Download Lesson 1.1.1

Introduction
Probability distributions are useful in making decisions in
many areas of life, including business and scientific
research. The normal distribution is one of many types of
probability distributions, and perhaps the one most
widely used. Learning how to use the properties of
normal distributions will be a valuable asset in many
careers and subjects, including economics, education,
finance, medicine, psychology, and sports.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Introduction, continued
Understanding a data set requires finding four key
components:
• the overall shape of the distribution
• a measure of central tendency or average
• a measure of variation
• a measure of population or sample size
The first three components are used in determining
proportions and probabilities associated with values in
normal distributions. The two main classes of data are
discrete and continuous. We will begin by focusing on
continuous distributions, particularly the normal distribution.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts
• Understanding a data set, and how an individual value
relates to the data set, requires information about the
overall shape of the distribution as well as measures
of center, measures of variation, and population (or
sample) size. There are two types of data: discrete
and continuous.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• Discrete data refers to a set of values with gaps
between successive values.
• For example, if you hire a bus with 65 seats for a field
trip, but 82 people sign up to go on the field trip, you
need more seats. You would increase the number of
buses from one bus to two buses, rather than from
one bus to a fraction of a second bus.
• When using discrete data, we can assign probabilities
to individual values. For example, the probability of
rolling a 6 on a fair die is 1:6 or 1/6 or 1 to 6.
.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• In contrast, continuous data is a set of values for
which there is at least one value between any two
given values—there are no gaps. For example, if a
car accelerates from 30 miles per hour to 40 miles per
hour, the car passes through every speed between 30
and 40 miles per hour. It does not skip instantly from
30 miles per hour to 40.
• When using continuous data, we need to assign
probabilities to an interval or range of values.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• For continuous data, the probability of an exact
value is essentially 0, so we must assign a range
or an interval of interest to calculate probability.
For example, a car will accelerate through a
series of speeds in miles per hour, including an
infinite number of decimals. Because there are an
infinite number of values between the starting
speed and the desired speed, the probability of
determining an exact speed is essentially 0.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• An interval is a range or a set of values that
starts with a specified value, ends with a specified
value, and includes every value in between. The
starting and ending values are the limits, or
boundaries, of the interval.
• In other words, an interval is a set of values
between a lower bound and an upper bound. The
size of the interval depends on the situation being
observed.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• The probability that a randomly selected student from
a given high school is exactly 64 inches tall is
effectively 0, since methods of measuring are not
completely precise. Measuring tapes and rulers can
vary slightly, and when we take measurements, we
often round to the nearest quarter inch or eighth of an
inch; it is impossible to determine a person’s height to
the exact decimal place. However, we can determine
the probability that a student’s height falls between
two values, such as 63.5 and 64.5 inches, since this
interval includes all of the infinite decimal values
between these two heights.
1.1.1: Normal Distributions and the 68–95–99.7 Rule
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Key Concepts, continued
• To determine the probability of an outcome using
continuous data, we use the proportion of the area
under the normal curve associated with the distribution
of that data.
• A normal curve is a symmetrical curve representing
the normal distribution.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• A probability distribution is a graph of the values of a
random variable with associated probabilities.
• A random variable is a variable with a numerical value
that changes depending on each outcome in a sample
space. A random variable can take on different values,
and the value that a random variable takes is
associated with chance.
• The area under a probability distribution is equal to 1;
that is, 100% of all possible data values within the
interval are represented under the curve.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• A continuous distribution is a graphed set of values
(a curve) in a continuous data set.
• We will examine two types of continuous distributions:
uniform and normal.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
Continuous Uniform Distributions
• A uniform distribution is a set of values that are
continuous, are symmetric to a mean, and have equal
frequencies corresponding to any two equally sized
intervals.
• In other words, the values are spread out uniformly
throughout the distribution.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• To determine the probability of an outcome using a
uniform distribution, we calculate the ratio of the width
of the interval of interest for the given outcome to the
overall width of the distribution:
width of the interval of interest
total width of the interval of distribution
• The result of this proportion is equal to the probability of
the outcome.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• In the uniform distribution below, the data values are
spread evenly from 1 to 9:
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
Continuous Normal Distributions
• Another type of a continuous distribution is a normal
distribution.
• A normal distribution is a set of values that are
continuous, are symmetric to the mean, and have
higher frequencies in intervals close to the mean than
equal-sized intervals away from the mean. When
graphed, data following a normal distribution forms a
normal curve.
