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 crea

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
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
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Normal Distribution (2 of 2)
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Formats Used for Finding Normal
Distribution Areas
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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:
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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).
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
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.
Copyright © 2018, 2014, 2012 Pearson Education, Inc. All Rights Reserved
Notation
• P(a a) denotes the probability that the z score is
greater than a.
• P(z