Sfs4e ppt 07

Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.
Chapter
The Normal
Probability
Distribution
7

Section
Properties of the
Normal
Distribution
7.1

Objectives
1. Use the uniform probability distribution
2. Graph a normal curve
3. State the properties of the normal curve
4. Explain the role of area in the normal density
function
5-3

Objective 1
• Use the Uniform Probability Distribution
5-4

Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc.7-5
Suppose that United Parcel Service is supposed to deliver a
package to your front door and the arrival time is somewhere
between 10 am and 11 am. Let the random variable X
represent the time from 10 am when the delivery is supposed
to take place. The delivery could be at 10 am (x = 0) or at 11
am (x = 60) with all 1-minute interval of times between x = 0
and x = 60 equally likely. That is to say your package is just
as likely to arrive between 10:15 and 10:16 as it is to arrive
between 10:40 and 10:41. The random variable X can be any
value in the interval from 0 to 60, that is, 0 < X < 60.
Because any two intervals of equal length between 0 and 60,
inclusive, are equally likely, the random variable X is said to
follow a uniform probability distribution.
EXAMPLE Illustrating the Uniform DistributionEXAMPLE Illustrating the Uniform Distribution

A probability density function (pdf) is an
equation used to compute probabilities of
continuous random variables. It must satisfy
the following two properties:
1.The total area under the graph of the
equation over all possible values of the
random variable must equal 1.
2.The height of the graph of the equation must
be greater than or equal to 0 for all possible
values of the random variable.
A probability density function (pdf) is an
equation used to compute probabilities of
continuous random variables. It must satisfy
the following two properties:
1.The total area under the graph of the
equation over all possible values of the
random variable must equal 1.
2.The height of the graph of the equation must
be greater than or equal to 0 for all possible
values of the random variable.
5-6

The graph below illustrates the properties for
the “time” example. Notice the area of the
rectangle is one and the graph is greater than or
equal to zero for all x between 0 and 60,
inclusive.
Because the area of a
rectangle is height
times width, and the
width of the
rectangle is 60, the
height must be 1/60.

Values of the random variable X less than 0
or greater than 60 are impossible, thus the
function value must be zero for X less than 0
or greater than 60.

The area under the graph of the density
function over an interval represents the
probability of observing a value of the
random variable in that interval.

The probability of choosing a time that is
between 15 and 30 minutes after the hour is the
area under the uniform density function.
15 30
Area
= P(15 < x < 30)
= 15/60
= 0.25
EXAMPLE Area as a ProbabilityEXAMPLE Area as a Probability

Objective 2
• Graph a Normal Curve
5-11

Relative frequency histograms that are
symmetric and bell-shaped are said to have the
shape of a normal curve.

If a continuous random variable is normally
distributed, or has a normal probability
distribution, then a relative frequency
histogram of the random variable has the shape
of a normal curve (bell-shaped and symmetric).

Objective 3
• State the Properties of the Normal Curve
5-15

Properties of the Normal Density Curve
1. It is symmetric about its mean, μ.
2. Because mean = median = mode, the
curve has a single peak and the highest point
occurs at x = μ.
•It has inflection points at μ – σ and μ – σ
•The area under the curve is 1.
•The area under the curve to the right of μ
equals the area under the curve to the left of μ,
which equals 1/2.
Properties of the Normal Density Curve
1. It is symmetric about its mean, μ.
2. Because mean = median = mode, the
curve has a single peak and the highest point
occurs at x = μ.
•It has inflection points at μ – σ and μ – σ
•The area under the curve is 1.
•The area under the curve to the right of μ
equals the area under the curve to the left of μ,
which equals 1/2.
5-16

