Lesson 2.2

Students Connie and Oscar from Exercises 7
and 8 in Lesson 2.1 had the same mean and
median scores, but the ranges and IQR for
their scores were very different. The range
and IQR both measure the spread of the data.
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
The IQR describes the spread of data relative
to the median.
It can also be useful to look at the spread
relative to the mean.
◦ One way to measure the spread of data relative to
the mean is to calculate the deviations, the signed
differences between the data values and the mean.
◦ Recall the mean score for each student was 84.
Here are the deviations for Connie’s and Oscar’s
scores.
What do the deviations tell you about each
student’s performance?
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
In this investigation you will attempt to
control the setup of an experiment in order to
limit the variability of your results.
Select and perform one of these experiments.
Make complete and careful notes about the
setup of your experiment.
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
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In this experiment you’ll roll a ball of
paper down a ramp and off the edge of
your desk.
Build your ramp from books, notebooks,
or a pad of paper.
Select the height and slope of your
ramp and the distance from the edge of
your desk, and determine any other
factors that might affect your results.
Make a ball by crumpling a piece of
paper, and roll it down the ramp.
Record the horizontal distance to the
place where the ball hits the floor.
Repeat this procedure with the same
ball, the same ramp setup, and the
same release another seven or eight
times.
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In this experiment you’ll use a ruler to
launch a rubber band.
Select the height and angle of your launch
and the length of your stretch, and
determine any other factors that might
affect your results.
Launch the rubber band into an area clear of
obstructions.
Record the horizontal distance of the flight.
Repeat this procedure as precisely as you
can with the same rubber band, the same
launch setup, and the same stretch another
seven or eight times.

Use your data from Experiment 1 or 2.
Calculate the mean distance for your trials
and then calculate the deviations.
Mean (band)=169.2428
Mean (ball) =28.4
Deviation of the band = 16.951
Deviation of the ball = 6.943

In general, how much do your data values
differ from the mean?
◦ How does the variability in your results relate to
how controlled your setup was?
◦ Determine a way to calculate a single value that
tells how accurate your group was at repeating the
procedure.
◦ Write a formula to calculate your statistic using the
deviations.
Deviation =average| sample – mean|

A value known as the standard deviation
helps measure the spread of data away from
the mean. Use your calculator to find this
value for your data.
Standard deviation for the band = 20.190332434349
Standard deviation for the ball = 8.8928810479694

What are the units of the statistic you
calculated in Step 2? The units of the
standard deviation you found in Step 3 are
the same as those of the original
measurements. How does your statistic
compare to the standard deviation?

If you were going to repeat the experiment,
how would you change your procedures to
minimize the standard deviation?
Connie: {82, 86, 82, 84, 85, 84, 85} mean: 84
Oscar: {72, 94, 76, 96, 90, 76, 84} mean:84
Consider Connie’s and Oscar’s scores and
their deviations from the mean score for each
student.
How can you combine the deviations into a
single value that reflects the spread in a data
set?
Finding the sum is a natural choice. However,
if you think of the mean as a balance point in
a data set, then the directed distances above
and below the mean should cancel out. Hence,
the deviation sum for both Connie and Oscar
is zero.
Connie: {82, 86, 82, 84, 85, 84, 85} mean: 84
Oscar: {72, 94, 76, 96, 90, 76, 84} mean:84
variance
Standard deviation
The standard deviation provides one way to judge the “average
difference” between data values and the mean. It is a measure of how
the data are spread around the mean.
In statistics, the mean is often referred to by the symbol x .
Another symbol, Σ , is used to indicate the sum of the data
values.
For example,
5
x
i 1
i
 x1  x2  x3  x4  x5
Where x1 , x2 , x3 , x4 , and x5 are individual data.
n
The mean of n data values is given by
x 
x
i 1
n
i
Sigma notation is useful for defining standard deviation.
The standard deviation, s, is a measure of the spread
of a data set.
 x
n
s
i 1
i
x

n 1
where xi represents the individual data values, n is the number
of values, and x is the mean. The standard deviation has the
same units as the data.
The larger standard deviation for Oscar
indicates that his scores generally lie much
farther from the mean than do Connie’s.
A large value for the standard deviation tells
you that the data values are not as tightly
packed around the mean.
As a general rule, a set with more data near the
mean will have less spread and a smaller
standard deviation.
You may wonder why you divide by (n -1) when
calculating standard deviation.
As you know, the sum of the deviations is zero.
So, if you know all but one of the deviations, you
can calculate the last deviation by making sure the
sum will be zero.
The last deviation depends on the rest, so the set
of deviations contains only (n -1) independent
pieces of data.
This table gives the student-to-teacher ratios for
public elementary and secondary schools in the
United States.
a. Calculate the mean and the standard deviation.
What do the statistics tell you about the spread of
the student-to-teacher ratios?
Mean=15.546
SD = 2.48
Mean + SD =18.03
Mean=15.55
Mean - SD =13.07
More than half the ratios are within 2.48
students of the mean.
Identify any values that are more than two
standard deviations from the mean.
What percentage of all values are more than two
standard deviations from the mean?
Mean + 2 SD = 20.51
Mean=15.55
Mean - 2 SD =10.58
There are no states that average fewer than 10.58 students per teacher.
There are four states that exceed this value: OR (20.6), CA (21.1), AZ (21.3), and
UT (22.4).
So 4 out of 50 or 8%, of the states are more than two standard deviations from the
mean.
1. The monthly averages of precipitation
for Tucson, AZ, in inches, in a given year
are 0.8, 0.8, 0.8, 0.2, 0.2, 0.4, 2.5, 2.7,
1.4, 1.1, 0.6, and1.1.
a. Find the standard deviation.
Mean = 1.05
Mean +2SD= 2.6
Mean = 1.05
Mean -2SD= -.5
1 out of 12 data are outside the 2 sd from the mean or 8.3% of
the data
1. The monthly averages of precipitation for
Tucson, AZ, in inches, in a given year are 0.8,
0.8, 0.8, 0.2, 0.2, 0.4, 2.5, 2.7, 1.4, 1.1, 0.6,
and1.1.
a. Find the standard deviation.
b. What percentage of the values are more than
two standard deviations away from the mean?
[8.3%]