example – Page 280, #22
On time Flights: Recently, American Airlines had 72.3% of its
flights arriving time (based on data from U.S. Department of
Transportation). In a check of 40 randomly selected American
Airlines flights, 19 arrived on time. Estimate the probability
of getting 19 or fewer on-time flights among 40, assuming
that the 72.3% rate is correct. Is it unusual to get 19 or fewer
on-time flights among 40 randomly selected American flights?
x = number that arrive on time
binomial: n = 40 and p = 0.723
Example – Page 280, #22
x = number that arrive on time
binomial: n = 40 and p = 0.723
normal approximation is appropriate since
np  40(0.723)  28.92  5
nq  40(0.277)  11.08  5
  np  40(0.723)  28.92
  npq  40(0.723)(0.277)  2.830
Example – Page 280, #22
P(x  19)
 Pc (x  19.5)
 0.00044
Yes; since
0.0004 < 0.05,
getting 19 or fewer
on-time would be
unusual event.
  28.92
  2.830
normalcdf (E 99,19.5,28.92,2.830)
0.00044
19.5
28.92
x
– 3.33
0
z
Lesson 5-7
Determining Normality
Procedure for Determining Whether
Data Have a Normal Distribution
1. Histogram: Construct a histogram. Reject
normality if the histogram departs dramatically
from a bell shape.
2. Outliers: Identify outliers. Reject normality if
there is more than one outlier present.
3. Normal Quantile Plot: If the histogram is
basically symmetric and there is at most one
outlier
Normal Quantile Plot
A Normal Quantile Plot is a graph of points (x,y), where
each x value is from the original set of sample data, and
each y value is a z score corresponding to a quantile value
of the standard normal distribution.
Procedure for Determining Whether
Data Have a Normal Distribution
Step 1 – Enter the Data into the List
Step 2 – Press 2nd Y = to access STAT PLOT.
Step 3 – Press ZOOM, and select 9:Zoomstat
Interpreting the Normal Quantile Plot

We can conclude that the given data is appears
to come from a normally distributed population if:


The points are roughly linear – the points appear to lie
reasonably close to a straight line.
We reject normality if:


The points do not lie close to a straight line.
The points exhibit some systematic pattern that is not
a straight line pattern.
Example – Page 286, #2 and 4
Examine the normal quantile plot and determine whether
it depicts the data that have a normal distribution.
#2 Not normal: It would require a broken line with four
segments of four different slopes to pass through
the data
#4 Normal: Form pattern can be approximate by a
single line.
Example – Page 286, #6
Determine whether the requirement of a normal distribution
is satisfied. Assume that the requirement is loose in the
sense that the population distribution need not be exactly
normal, but it must be a distribution that is basically
symmetric and with only one mode.
Head Circumference: The head circumference of males, as
listed in Data Set 3 in Appendix B
Example – Page 286, #6
The circumference appear approximately normally
distributed, with the frequencies tapering off in both
directions about the modal class.
Example – Page 286, #10
Generating Normal Quantile Plots: Use a TI-83 Plus
calculator to generate a normal Quantile Plots or normal
probability plots. Generate the graph then determine
whether the data came from a normally distributed
population
Exercise #6
Example – Page 286, #10
Yes, since the points lie close to a straight line, conclude
that the population distribution is approximately normal.
Example – Page 287, #16
On the days immediately following the destruction caused by the terrorist
attacks on September 11, 2001, lead amounts (in micrograms per cubic
meter) in the air were recorded at Building 5 of the World Trade Center
site, and these values were obtained: 5.40, 1.10, 0.42, 0.73, 0.48,
1.10. Determine whether the data appear to be from a population with a
normal distribution.
The resulting normal quantile plot
at the right indicates the data do not
appear to come from a population
with a normal distribution.