A Closer Look at the NEW
High School Statistics
Standards
Focus on A.9 and AII.11
K-12 Mathematics Institutes
Fall 2010
Vertical Articulation
5.16 The student will
b) describe the mean as fair share
6.15 The student will
a) describe mean as balance point
Algebra I SOL A.9 The student, given
a set of data, will interpret variation
in real-world contexts and interpret
mean absolute deviation, standard
deviation, and z-scores.
Fall 2010
2
Vertical Articulation
AFDA.7 The student will analyze the
normal distribution.
Algebra II SOL A.11 The student will
identify properties of the normal
distribution and apply those
properties to determine
probabilities associated with areas
under the standard normal curve.
Fall 2010
3
Before we start – just a little
reminder about sigma notation
and subscript notation
8
i  1 2  3  4  5  6  7  8
i 1
6
x
i 1
Fall 2010
i
 x1  x 2  x3  x 4  x5  x6
4
Mean of a Data Set Containing
n Elements = µ
n

x
i 1
n
i
x1  x2  x3  x4  ...  xn

n
x = Sample mean
µ = Population mean
Fall 2010
5
Mean Problem
Joe has the following test grades:
85, 80, 83, 91, 97 and 72. In order
to make the academic team he
needs to have an 85 average. With
one test yet to take, he wants to
know what score he will need on
that to have an 85 average.
Fall 2010
6
Solve for x:
85  80  83  91  97  72  x
 85
7
What score will “balance” the number line ?
13
72
5 2
80
2
6
87
83
12
91
97
85
Fall 2010
7
A student counted the number of
players playing basketball in the
Central Tendency Tournament
each day over its two week period.
Data Set#1
10, 30, 50, 60, 70, 30, 80,
90, 20, 30, 40, 40, 60, 20
Fall 2010
8
A student counted the number of
players playing basketball in the
Dispersion Tournament each day
over its two week period.
Data Set#2
50, 30, 40, 50, 40, 60, 50,
40, 30, 50, 30, 50, 60, 50
Fall 2010
9
How are the
two data
sets similar
and how are
they
different?
Mean
Fall 2010
Data Set #1
  45
Data Set #2
  45
10
How are the
two data
sets similar
and how are
they
different?
Frequency
Fall 2010
Frequency (x)
X
Data Set #1
Data Set #2
10
1
0
20
2
0
30
3
3
40
2
3
50
1
6
60
2
2
70
1
0
80
1
0
90
1
0
11
Line Plot
x
0
x
x
x
x
x
x
x
x
x
x
10
20
30
40
50
60
70
80
70
80
x
x
x
90
100
Data Set #1
x
x
0
10
20
x
x
x
x
x
x
x
x
x
x
x
30
40
50
60
90
Data Set #2
Fall 2010
12
100
Data Set#1
90
80
Distance from the mean
70
60
60
50
Mean = 45
40 40
30
Fall 2010
30
30
20
10
14
20
90
What if we find the
average of the difference
between each data value
and the mean?
x  
60
i
15
n
80
45
35
70
25
60
15
50
5
Mean = 45
-5 -5
-15
-35
Fall 2010
30
-15
-25
30
-15
-25
30
20
10
40 40
15
20
What if we find the
average of the difference
between each data value
and the mean?
x  
i
n
-35-15+5+15+25-15+35+45-25-15-5-5+15-25
14
Fall 2010
16
=0
90
What if we find the
average of the DISTANCES
from each data value to 70
the mean?
 x 
60
i
15
n
80
45
35
25
60
15
50
5
Mean = 45
5 5
15
35
Fall 2010
30
15
25
30
15
25
30
20
10
40 40
17
20
What if we find the
average of the DISTANCES
from each data value to
the mean?
 x 
i
n
35+15+5+15+25+15+35+45+25+15+5+5+15+25=
14
280
= 20
14
Fall 2010
18
Mean Absolute Deviation
n

i 1
xi  
n
Fall 2010
19
X
|X-μ|
50
5
30
15
40
5
50
5
40
5
60
15
50
5
40
5
30
15
50
5
μ=45
30
15
50
5
Sum = 120
60
15
50
5
Calculate the
Mean
Absolute
Deviation of
Data Set #2
Fall 2010
20
120
Mean Abs. Dev. =
 8.57
14
Fall 2010
21
What if we find the
average of the squares of
the difference from each
data value to the mean?
 x
 
