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Chapter
37
The Normal
Probability
Distribution
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Pearson Prentice Hall. All rights reserved
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Section 7.1 Properties of the Normal Distribution
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EXAMPLE
Illustrating the Uniform Distribution
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.
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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.
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Values of the random variable X less than 0 or greater than
60 are impossible, thus the equation must be zero for X
less than 0 or greater than 60.
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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.
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EXAMPLE
Area as a Probability
The probability of choosing a time that is between 15 and 30
seconds after the minute is the area under the uniform density
function.
Area
= P(15 < x < 30)
= 15/60
= 0.25
15
30
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Relative frequency histograms that are symmetric and
bell-shaped are said to have the shape of a normal
curve.
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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).
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Use the graph of the given normal distribution
to identify μ and σ.
62
68
74
80
86
A. μ = 80 and σ = 36
B. μ = 80 and σ = 6
C. μ = 62 and σ = 98
D. μ = 6 and σ = 80
Slide 7- 17
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92
98
X
Use the graph of the given normal distribution
to identify μ and σ.
62
68
74
80
86
A. μ = 80 and σ = 36
B. μ = 80 and σ = 6
C. μ = 62 and σ = 98
D. μ = 6 and σ = 80
Slide 7- 18
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92
98
X
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EXAMPLE
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.
(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.
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Section 7.2 The Standard Normal Distribution
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The table gives the area under the standard normal curve for
values to the left of a specified Z-score, zo, as shown in the
figure.
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EXAMPLE
Finding the Area Under the Standard Normal Curve
Find the area under the standard normal curve to the left of z = -0.38.
Area left of z = -0.38 is 0.3520.
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Area under the normal curve to the
right of zo = 1 – Area to the left of zo
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EXAMPLE
Finding the Area Under the Standard Normal Curve
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 © 2010 Pearson Prentice Hall. All rights reserved
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EXAMPLE
Finding the Area Under the Standard Normal Curve
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
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EXAMPLE
Finding a z-score from a Specified Area to the Left
Find the z-score such that the area to the left of the z-score is 0.7157.
The z-score such that the area to the left of the z-score is 0.7157 is z = 0.57.
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EXAMPLE
Finding a z-score from a Specified Area to the Right
Find the z-score such that the area to the right of the z-score is 0.3021.
The area left of the z-score is 1 – 0.3021 = 0.6979.
The approximate z-score that corresponds to an area of 0.6979 to the left (0.3021
to the right) is 0.52. Therefore, z = 0.52.
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EXAMPLE
Finding a z-score from a Specified Area
Find the z-scores that separate the middle 80% of the area under the normal curve
from the 20% in the tails.
Area = 0.8
Area = 0.1
Area = 0.1
z1 is the z-score such that the area left is 0.1, so z1 = -1.28.
z2 is the z-score such that the area left is 0.9, so z2 = 1.28.
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Notation for the Probability of a Standard Normal Random
Variable
P(a < Z < b)
represents the probability a standard
normal random variable is between
a and b
P(Z > a)
represents the probability a standard
normal random variable is greater
than a.
P(Z < a)
represents the probability a standard
normal random variable is less than a.
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Find the probability of the standard normal
random variable Z.
P(Z < 1.49)
A. 0.9319
B. 0.0681
C. 0.6879
D. 0.3121
Slide 7- 41
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Find the probability of the standard normal
random variable Z.
P(Z < 1.49)
A. 0.9319
B. 0.0681
C. 0.6879
D. 0.3121
Slide 7- 42
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EXAMPLE
Finding Probabilities of Standard Normal Random Variables
Find each of the following probabilities:
(a) P(Z < -0.23)
(b) P(Z > 1.93)
(c) P(0.65 < Z < 2.10)
(a) P(Z < -0.23) = 0.4090
(b) P(Z > 1.93) = 0.0268
(c) P(0.65 < Z < 2.10) = 0.2399
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Find the probability of the standard normal