• Normal distributions are symmetric to the mean. This
means that 50% of the data is to the right of the mean
and 50% of the data is to the left of the mean.
1.1.1: Normal Distributions and the 68–95–99.7 Rule
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Key Concepts, continued
• The mean is a measure of center in a set of numerical
data, computed by adding the values in a data set and
then dividing the sum by the number of values in the
data set.
• The mean is denoted by the Greek lowercase letter
mu, μ.
• The Greek letter μ is also used when reporting the
mean of a population.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• A population is made up of all of the people, objects,
or phenomena of interest in an investigation. A sample
is a subset of the population—that is, a smaller portion
that represents the whole population.
• The standard deviation is a measure of average
variation about a mean.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• Technically, the standard deviation is the square root of
the average squared difference from the mean, and is
denoted by the lowercase Greek letter sigma, .
Steps to Find the Standard Deviation
1. Calculate the difference between the mean
and each number in the data set.
2. Square each difference.
3. Find the mean of the squared differences.
4. Take the square root of the resulting number.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• Approximately 68% of the values in a normal
distribution are within one standard deviation of the
mean. Written as an equation, this is m ± 1s » 68%. In
other words, the mean, μ, plus or minus the standard
deviation σ times 1 is approximately equal to 68% of
the values in the distribution.
• In the graph that follows, the shading represents these
68% of values that fall within one standard deviation of
the mean.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
Data Within One Standard Deviation of the Mean
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• Approximately 95% of the values in a normal
distribution are within two standard deviations of the
mean, as shown by the shading in the graph on the
next slide.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
Data Within Two Standard Deviations of the Mean
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• Approximately 99.7% of the values in a normal
distribution are within three standard deviations of the
mean, as shaded in the graph on the next slide.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
Data Within Three Standard Deviations of the Mean
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• These percentages of data under the normal curve
(m ± 1s » 68%, m ± 2s » 95%, and m ± 3s » 99.7%)
follow what is called the 68–95–99.7 rule. This rule is
also known as the Empirical Rule.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Key Concepts, continued
• The standard normal distribution has a mean of 0
and a standard deviation of 1. A normal curve is often
referred to as a bell curve, since its shape resembles
the shape of a bell. Normal distribution curves are a
common tool for teachers who want to analyze how
their students performed on a test. If a test is “fair,” you
can expect a handful of students to do very well or very
poorly, with most scores being near average—a normal
curve. If the curve is shifted strongly toward the lower
or higher ends of the scores, then the test was too hard
or too easy.
1.1.1: Normal Distributions and the 68–95–99.7 Rule
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Common Errors/Misconceptions
• applying the 68–95–99.7 rule to distributions that are
not normally distributed
• assuming that all normal distributions have a mean of 0
and/or a standard deviation of 1
• not applying symmetry in a normal distribution to
calculate probabilities
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice
Example 2
Madison needs to ride a shuttle bus to reach an airport
terminal. Shuttle buses arrive every 15 minutes, and the
arrival times for buses are uniformly distributed. What is
the probability that Madison will need to wait more than
6 minutes for the bus?
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 2, continued
1. Sketch a uniform distribution and shade
the area of the interval of interest.
Start by drawing a number line.
The interval of the distribution goes from 0 minutes
to 15 minutes, and the interval of interest is from 6 to
15. Shade the region between 6 and 15.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 2, continued
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 2, continued
2. Determine the total width of the
distribution.
We can see that the total width of the distribution is
15 minutes.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 2, continued
3. Determine the width of the interval of
interest.
Find the absolute value of the difference of the
endpoints of the interval of interest.
15 - 6 = 9 = 9
The width of the interval of interest is 9 minutes.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 2, continued
4. Determine the proportion of the area of
the interval of interest to the total area of
the distribution.
Create a ratio comparing the area that corresponds
to arrival times between 6 and 15 minutes to the area
of the total time frame of 15 minutes between buses.
The proportion of the area of interest to the total area
of the distribution is equal to the area of interest
divided by the total area of the distribution.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 2, continued
width of the interval of interest
total width of the interval of distribution
=
9
15
=
3
5
= 0.6
The proportion of the area of interest to the total area
of the distribution is 0.6.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 2, continued
5. Interpret the proportion in terms of the
context of the problem.
The probability that Madison will wait more than 6
minutes for the bus is 0.6.