6. As x increases without bound (gets larger and
larger), the graph approaches, but never
reaches, the horizontal axis. As x decreases
without bound (gets more and more
negative), the graph approaches, but never
reaches, the horizontal axis.
7. The Empirical Rule: Approximately 68% of
the area under the normal curve is between
x = μ – σ and x = μ + σ; approximately 95%
of the area is between x = μ – 2σ and x = μ +
2σ; approximately 99.7% of the area is
between
x = μ – 3σ and x = μ + 3σ.
6. As x increases without bound (gets larger and
larger), the graph approaches, but never
reaches, the horizontal axis. As x decreases
without bound (gets more and more
negative), the graph approaches, but never
reaches, the horizontal axis.
7. The Empirical Rule: Approximately 68% of
the area under the normal curve is between
x = μ – σ and x = μ + σ; approximately 95%
of the area is between x = μ – 2σ and x = μ +
2σ; approximately 99.7% of the area is
between
x = μ – 3σ and x = μ + 3σ.5-17

Objective 4
• Explain the Role of Area in the Normal
Density Function
5-19

The data on the next slide represent the heights
(in inches) of a random sample of 50 two-year
old males.
(a) Draw a histogram of the data using a lower
class limit of the first class equal to 31.5 and a
class width of 1.
(b) Do you think that the variable “height of 2-
year old males” is normally distributed?
EXAMPLE A Normal Random VariableEXAMPLE A Normal Random Variable

36.0 36.2 34.8 36.0 34.6 38.4 35.4 36.8
34.7 33.4 37.4 38.2 31.5 37.7 36.9 34.0
34.4 35.7 37.9 39.3 34.0 36.9 35.1 37.0
33.2 36.1 35.2 35.6 33.0 36.8 33.5 35.0
35.1 35.2 34.4 36.7 36.0 36.0 35.7 35.7
38.3 33.6 39.8 37.0 37.2 34.8 35.7 38.9
37.2 39.3

In the next slide, we have a normal density
curve drawn over the histogram. How does
the area of the rectangle corresponding to a
height between 34.5 and 35.5 inches relate
to the area under the curve between these
two heights?

Area under a Normal Curve
Suppose that a random variable X is normally
distributed with mean μ and standard deviation
σ. The area under the normal curve for any
interval of values of the random variable X
represents either
• the proportion of the population with the
characteristic described by the interval of
values or
• the probability that a randomly selected
individual from the population will have the
values.
Area under a Normal Curve
Suppose that a random variable X is normally
σ. The area under the normal curve for any
interval of values of the random variable X
represents either
• the proportion of the population with the
values or
• the probability that a randomly selected
individual from the population will have the
values.
5-26

EXAMPLE Interpreting the Area Under a Normal CurveEXAMPLE Interpreting the Area Under a Normal Curve
The weights of giraffes are approximately normally
distributed with mean μ = 2200 pounds and standard
deviation σ = 200 pounds.
(a)Draw a normal curve with the parameters labeled.
(b)Shade the area under the normal curve to the left of x
= 2100 pounds.
(c)Suppose that the area under the normal curve to the
left of x = 2100 pounds is 0.3085. Provide two
interpretations of this result.

EXAMPLE Interpreting the Area Under a Normal CurveEXAMPLE Interpreting the Area Under a Normal Curve
μ = 2200 pounds and σ = 200 pounds
(a), (b)
(c)
•The proportion of giraffes whose weight is less than
2100 pounds is 0.3085
•The probability that a randomly selected giraffe
weighs less than 2100 pounds is 0.3085.

Section
Applications of
the Normal
Distribution
7.2

Objectives
1. Find and interpret the area under a normal
curve
2. Find the value of a normal random variable
5-30

Objective 1
• Find and Interpret the Area Under a Normal
Curve
5-31

Standardizing a Normal Random Variable
Suppose that the random variable X is normally
σ. Then the random variable
is normally distributed with mean μ = 0 and
standard deviation σ = 1.The random variable Z
is said to have the standard normal distribution.
Standardizing a Normal Random Variable
Suppose that the random variable X is normally
σ. Then the random variable
is normally distributed with mean μ = 0 and
standard deviation σ = 1.The random variable Z
is said to have the standard normal distribution.
5-32
Z =
X − µ
σ

Standard Normal Curve

The table gives the area under the standard
normal curve for values to the left of a
specified Z-score, z, as shown in the figure.

