STAT13T UC Davis 900 Filipino Students Population in High Schools in Mesquite Texas Probability Study Paper What would you want to study? If you could create a poll or survey, what would your topic be?Describe your population (the type of person you want), sample (n = ?), one to three questions you would ask, and how you would graph the results.Please remember to use APA formatting if you use help from the textbook or a website. Try to use your own words! Elementary Statistics
Thirteenth Edition
Chapter 6
Normal
Probability
Distributions
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Normal Probability Distributions
6-1 The Standard Normal Distribution
6-2 Real Applications of Normal Distributions
6-3 Sampling Distributions and Estimators
6-4 The Central Limit Theorem
6-5 Assessing Normality
6-6 Normal as Approximation to Binomial
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Key Concept
In this section we present the standard normal distribution, which
is a specific normal distribution having the following three
properties:
1. Bell-shaped: The graph of the standard normal distribution is
bell-shaped.
2. µ = 0: The standard normal distribution has a mean equal to 0.
3. σ = 1: The standard normal distribution has a standard deviation
equal to 1.
In this section we develop the skill to find areas (or probabilities or
relative frequencies) corresponding to various regions under the
graph of the standard normal distribution. In addition, we find z
scores that correspond to areas under the graph.
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Normal Distribution (1 of 2)
• Normal Distribution
– If a continuous random variable has a distribution
with a graph that is symmetric and bell-shaped, we
say that it has a normal distribution.
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Normal Distribution (2 of 2)
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Uniform Distribution (1 of 2)
Properties of uniform distribution:
1. The area under the graph of a continuous probability
distribution is equal to 1.
2. There is a correspondence between area and
probability, so probabilities can be found by
identifying the corresponding areas in the graph
using this formula for the area of a rectangle:
Area = height × width
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Uniform Distribution (2 of 2)
• Uniform Distribution
– A continuous random variable has a uniform
distribution if its values are spread evenly over
the range of possibilities. The graph of a uniform
distribution results in a rectangular shape.
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Density Curve
• Density Curve
– The graph of any continuous probability distribution
is called a density curve, and any density curve
must satisfy the requirement that the total area
under the curve is exactly 1.
Because the total area under any density curve is
equal to 1, there is a correspondence between area
and probability.
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Example: Waiting Times for Airport
Security (1 of 7)
During certain time periods at JFK airport in New York
City, passengers arriving at the security checkpoint
have waiting times that are uniformly distributed
between 0 minutes and 5 minutes, as illustrated in the
figure on the next page.
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Example: Waiting Times for Airport
Security (2 of 7)
Refer to the figure to see these properties:
• All of the different possible waiting times are equally
likely.
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Example: Waiting Times for Airport
Security (3 of 7)
Refer to the figure to see these properties:
• Waiting times can be any value between 0 min and 5
min, so it is possible to have a waiting time of
1.234567 min.
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Example: Waiting Times for Airport
Security (4 of 7)
Refer to the figure to see these properties:
• By assigning the probability of 0.2 to the height of the
vertical line in the figure, the enclosed area is
exactly 1.
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Example: Waiting Times for Airport
Security (5 of 7)
Given the uniform distribution illustrated in the figure,
find the probability that a randomly selected passenger
has a waiting time of at least 2 minutes.
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Example: Waiting Times for Airport
Security (6 of 7)
Solution
The shaded area represents waiting times of at least 2
minutes. Because the total area under the density curve
is equal to 1, there is a correspondence between area
and probability. We can easily find the desired
probability by using areas.
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Example: Waiting Times for Airport
Security (7 of 7)
Solution
P(wait time of at least 2 min) = height × width of shaded
area in the figure = 0.2 × 3 = 0.6
The probability of
randomly selecting a
passenger with a
waiting time of at least
2 minutes is 0.6.
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Standard Normal Distribution
• Standard Normal Distribution
– The standard normal distribution is a normal
distribution with the parameters of µ = 0 and σ = 1.
The total area under its density curve is equal to 1.
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Finding Probabilities When Given z
Scores (1 of 3)
• We can find areas (probabilities) for different regions
under a normal model using technology or Table A-2.
• Technology is strongly recommended.
Because calculators and software generally give more
accurate results than Table A-2, we strongly
recommend using technology.
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Finding Probabilities When Given z
Scores (2 of 3)
If using Table A-2, it is essential to understand these
points:
1. Table A-2 is designed only for the standard normal
distribution, which is a normal distribution with a
mean of 0 and a standard deviation of 1.
2. Table A-2 is on two pages, with the left page for
negative z scores and the right page for positive z
scores.
3. Each value in the body of the table is a cumulative
area from the left up to a vertical boundary above
a specific z score.
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Finding Probabilities When Given z
Scores (3 of 3)
4. When working with a graph, avoid confusion
between z scores and areas.
z score: Distance along the horizontal scale of the
standard normal distribution (corresponding to the
number of standard deviations above or below the
mean); refer to the leftmost column and top row of
Table A-2.