✔
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 2, continued
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice
Example 4
The scores of a particular college admission test are
normally distributed, with a mean score of 30 and a
standard deviation of 2. Erin scored a 34 on her test. If
possible, determine the percent of test-takers whom Erin
outperformed on the test.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 4, continued
1. Sketch a normal curve and shade the
area of the interval of interest.
To sketch the normal curve, start by drawing a
number line with a range of values that includes
Erin’s score, 34. The mean of the scores is 30, so
the middle number on the line will be 30. A range of
24 to 36 will give us a number line that has the
mean, 30, in the middle, and an even number of data
points on either side of the mean.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 4, continued
We want to know how many test-takers had scores
lower than Erin’s. Erin scored a 34; therefore, the
area of interest is the area to the left of 34.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 4, continued
2. Determine how many standard deviations
away from the mean Erin’s score is.
From the problem statement, we know that Erin
scored a 34, the mean is 30, and the standard
deviation is 2. Erin’s score is greater than the mean.
Also, we can determine that Erin scored two standard
deviations above the mean.
μ + 1s = 30 + 1(2) = 32
μ + 2s = 30 + 2(2) = 34
Erin’s score
μ + 3s = 30 + 2(3) = 36
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 4, continued
3. Use symmetry and the 68–95–99.7 rule to
determine the area of interest.
We know that the data in a normal curve is
symmetrical about the mean. Since the area under
the curve is equal to 1, the area to the left of the
mean is 0.5, as shaded in the graph on the next
slide.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 4, continued
Erin’s score is above the mean; therefore, we need to
determine the area between the mean and Erin’s
score and add it to the area below the mean to find
the total area of interest.
1.1.1: Normal Distributions and the 68–95–99.7 Rule
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Guided Practice: Example 4, continued
Recall that the 68–95–99.7 rule states the
percentages of data under the normal curve are as
follows: m ± 1s » 68%, m ± 2s » 95%, and
m ± 3s » 99.7%. We know that m ± 2s » 95%. We
have already accounted for the area to the left of the
mean, which includes from the mean down to –2σ.
Since we found that Erin’s score is two standard
deviations from the mean, we need to determine the
area from the mean up to +2σ.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 4, continued
Since data is symmetric about the mean, we know
that half of the area encompassed between μ ± 2s is
above the mean. Therefore, divide 0.95 by 2.
The following graph shows the shaded area of
interest to the right of the mean up until Erin’s score
of 34.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 4, continued
Add the two areas together to get the total area below
2σ, which is equal to Erin’s score of 34.
0.50 + 0.475 = 0.975
The total area of interest for this data is 0.975.
1.1.1: Normal Distributions and the 68–95–99.7 Rule
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Guided Practice: Example 4, continued
A graphing calculator can also be used to calculate
the area of interest.
On a TI-83/84:
Step 1: Press [2ND][VARS] to bring up the
distribution menu.
Step 2: Arrow down to 2: normalcdf. Press [ENTER].
Step 3: Enter the following values for the lower bound,
upper bound, mean (μ), and standard deviation
(σ). Press [,] after typing each value. Lower: [(–
)][99]; upper: [34]; μ: [30]; σ: [2].
Step 4: Press [ENTER] to calculate the area of
interest.
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 4, continued
On a TI-Nspire:
Step 1: Press the [home] key.
Step 2: Arrow over to the spreadsheet icon and
press [enter].
Step 3: Press the [menu] key. Arrow down to 4:
Statistics, then arrow right to bring up the
sub-menu. Arrow down to 2: Distributions
and press [enter].
Step 4: Arrow down to 2: Normal Cdf. Press [enter].
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 4, continued
Step 5: Enter the values for the lower bound, upper
bound, mean (μ), and standard deviation
(σ), using the [tab] key to navigate between
fields. Lower Bound: [(–)][99]; Upper Bound:
[34]; μ: [30]; σ: [2]. Tab down to “OK” and
press [enter].
Step 6: The values entered will appear in the
spreadsheet. Press [enter] again to
calculate the probability.
The result from the graphing calculator verifies the
area of interest is 0.975.
1.1.1: Normal Distributions and the 68–95–99.7 Rule
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Guided Practice: Example 4, continued
4. Interpret the proportion in terms of the
context of the problem.
Convert the area of interest to a percent.
0.975 = 97.5%
Erin outperformed 97.5% of the students who also
took the exam.
✔
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1.1.1: Normal Distributions and the 68–95–99.7 Rule
Guided Practice: Example 4, continued
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1.1.1: Normal Distributions and the 68–95–99.7 Rule