IQ scores can be modeled by a normal
distribution with μ = 100 and σ = 15.
An individual whose IQ score is 120, is 1.33
standard deviations above the mean.
z =
x − µ
σ
=
120 −100
15
= 1.33

The area under the standard normal curve to
the left of z = 1.33 is 0.9082.

Use the Complement Rule to find the area to
the right of z = 1.33.

Areas Under the Standard Normal Curve

Find the area under the standard normal curve to the left
of z = –0.38.
EXAMPLE Finding the Area Under the Standard Normal CurveEXAMPLE Finding the Area Under the Standard Normal Curve
Area to the left of z = –0.38 is 0.3520.

Area under the normal curve to the right
of zo = 1 – Area to the left of zo

Find the area under the standard normal curve to the
right of z = 1.25.
Area right of 1.25 = 1 – area left of 1.25
= 1 – 0.8944 = 0.1056

Find the area under the standard normal curve between
z = –1.02 and z = 2.94.
Area between –1.02 and 2.94
= (Area left of z = 2.94) – (area left of z = –1.02)
= 0.9984 – 0.1539
= 0.8445

Problem: Find the area to the left of x.
Approach: Shade the area to the left of x.
Solution:
• Convert the value of x to a z-score. Use
Table V to find the row and column that
correspond to z. The area to the left of x is the
value where the row and column intersect.
• Use technology to find the area.

Problem: Find the area to the right of x.
Approach: Shade the area to the right of x.
Solution:
• Convert the value of x to a z-score. Use
Table V to find the area to the left of z (also is
the area to the left of x). The area to the right of
z (also x) is 1 minus the area to the left of z.

Problem: Find the area between x1 and x2.
Approach: Shade the area between x1 and x2.
Solution:
• Convert the values of x to a z-scores. Use
Table V to find the area to the left of z1 and to
the left of z2. The area between z1 and z2 is (area
to the left of z2) – (area to the left of z1).

Objective 2
• Find the Value of a Normal Random Variable
5-46

Procedure for Finding the Value of a Normal
Random Variable
Step 1: Draw a normal curve and shade the
area corresponding to the proportion, probability,
or percentile.
Step 2: Use Table V to find the z-score that
corresponds to the shaded area.
Step 3: Obtain the normal value from the
formula x = μ + zσ.
Procedure for Finding the Value of a Normal
Random Variable
Step 1: Draw a normal curve and shade the
area corresponding to the proportion, probability,
or percentile.
Step 2: Use Table V to find the z-score that
corresponds to the shaded area.
Step 3: Obtain the normal value from the
formula x = μ + zσ.
5-47

The combined (verbal + quantitative reasoning) score on
the GRE is normally distributed with mean 1049 and
standard deviation 189.
(Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.)
EXAMPLE Finding the Value of a Normal Random
Variable
Variable
What is the score of a student whose percentile rank is at
the 85th
percentile?

Variable
Variable
The z-score that corresponds to the 85th
percentile is
the z-score such that the area under the standard
normal curve to the left is 0.85. This z-score is 1.04.
x = µ + zσ
= 1049 + 1.04(189)
= 1246
Interpretation: A person who scores 1246 on the GRE
would rank in the 85th
percentile.

It is known that the length of a certain steel rod is
normally distributed with a mean of 100 cm and a
standard deviation of 0.45 cm. Suppose the
manufacturer wants to accept 90% of all rods
manufactured. Determine the length of rods that make
up the middle 90% of all steel rods manufactured.
Variable
Variable

Variable
Variable
Area = 0.05Area = 0.05 z1 = –1.645 and z2 = 1.645
x1 = µ + z1σ
= 100 + (–1.645)(0.45)
= 99.26 cm
x2 = µ + z2σ
= 100 + (1.645)(0.45)
= 100.74 cm
Interpretation: The length of steel rods that make up the
middle 90% of all steel rods manufactured would have
lengths between 99.26 cm and 100.74 cm.