Area: Region under the curve; refer to the values in
the body of Table A-2.
5. The part of the z score denoting hundredths is
found across the top row of Table A-2.
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Formats Used for Finding Normal
Distribution Areas
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Example: Bone Density Test (1 of 7)
A bone mineral density test can be helpful in identifying
the presence or likelihood of osteoporosis. The result of
a bone density test is commonly measured as a z
score. The population of z scores is normally distributed
with a mean of 0 and a standard deviation of 1, so these
test results meet the requirements of a standard normal
distribution.
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Example: Bone Density Test (2 of 7)
The graph of the bone density test scores is as shown
in the figure.
A randomly selected adult undergoes a bone density
test. Find the probability that this person has a bone
density test score less than 1.27.
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Example: Bone Density Test (3 of 7)
Solution
Note that the following are the same (because of the
aforementioned correspondence between probability
and area):
• Probability that the bone density test score is less than
1.27
• Shaded area shown in the figure
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Example: Bone Density Test (4 of 7)
Solution
So we need to find the area in the figure to the left of
z = 1.27.
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Example: Bone Density Test (5 of 7)
Solution
Using Table A-2, begin with the z score of 1.27 by
locating 1.2 in the left column; next find the value in the
adjoining row of probabilities that is directly below 0.07,
as shown:
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Example: Bone Density Test (6 of 7)
Solution
Table A-2 shows that there is an area of 0.8980
corresponding to z = 1.27. We want the area below
1.27, and Table A-2 gives the cumulative area from the
left, so the desired area is 0.8980.
Because of the correspondence between area and
probability, we know that the probability of a z score
below 1.27 is 0.8980.
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Example: Bone Density Test (7 of 7)
Interpretation
The probability that a randomly selected person has a
bone density test result below 1.27 is 0.8980, shown as
the shaded region. Another way to interpret this result is
to conclude that 89.80% of people have bone density
levels below 1.27.
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Example: Bone Density Test: Finding
the Area to the Right of a Value (1 of 4)
Using the same bone density test, find the
probability that a randomly selected person has a
result above −1.00 (which is considered to be in the
“normal” range of bone density readings).
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Example: Bone Density Test: Finding
the Area to the Right of a Value (2 of 4)
Solution
If we use Table A-2, we should know that it is designed
to apply only to cumulative areas from the left.
Referring to the page with negative z scores, we find
that the cumulative area from the left up to z = −1.00 is
0.1587, as shown in the figure.
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Example: Bone Density Test: Finding
the Area to the Right of a Value (3 of 4)
Solution
Because the total area under the curve is 1, we can find
the shaded area by subtracting 0.1587 from 1. The
result is 0.8413.
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Example: Bone Density Test: Finding
the Area to the Right of a Value (4 of 4)
Interpretation
Because of the correspondence between probability
and area, we conclude that the probability of
randomly selecting someone with a bone density
reading above −1 is 0.8413 (which is the area to the
right of z = −1.00). We could also say that 84.13% of
people have bone density levels above −1.00.
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Example: Bone Density Test: Finding
the Area Between Two Values (1 of 3)
A bone density reading between −1.00 and −2.50
indicates the subject has osteopenia, which is some
bone loss. Find the probability that a randomly selected
subject has a reading between −1.00 and −2.50.
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Example: Bone Density Test: Finding
the Area Between Two Values (2 of 3)
Solution
1. The area to the left of z = −1.00 is 0.1587.
2. The area to the left of z = −2.50 is 0.0062.
3. The area between z = −1.00 and z = −2.50 is the
difference between the areas found above.
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Example: Bone Density Test: Finding
the Area Between Two Values (3 of 3)
Interpretation
Using the correspondence between probability and
area, we conclude that there is a probability of 0.1525
that a randomly selected subject has a bone density
reading between −1.00 and −2.50.
Another way to interpret this result is to state that
15.25% of people have osteopenia, with bone density
readings between −1.00 and −2.50.
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Generalized Rule
The area corresponding to the region between two z
scores can be found by finding the difference between
the two areas found in Table A-2.
Don’t try to memorize a rule or formula for this case.
Focus on understanding by using a graph. Draw a
graph, shade the desired area, and then get creative to
think of a way to find the desired area by working with
cumulative areas from the left.
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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.
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Finding z Scores from Known Areas
1. Draw a bell-shaped curve and identify the region
under the curve that corresponds to the given
probability. If that region is not a cumulative region
from the left, work instead with a known region that
is a cumulative region from the left.
2. Use technology or Table A-2 to find the z score.
With Table A-2, use the cumulative area from the
left, locate the closest probability in the body of the
table, and identify the corresponding z score.
