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Two Trig Honors classes took a recent
probability quiz. There were 10 students
in each class, and each class had an
average score of 81.5
Since the averages are the same, can we
assume that the students in both classes all
did pretty much the same on the exam?
Why or why not?
The average
(mean) does
not tell us
anything about
the distribution
or variation in
the grades!
Observe the
dot plots!
Which has the
Greater spread?
Mean
We need a way to measure the
distribution of the data (not just the
average)
Why not just give the range of the data
(different between the highest and lowest
values?)
Observe a set of data with an average of
17.95 and the range is 23.
Here is the average
And here is the range
Most of the numbers are
in this area, and are not
evenly distributed
throughout the range.
The Standard Deviation is a number that
measures how far away each number in a
set of data is from their mean
If the Standard Deviation
is large, it means the
numbers are spread out
from their mean.
If the Standard Deviation
is small, it means the
numbers are close to
their mean.
scores on the
Quiz for Team A:
72
76
80
80
81
83
84
85
85
89
The Standard Deviation measures how far away
each number in a set of data is from their mean.
Let’s start with the lowest score, 72.
How far away is 72 from the mean of 81.5?
72 - 81.5 = - 9.5
Or look at the highest score, 89.
How far away is 89 from the mean of 81.5?
89 - 81.5 = 7.5
Mean:
81.5
Distance
from Mean
The first step
in finding the
Standard
Deviation is to
find all the
distances from
the mean
72
76
80
80
81
83
84
85
85
89
-9.5
7.5
Distance
from Mean
72
76
80
80
81
83
84
85
85
89
- 9.5
- 5.5
- 1.5
- 1.5
- 0.5
1.5
2.5
3.5
3.5
7.5
Next you need to
square each of
the distances to
turn them into
positive
numbers.
72
76
80
80
81
83
84
85
85
89
Distance
from Mean
Distances
Squared
- 9.5
- 5.5
- 1.5
- 1.5
- 0.5
1.5
2.5
3.5
3.5
7.5
90.25
30.25
2.25
2.25
0.25
2.25
6.25
12.25
12.25
56.25
Then sum up all
the distances
squared
72
76
80
80
81
83
84
85
85
89
Distance
from Mean
Distances
Squared
- 9.5
- 5.5
- 1.5
- 1.5
- 0.5
1.5
2.5
3.5
3.5
7.5
90.25
30.25
2.25
2.25
0.25
2.25
6.25
12.25
12.25
56.25
Sum:
214.5
Divide by n
(where n
represents the
number of data
points)
72
76
80
80
81
83
84
85
85
89
Distance
from Mean
Distances
Squared
- 9.5
- 5.5
- 1.5
- 1.5
- 0.5
1.5
2.5
3.5
3.5
7.5
90.25
30.25
2.25
2.25
0.25
2.25
6.25
12.25
12.25
56.25
Sum:
214.5
10
= 21.45
Finally, take the
square root of the
average distances
AND YOU HAVE THE
STANDARD
DEVIATION!!!
72
76
80
80
81
83
84
85
85
89
Distance
from Mean
Distances
Squared
- 9.5
- 5.5
- 1.5
- 1.5
- 0.5
1.5
2.5
3.5
3.5
7.5
90.25
30.25
2.25
2.25
0.25
2.25
6.25
12.25
12.25
56.25
Sum:
214.5
10
= 21.45
= 4.63
If we followed the
same process, we’d
get a STANDARD
DEVIATION OF
15.91
57
65
83
94
95
96
98
93
71
63
Distance
from Mean
Distances
Squared
- 24.5
- 16.5
1.5
12.5
13.5
14.5
16.5
11.5
- 10.5
-18.5
600.25
272.25
2.25
156.25
182.25
210.25
272.25
132.25
110.25
342.25
Sum:
2280.5
10
= 228.05
= 15.10
Team A
Team B
Average on
the Quiz
81.5
81.5
Standard
Deviation
4.63
15.10
Mean
We can see the standard
deviation gives us a
much better
understanding of the
distribution/spread 
Let’s recap that formula
(include on your notesheet )
= mean
x1, x2, … xn,…  Data Values
n = # of data values in set
Use the formula to find the standard
Deviation of the data set:
43, 48, 41, 51, 42
Use the formula to find the standard
Deviation of the data set (on notesheet):
43, 48, 41, 51, 42
Thinking Questions:
1. What kind of data set would have a
Standard deviation of zero?
2. Can a standard deviation be negative?
Explain.
Thinking Questions:
1. What kind of data set would have a
Standard deviation of zero?
2. Can a standard deviation be negative?
Explain.
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