random variable Z.
P(Z ≥ –2.31)
A. 0.0104
B. 0.0087
C. 0.9896
D. 0.9913
Slide 7- 44
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Find the probability of the standard normal
random variable Z.
P(Z ≥ –2.31)
A. 0.0104
B. 0.0087
C. 0.9896
D. 0.9913
Slide 7- 45
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Find the probability of the standard normal
random variable Z.
P(–2.14 < Z < 0.95)
A. 0.1170
B. 0.0681
C. 0.1873
D. 0.8127
Slide 7- 46
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Find the probability of the standard normal
random variable Z.
P(–2.14 < Z < 0.95)
A. 0.1170
B. 0.0681
C. 0.1873
D. 0.8127
Slide 7- 47
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Find the Z-score such that the area under
the standard normal curve to the right is
0.10.
A. 1.28
B. -1.28
C. 0.5398
D. 0.8159
Slide 7- 48
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Find the Z-score such that the area under
the standard normal curve to the right is
0.10.
A. 1.28
B. -1.28
C. 0.5398
D. 0.8159
Slide 7- 49
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For any continuous random variable, the probability of observing a specific
value of the random variable is 0. For example, for a standard normal
random variable, P(a) = 0 for any value of a. This is because there is no area
under the standard normal curve associated with a single value, so the
probability must be 0. Therefore, the following probabilities are equivalent:
P(a < Z < b) = P(a < Z < b) = P(a < Z < b) = P(a < Z < b)
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IQ scores are normally distributed with a
mean of 100 and a standard deviation of
15. Find the probability a randomly
selected person has an IQ score greater
than 120.
A. 0.9082
B. 0.6293
C. 0.0918
D. 0.3707
Slide 7- 51
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IQ scores are normally distributed with a
mean of 100 and a standard deviation of
15. Find the probability a randomly
selected person has an IQ score greater
than 120.
A. 0.9082
B. 0.6293
C. 0.0918
D. 0.3707
Slide 7- 52
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Section 7.4 Assessing Normality
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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.
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The idea behind finding the expected Z-score is that if the data
comes from a population that is normally distributed, we should
be able to predict the area left of each of the data values. The
value of fi represents the expected area left of the ith data value
assuming the data comes from a population that is normally
distributed. For example, f1 is the expected area left of the
smallest data value, f2 is the expected area left of the second
smallest data value, and so on.
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A discrete random variable is given. Assume
the probability will be approximated using a
normal distribution. Describe the area under
the normal curve that will be computed.
The probability of getting at least 80
successes
A. Right of X = 80.5
B. Right of X = 79.5
C. Left of X = 80.5
D. Left of X = 79.5
Slide 7- 56
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A discrete random variable is given. Assume
the probability will be approximated using a
normal distribution. Describe the area under
the normal curve that will be computed.
The probability of getting at least 80
successes
A. Right of X = 80.5
B. Right of X = 79.5
C. Left of X = 80.5
D. Left of X = 79.5
Slide 7- 57
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Section 7.5 The Normal Approximation to the Binomial
Probability Distribution
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Criteria for a Binomial Probability Experiment
An experiment is said to be a binomial experiment
provided:
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.
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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 general rule of
thumb, if np(1 – p) > 10, the probability
distribution will be approximately symmetric
and bell-shaped.
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P(X = 18) ≈ P(17.5 < X < 18.5)
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P(X < 18) ≈ P(X < 18.5)
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EXAMPLE
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?
np(1  p)  400(0.35)(1  0.35)  X  400(0.35)
 91  10
 140
 X  400(0.35)(1  0.35)
 9.54
P( X  150)  P( X  149)
 P( X  149.5)
149.5  140 

 PZ 

9.54


 0.8413
(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)
159.5  140 

 PZ 

9.54


 0.0207
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Forty-three percent of marriages end in
divorce. You randomly select 150 married
couples. Find the probability that fewer than
70 of the marriages will end in divorce.
A. 0.8389
B. 0.8186
C. 0.8251
D. 0.7967
Slide 7- 67
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Forty-three percent of marriages end in
divorce. You randomly select 150 married
couples. Find the probability that fewer than
70 of the marriages will end in divorce.
A. 0.8389
B. 0.8186
C. 0.8251
D. 0.7967
Slide 7- 68
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