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Critical Value (1 of 2)
• Critical Value
– For the standard normal distribution, a critical
value is a z score on the borderline separating
those z scores that are significantly low or
significantly high.
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Critical Value (2 of 2)
Notation
The expression zα denotes the z score with an area of α
to its right.
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Example: Finding the Critical Value
zα (1 of 3)
Find the value of z0.025. (Let α = 0.025 in the
expression zα.)
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Example: Finding the Critical Value
zα (2 of 3)
Solution
The notation of z0.025 is used to represent the z score
with an area of 0.025 to its right. Refer to the figure and
note that the value of z = 1.96 has an area of 0.025 to
its right, so z0.025 = 1.96. Note that z0.025 corresponds to
a cumulative left area of 0.975.
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Example: Finding the Critical Value
zα (3 of 3)
CAUTION
When finding a value of zα for a particular value of α,
note that α is the area to the right of zα, but Table A-2
and some technologies give cumulative areas to the left
of a given z score.
To find the value of zα, resolve that conflict by using the
value of 1 − α. For example, to find z0.1, refer to the z
score with an area of 0.9 to its left.
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Elementary Statistics
Thirteenth Edition
Chapter 6
Normal
Probability
Distributions
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Normal Probability Distributions
6-1 The Standard Normal Distribution
6-2 Real Applications of Normal Distributions
6-3 Sampling Distributions and Estimators
6-4 The Central Limit Theorem
6-5 Assessing Normality
6-6 Normal as Approximation to Binomial
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Key Concept
This section presents methods for working with normal
distributions that are not standard. That is, the mean is
not 0 or the standard deviation is not 1, or both.
The key is that we can use a simple conversion that
allows us to “standardize” any normal distribution so
that the same methods of the previous section can be
used.
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Conversion Formula
(round z scores to 2 decimal places)
The formula allows us to “standardize” any normal
distribution so that x values can be transformed to z
scores.
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Converting to a Standard Normal
Distribution
The figures illustrate the conversion from a nonstandard
to a standard normal distribution. The area in any normal
distribution bounded by some score x (as in Figure a) is
the same as the area bounded by the corresponding z
score in the standard normal distribution (as in Figure b).
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Procedure for Finding Areas with a
Nonstandard Normal Distribution
1. Sketch a normal curve, label the mean and any specific
x values, and then shade the region representing the
desired probability.
2. For each relevant value x that is a boundary for the
shaded region, use the formula
to convert that value to the equivalent z score. (With
many technologies, this step can be skipped.)
3. Use technology (software or a calculator) or Table A-2 to
find the area of the shaded region. This area is the
desired probability.
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Example: What Proportion of Men Are Taller
Than the 72 in. Height Requirement for
Showerheads? (1 of 6)
Heights of men are normally distributed with a mean
of 68.6 in. and a standard deviation of 2.8 in. Find the
percentage of men who are taller than a showerhead
at 72 in.
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Example: What Proportion of Men Are Taller
Than the 72 in. Height Requirement for
Showerheads? (2 of 6)
Solution
Step 1: Men have heights that are normally distributed
with a mean of 68.6 in. and a standard deviation of 2.8
in. The shaded region represents the men who are taller
than the showerhead height of 72 in.
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Example: What Proportion of Men Are Taller
Than the 72 in. Height Requirement for
Showerheads? (3 of 6)
Solution
Step 2: We can convert the showerhead height of 72 in.
to the z score of 1.21 by using the conversion formula
as follows:
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Example: What Proportion of Men Are Taller
Than the 72 in. Height Requirement for
Showerheads? (4 of 6)
Solution
Step 3: Technology: Technology can be used to find that
the area to the right of 72 in. in the figure is 0.1123
rounded. (With many technologies, Step 2 can be skipped.)
The result of 0.1123 from technology is more accurate than
the result of 0.1131 found by using Table A-2.
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Example: What Proportion of Men Are Taller
Than the 72 in. Height Requirement for
Showerheads? (5 of 6)
Solution
Table A-2: Use Table A-2 to find that the cumulative area
to the left of z = 1.21 is 0.8869. (Remember, Table A-2 is
designed so that all areas are cumulative areas from the
left.) Because the total area under the curve is 1, it
follows that the shaded area is 1 − 0.8869 = 0.1131.
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Example: What Proportion of Men Are Taller
Than the 72 in. Height Requirement for
Showerheads? (6 of 6)
Interpretation
The proportion of men taller than 72 in. is 0.1123, or
11.23%. About 11% of men may find the design to be
unsuitable.
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Finding Values From Known Areas (1 of 3)
Here are helpful hints for those cases in which the area
(or probability or percentage) is known and we must find
the relevant value(s):
1. Graphs are extremely helpful in visualizing,
understanding, and successfully working with normal
provability distributions, so they should always be used.
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Finding Values From Known Areas (2 of 3)
2. Don’t confuse z scores and areas. z scores are
distan...
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