The notation zα (pronounced “z sub alpha”) is the
z-score such that the area under the standard
normal curve to the right of zα is α.
The notation zα (pronounced “z sub alpha”) is the
z-score such that the area under the standard
normal curve to the right of zα is α.
5-52

Section
Assessing
Normality
7.3

Objectives
1. Use normal probability to assess normality
5-54

Suppose that we obtain a simple random sample from a
population whose distribution is unknown. Many of the
statistical tests that we perform on small data sets (sample
size less than 30) require that the population from which the
sample is drawn be normally distributed. Up to this point,
we have said that a random variable X is normally
distributed, or at least approximately normal, provided the
histogram of the data is symmetric and bell-shaped. This
method works well for large data sets, but the shape of a
histogram drawn from a small sample of observations does
not always accurately represent the shape of the population.
For this reason, we need additional methods for assessing
the normality of a random variable X when we are looking at
sample data.

Objective 1
• Use Normal Probability Plots to Assess
Normality
5-56

A normal probability plot plots observed data
versus normal scores.
A normal score is the expected Z-score of the
data value if the distribution of the random
variable is normal. The expected Z-score of an
observed value will depend upon the number of
observations in the data set.

Drawing a Normal Probability Plot
Step 1 Arrange the data in ascending order.
Step 2 Compute
where i is the index (the position of the data
value in the ordered list) and n is the number of
observations. The expected proportion of
observations less than or equal to the ith data
value is f.
Step 1 Arrange the data in ascending order.
Step 2 Compute
where i is the index (the position of the data
value in the ordered list) and n is the number of
observations. The expected proportion of
observations less than or equal to the ith data
value is f.
5-58
fi =
i − 0.375
n + 0.25

Step 3 Find the z-score corresponding to fi
from Table V.
Step 4 Plot the observed values on the
horizontal axis and the corresponding expected
z-scores on the vertical axis.
Step 3 Find the z-score corresponding to fi
from Table V.
Step 4 Plot the observed values on the
horizontal axis and the corresponding expected
z-scores on the vertical axis.
5-59

The idea behind finding the expected z-score is that, if
the data comes from normally distributed population,
we could predict the area to the left of each of the data
value. The value of fi represents the expected area left
of the ith
observation when the data come from a
population that is normally distributed. For example, f1
is the expected area to the left of the smallest data value,
f2 is the expected area to the left of the second smallest
data value, and so on.

If sample data is taken from a population that is
normally distributed, a normal probability plot
of the actual values versus the expected Z-
scores will be approximately linear.

We will be content in reading normal
probability plots constructed using the
statistical software package, MINITAB. In
MINITAB, if the points plotted lie within the
bounds provided in the graph, then we have
reason to believe that the sample data comes
from a population that is normally distributed.

The following data represent the time between eruptions
(in seconds) for a random sample of 15 eruptions at the
Old Faithful Geyser in California. Is there reason to
believe the time between eruptions is normally
distributed?
728 678 723 735 735
730 722 708 708 714
726 716 736 736 719
EXAMPLE Interpreting a Normal Probability PlotEXAMPLE Interpreting a Normal Probability Plot

The random variable “time between eruptions” is
likely not normal.

Suppose that seventeen randomly selected workers at a
detergent factory were tested for exposure to a Bacillus
subtillis enzyme by measuring the ratio of forced
expiratory volume (FEV) to vital capacity (VC).
NOTE: FEV is the maximum volume of air a person
can exhale in one second; VC is the maximum volume
of air that a person can exhale after taking a deep
breath. Is it reasonable to conclude that the FEV to VC
(FEV/VC) ratio is normally distributed?
Source: Shore, N.S.; Greene R.; and Kazemi, H. “Lung Dysfunction in Workers Exposed to Bacillus subtillis
Enzyme,” Environmental Research, 4 (1971), pp. 512 - 519.

