Download Describing Quantitative Data Numerically

Describing
Quantitative Data
Numerically
Symmetric Distributions
Mean, Variance, and
Standard Deviation
Symmetric Distributions


Describing a “typical” value for a set of
data when the distribution is at least
approximately symmetric allows us to
choose our measure of center:
We can use either


Mean
Median
Finding the Mean of a Distribution

The mean of a set of
numbers is the
arithmetic average.
We find this value by
adding together each
value and then
dividing by the
number of values we
added together

The formula for the
mean is:
xi
x
n
Let’s see the Formula in Action

Consider Babe
Ruth’s HR data
54 59 35 41 46 25 47 60
54 46 49 46 41 34 22

A check of a dotplot
indicates that the
distribution is
approximately
symmetric
Dot Plot
Babe Ruth
20
25
30
35
40
45
HR
50
55
60
65
xi
x
n

So… the first step is to add all the values
54 + 59 + 35 + 41 + 46 + 25 + 47 + 60 + 54 +
46 + 49 + 46 + 41 + 34 + 22 =
659
 Now we need to divide that sum by the number
of values we added together.
659
 43.9333
15

So the mean of the data is 43.9333. Now, if we wish to
talk about the “typical” number of home runs for Babe
Ruth (and we ALWAYS wish to talk about the context of
our data!), we could say something like…
On average, Babe Ruth hit approximately 44 home runs
per season during the 15 seasons he played.




Remember that although the center is a
very important part of our description, we
also need to look at the spread of the
distribution.
When we use the mean as our measure of
center, we use the standard deviation as
our measure of spread.
We can think of standard deviation as “an
average distance of values from the mean”
To calculate the standard deviation by
hand, we’ll make a data table…
Calculating Standard Deviation
 x  x 
n 1
2
S
=
SUM
X
X
X-X
(X – X)2
54
43.9333
10.0667
101.3384
59
43.9333
15.0667
227.0054
35
43.9333
-8.9333
79.8038
41
43.9333
-2.9333
8.6042
46
43.9333
2.0667
4.2712
25
43.9333
-18.9333
358.4698
47
43.9333
3.0667
9.4046
60
43.9333
16.0667
258.1388
54
43.9333
10.0667
101.3384
46
43.9333
2.0667
4.2712
49
43.9333
5.0667
25.6714
46
43.9333
2.0667
4.2712
41
43.9333
-2.9333
8.6042
34
43.9333
-9.9333
98.6704
22
43.9333
-21.9333
481.0696
.0005 (essentially 0)
1770.9333
Creating the Data Table
X-X

The first part of our
formula indicates that
we need to find the
distance from the
mean for each of our
values (x – x)
54 – 43.9333 =
10.0667
15.0667
-8.9333
-2.9333
2.0667
-18.9333
3.0667
16.0667
10.0667
2.0667
5.0667
2.0667
-2.9333
-9.9333
-21.9333





Now that we know the individual distances for
each value, we want to find an “average” of
those distances.
To find an average we have to add all the values
together
We find, though, that the sum of those values is
always zero.
Why? Because some of the values are above
the mean (positive values) and some are below
(negative). The positives and negatives cancel
each other out.
So what values can we use to find the “average”
distance from the mean for a set of values?

One way to get rid of
the negative values in
these distances is to
square each of the
values. That’s exactly
what our formula tells
us to do. (x – x)2
(X – X)2
101.3384
227.0054
79.8038
8.6042
4.2712
358.4698
9.4046
258.1388

101.3384
Once we have these
values, to find the
average we must add
them together
4.2712
25.6714
4.2712
8.6042
98.6704
481.0696
SUM =
1770.9333
 x  x 
n 1
The final step in finding
2
an average is to divide by
the number of values we
added together, but our
formula is a little different
here.
•Instead of dividing by the total number of values we added
together, we divide by 1 less than the total.

•Why? We have taken a “sample” of the data instead of
every piece of data in the population. Since another
“sample” would produce a slightly different mean, it would
also produce a slightly different standard deviation. Dividing
by 1 less than the total number of values added together will
give us a slightly larger spread to account for this sampling
variation.



So, we divide the
“sum of the squared
deviations” by n-1
We have now
calculated
everything inside the
square root sign
This value is an
important one—It is
called the
Variance --S2
 x  x 
n 1
2
1770.9333

15  1
1770.9333

14
126.4952 square HR



Since the units of the
variance are not the
same as our original
units, we have one
more calculation we
must make.
The square root of the
variance will restore
the original units and
give us the “average
distance from the
mean”—the standard
deviation
S = 11.2470
 x  x 

n 1
2
126.4952 
11.2470
TI-Tips
Mean, Variance, &
Standard Deviation
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Find the
MEAN
Enter the data into a list
2nd STAT
MATH
3:mean(list name)

If you have used a
frequency list,
3:mean(data list, freq list)
TI-Tips
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


Find the Variance
Enter the data in a list
2nd STAT
MATH
8:variance(list name)

If you have used a
frequency list,
8:variance(data list,
freq list)
TI-Tips




Find
Standard Deviation
Enter the data in a list
2nd STAT
Math
7:stdDev(list name)

If you have used a
frequency list,
7:stdDev(data list, freq list)
Additional Resources
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
Practice of Statistics: Pg 30-34, 43-46
Homework: HW 1.2: 1a-d