90
80
45
35
70
2
i
60
15
n
25
60
15
50
5
Mean = 45
5 5
15
35
Fall 2010
30
15
25
30
15
25
30
20
10
40 40
22
20
What if we find the
average of the squares of
the difference from each
data value to the mean?
 x
 
2
i
n
352+152+52+152+252+152+352
2
2
2
2
2
2
2
+45 +25 +15 +5 +5 +15 +25
Called the VARIANCE
=7550
7550
= 539.286
14
Fall 2010
23
Standard Deviation of a
Population Data Set
n

Fall 2010
x  
i 1
2
i
n
24
Standard Deviation of
Data Set #1
539.286  23.222
Fall 2010
25
One Standard Deviation on80
either side of the Mean
90
70
 
60
60
50
Mean = 45
40 40
30
 
Fall 2010
30
30
20
10
26
20
Population vs. Sample Standard
Deviation for Data Set #1
Casio
Texas Instruments

This is
if the data set is the population.
Population Standard Deviation
Sample Standard Deviation
Fall 2010
27
“Sample Standard Deviation”
and Bessel Adjustment
n
s
Fall 2010


x

x
 i
i 1
n 1
28
2
Standard Deviation Notation
Recap
µ = mean of a population
σ = population standard deviation
s = sample standard deviation
(estimation of a population
standard deviation based upon a
sample)
Fall 2010
29
How do the 2 data sets compare?
Data Set #1
Fall 2010
Data Set #2
30
Describing the position of data
relative to the mean.
- Can measure in terms of actual
data distance units from the mean.
xi  
- Measure in terms of standard
deviation units from the mean.
xi  

Fall 2010
 z-score  standard measure
31
Why do that?
So we can compare elements
from two different data sets
relative to the position within
their own data set.
Fall 2010
32
Consider this problem…
Amy scored a 31 on the
mathematics portion of her
2009 ACT® (µ=21 σ=5.3).
Stephanie scored a 720 on the
mathematics portion of her
2009 SAT® (µ=515 σ=116.0).
Fall 2010
33
Consider this problem…
Whose achievement was higher
on the mathematics portion of
their national achievement test?
Fall 2010
34
Using z-scores to compare
Amy
Stephanie
1.89 vs. 1.77
What Does This Mean?
Fall 2010
35
By the end of Algebra I, we
have asked and answered the
following BIG questions….
How do we quantify the central
tendency of a data set?
How do we quantify the spread of a
data set?
How do we quantify the relative
position of a data value within a
data set?
Fall 2010
36
So what do Algebra I student need to be able to do?
A.9 DOE ESSENTIAL KNOWLEDGE AND SKILLS
The student will use problem solving, mathematical communication, mathematical reasoning,
connections, and representations to
- Analyze descriptive statistics to determine the implications
for the real-world situations from which the data derive.
- Given data, including data in a real-world context, calculate
and interpret the mean absolute deviation of a data set.
- Given data, including data in a real-world context, calculate
variance and standard deviation of a data set and interpret
the standard deviation.
- Given data, including data in a real-world context, calculate
and interpret z-scores for a data set.
- Explain ways in which standard deviation addresses
dispersion by examining the formula for standard
deviation.
- Compare and contrast mean absolute deviation and
standard deviation in a real-world context.
Fall 2010
37
Let’s gather some data
and calculate some statistics.
Report your height
to the nearest inch.
Fall 2010
38
Fall 2010
freq
1
0
2
1
3
10
4
71
5
137
6
153
7
89
8
26
9
9
10
2
11
2
12
0
13
0
14
0
total
500
39
http://www.ssa.gov/OACT/babynames/
Length of
Boys’
Name
Summary
#letters
Statistics
Mean = 5.746
Population Standard
Deviation = 1.3044
Sample Standard
Deviation=1.3057
Fall 2010
40
Distribution
Length of most popular Boy names in 2009
180
160
153
137
140
Number of babies
120
100
89
80
71
60
40
26
20
10
0
1
1
2
9
2
2
0
0
0
10
11
12
13
14
0
Fall 2010
3
4
5
6
7
8
Number of letters
9
41
Length of most popular Boy names in 2009
180
160
is the probability of selecting a name
What is the probability of selecting What
a
name between 1 and 13 letters? greater than or equal to 3 letters, but less
than or equal to 9 letters?
What is the probability of selecting
a name with exactly 6 letters?
140
137
500
Number of babies
120
153
500
Make up a problem:
100
89
500
80
60
What is the probability of
________________?
71
500
40
20
1
500
0
1
2
10
500
3
26
500
4
5
6
7
8
Number of letters
Fall 2010
9
500
9
2
500
2
500
10
11
12
13
14
Let’s look at a distribution of
heights for a population.
probability
0.1995
0.0648
μ=68”
71”
height
Fall 2010
43
Height as Continuous Data
0.1995
0.0648
μ=68”
Fall 2010
71”
44
Algebra II &
Normal Distributions
Fall 2010
45
5 Characteristics of a
Normal Distribution
1. The mean, median and mode are equal.
2. The graph of a normal distribution is
called a NORMAL CURVE.
3. A normal curve is bell-shaped and
symmetrical about the mean.
4. A normal curve never touches, but gets
closer and closer to the x-axis as it gets
farther from the mean.
5. The total area under the curve is equal to
one.
Fall 2010
46
Examples of Normally
Distributed Data
SAT scores
Height of 10-year-old boys
Weight of cereal in each
24 ounce box
Tread life of tires
Time it takes to tie your shoes
Fall 2010
47
The probability density function for normally
distributed data can be written as a function of
the mean, standard deviation, and data values.
y
( x )2
1
2
2
e
2 2