It is reasonable to believe that FEV/VC is
normally distributed.

Section
The Normal
Approximation to
the Binomial
Probability
Distribution
7.4

Objectives
1. Approximate binomial probabilities using the
normal distribution
5-70

Objective 1
• Approximate Binomial Probabilities Using the
Normal Distribution
5-71

Criteria for a Binomial Probability Experiment
An experiment is said to be a binomial experiment if all
of the following are true:
1. The experiment is performed n independent times.
Each repetition of the experiment is called a trial.
Independence means that the outcome of one trial will
not affect the outcome of the other trials.
2. For each trial, there are two mutually exclusive
outcomes - success or failure.
3. The probability of success, p, is the same for each trial
of the experiment.
Criteria for a Binomial Probability Experiment
An experiment is said to be a binomial experiment if all
of the following are true:
1. The experiment is performed n independent times.
Each repetition of the experiment is called a trial.
Independence means that the outcome of one trial will
not affect the outcome of the other trials.
2. For each trial, there are two mutually exclusive
outcomes - success or failure.
3. The probability of success, p, is the same for each trial
of the experiment.
5-72

For a fixed p, as the number of trials n in a
binomial experiment increases, the probability
distribution of the random variable X becomes
more nearly symmetric and bell-shaped. As a
rule of thumb, if np(1 – p) > 10, the probability
distribution will be approximately symmetric
and bell-shaped.

The Normal Approximation to the Binomial
Probability Distribution
If np(1 – p) ≥ 10, the binomial random variable
X is approximately normally distributed, with
mean μX = np and standard deviation
The Normal Approximation to the Binomial
Probability Distribution
If np(1 – p) ≥ 10, the binomial random variable
X is approximately normally distributed, with
mean μX = np and standard deviation
5-74
σX = np 1− p( ).

P(X = 18) ≈ P(17.5 < X < 18.5)

P(X < 18) ≈ P(X < 18.5)

Exact Probability Using Binomial: P(a)
Approximate
Probability
Using Normal: P(a – 0.5 ≤ X ≤ a + 0.5)
Graphical Depiction

Exact Probability Using Binomial: P(X ≤ a)
Approximate Probability
Using Normal: P(X ≤ a + 0.5)
Graphical Depiction

Exact Probability Using Binomial: P(X ≥ a)
Using Normal: P(X ≥ a – 0.5)
Graphical Depiction

Exact Probability
Using Binomial: P(a ≤ X ≤ b)
Using Normal: P(X ≥ a – 0.5)
Graphical Depiction

EXAMPLE Using the Binomial Probability Distribution FunctionEXAMPLE Using the Binomial Probability Distribution Function
According to the Experian Automotive, 35% of all
car-owning households have three or more cars.
(a)In a random sample of 400 car-owning
households, what is the probability that fewer than
150 have three or more cars?
(1 ) 400(0.35)(1 0.35)
91 10
np p− = −
= ≥
400(0.35)
140
Xµ =
=
400(0.35)(1 0.35)
9.54
Xσ = −
=
( 150) ( 149)
( 149.5)
149.5 140
9.54
0.8413
P X P X
P X
P Z
< = ≤
≈ ≤
− 
= ≤ ÷
 
=

EXAMPLE Using the Binomial Probability Distribution FunctionEXAMPLE Using the Binomial Probability Distribution Function
According to the Experian Automotive, 35% of all
car-owning households have three or more cars.
(b) In a random sample of 400 car-owning
households, what is the probability that at least 160
have three or more cars?
P(X ≥ 160) ≈ P(X ≥ 159.5)
= P Z ≥
159.5 −140
9.54




= 0.0207

Sfs4e ppt 07

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Sfs4e ppt 07