(x,y)=(data value, relative likelihood for that data value to occur)
Fall 2010
48
Area under curve – up to a data value

P( x   )  0.5
Area under curve – from a data value to
 x1
P( x  x1 )  ???
∞
Area under curve – between two data values.
x1
x2
P( x1  x  x2 )  ??
68-95-99.7 Rule – Empirical Rule
Do not underestimate the power of the quick sketch.
Fall 2010
52
68-95-99.7 Rule – Empirical Rule
A normally distributed data set has
µ=50 and σ=5. What percent of the
data falls between 45 and 55?
A normally distributed data set has
µ=22 and σ=1.5. What would be the
value of an element of this data set
with z-score = 2? z-score = -2?
Fall 2010
53
A machine fills 12 ounce Potato Chip bags.
It places chips in the bags. Not all bags
weigh exactly 12 ounces. The weight of the
chips placed is normally distributed with a
mean of 12.4 ounces and with a standard
deviation 0.2 ounces. If you purchase a bag
filled by this dispenser what is the likelihood
it has less than 12 ounces?
Fall 2010
54
Can you represent this as area
under a normal curve?
Area=0.02275
Fall 2010
12
12.4
55
What fraction of the bags have between
12.1 and 12.5 ounces? Shade the region
that represents that amount.
Fall 2010
56
Standard Normal Distribution
 0
 1
Fall 2010
57
Standard
Normal
Curve
0
Fall 2010
58
Normal Distributions can be
transformed into a Standard Normal
Distribution using the z-score of
corresponding data values.
Example: 2010 SAT math scores
for college bound seniors in VA
Mean=512 Standard
Deviation=110
Fall 2010
College Board State Profile Report – Virginia
(college bound seniors March 2010)
59
Mapping
SAT Score
512 (
)
0
  )
512 –110 (    )
512+(2)110 (   2 )
512 – (2)110 (   2 )
512+110 (
Xi
Fall 2010
Standard score
1
-1
2
-2
xi – μ
z-score =
σ
60
z-scores below the mean
Fall 2010
61
Given the height of a
population is normally
distributed with a
mean height = 68”
with a standard
deviation = 3.2”,
what percent of the
population is less than
61”?
z-score=
61  68
 2.1875
3.2
So, 1.43% of the
population will be less
than to 61”
Round to -2.19 for the
z-table lookup.
Fall 2010
62
z-scores above the mean
Fall 2010
63
Using z-scores to compare
(revisited)
Amy
0.9786
97th percentile
Stephanie
0.9616
96th percentile
Fall 2010
64
So what do Algebra II students need to be able to do?
A2.11 DOE ESSENTIAL KNOWLEDGE AND SKILLS
The student will use problem solving, mathematical communication, mathematical reasoning,
connections, and representations to
- Identify the properties of a normal probability
distribution.
- Describe how the standard deviation and the mean
affect the graph of the normal distribution.
- Compare two sets of normally distributed data using a
standard normal distribution and z-scores.
- Represent probability as area under the curve of a
standard normal probability distribution.
- Use the graphing calculator or a standard normal
probability table to determine probabilities or
percentiles based on z-scores.
Fall 2010
65
Resources
2009 Mathematics SOL and related resources
http://www.doe.virginia.gov/testing/sol/stan
dards_docs/mathematics/review.shtml
Instructional docs including the technical
assistance documents for A.9 and AII.11
http://www.doe.virginia.gov/instruction/high
_school/mathematics/index.shtml
Fall 2010
66