Math 11 – Statistics
Chapter 3 – Measures of Central Tendency for ungrouped data
3 different measures
Mean (what we normally think of as average)
x
x
:
population  
sample x 
N
n
n 1
Median – the middle term
(This gives the position of middle term)
2
Mode: The term that occurs the most often.
Unimodal – a single value occurs most times
Bimodal – two terms occur the maximum number of times
Multimodal – many values occur the maximum number of times
Shapes of distributions and the affect on position of mean, median and mode
Median
Mean
Median
Mode
Mode
Median
Mean
Mode
Measures of Dispersion for ungrouped data
The Range  Largest Value  Smallest Value
Mean
Standard Deviation – the positive square root of the variance
Variance -
x   or
population:
x  
 
N
Sample:
s2 
2
2
  x  x 2
n 1
x  x is called the deviation of the x value from the mean.
If we add all the deviations from the mean we get zero.
 x     0
 x  x   0
Therefore, we standardize by squaring and then taking the square root.
  2
s  s2
There are shortcut formulas for the variance – you can use whichever is most convenient for the problem.
x  
 
N
2
s2 
x  x 
n 1
2
2


x
2
 x

N
The variance and the standard deviation are both
positive.
N
x
2

 x
n 1
n
2
2
 2 and s 2 because they are squared
 and s because they are the positive square roots
The numerical measure ( mean, median, mode, variance, or standard deviation ) of a population is
called a population parameter or just a parameter. (  ,  2 ,  )
The numerical measure of a sample is called a sample statistic or just a statistic ( x , s 2 , s )
3.3 Measures of Central Tendency for Grouped Data
Mean, Variance and Standard Deviation of Grouped Data
When data is already grouped, we don’t know the individual values of the variables.
ex The following table gives the grouped data on the weights of all 100 babies born at a hospital in
2005. (Notice, this is the total population)
Weight (pounds)
3 to less than 5
5 to less than 7
7 to less than 9
9 to less than 11
11 to less than 13
# of babies
5
30
40
20
5
Notice that this is continuous
data, therefore the class
limits are the same as the
class boundaries.
To calculate the mean we will have to assume that observations in each class are evenly distributed
throughout the class. Then the average value of the class would be the midpoint.
So, find the midpoint of each class and multiply it by the frequency of the class (i.e. the number of
observations in that class) This gives us the approximate total of that class. Next – add the totals
together to get the sum of all values in all classes.
Weight (pounds)
3 to less than 5
5 to less than 7
7 to less than 9
9 to less than 11
11 to less than 13
The mean now becomes  
 mf
Frequency (# of
babies)
f
5
30
40
20
5
N  100 .
 mf
N
Midpoint of class
m
4
6
8
10
12
for a population and x 
 mf
n
780
 7.8 pounds
N
100
The mean weight of babies born in 2005 was about 7.8 pounds.


Variance and Standard Deviation for Grouped Data

s
2
2
 f (m   )

2
N
 f m  x 

n 1
or shortcut  
2
2
or shortcut s 2 
m
m
2
  mf 
f
N
N
2
  mf 
f
n 1
n
2
2
Approx total of
values in class ( mf )
20
180
320
200
60
 mf  780
.
Back to babies
We need to find
 mf
weight
3 to less than 5
5 to less than 7
7 to less than 9
9 to less than 11
11 to less than 13
m
2
and
f
f
5
30
40
20
5
 f  100
m
4
6
8
10
12
mf
20
180
320
200
60
 mf  780
m2f
80
1080
2560
2000
720
2
 m f  6440
Substitute into the formula
2 
m
2
  mf 
f
N
N
2


6440  780
100  6440  6084  356  3.56 This is the variance.

100
100
100
  3.56  1.88679  1.89 pounds
When using single valued classes, the value of the class will be used as m.
ex
This table gives the number of errors committed by a college baseball team in all 45 games played
in the 2005-2006 season. (Note – this is a population)
Find the mean, variance and standard deviation
# of errors
0
1
2
3
4
5
Mean:
Variance:
# of games (f)
11
14
9
7
3
1
 f  45

mf
N
2 
Standard Deviation:

mf
0
14
18
21
12
5
 mf  70
70
 1.56 errors per game.
45
 m2 f 
N
  mf 
N
 2
186  70
45  77.11  1.71

45
45
  1.71  1.309
m2f
0
14
36
63
48
25
2
 m f  186
3.4
How do we use Standard Deviation?
When we have a symmetric bell-shaped curve (i.e. unimodal) then we can use The Empirical Rule.
1. 68% of the observations lie within one standard deviation of the mean.
2. 95% of the observations lie within two standard deviations of the mean.
3. 99.7% of the observations lie within three standard deviations of the mean
95%
68%
  3   2
 

99.7%
     2   3
3.5
Measures of Position
Position means where in the data set a specific value falls.
Quartiles are summary measures that divide the data set into 4 equal parts.
(The data must be ranked first.)
25%
25%
25%
25%
1st Quartile
2nd Quartile
3rd Quartile
Q1
Q2
Q3
Q2 = the median
IQR – Interquartile Range  Q3  Q1
Let’s look at the kitten data again. The data must be ranked first
37 38 40 41 44 45 49 51 52 52 54 56 57 57 58 60 63 63 65 69 72 76 77 84 91
25  1
 13
2
entry is 57
Q2 = the median which is
So the 13th
To find Q1 – find the median of the lower half:
Q1 
45  49 94

 47
2
2
Q3 
65  69 134

 67
2
2
12  1
 6.5 So halfway between the 6th and 7th items
2
The interquartile range is Q3  Q1  67  47  20
5 number summary: min, Q1 , Q2 , Q3 , max
We can see that 50% of the kittens weighed between 47 and 67 grams
25% weighed less than 47 grams, and 25% weighed more than 67 grams.
Where is 60 grams located?
Between 50% and 75% or in the 3rd quartile.
Box and Whisker Plots
Definition: A plot that shows the center, the spread, and the skewness of a data set.
The following 5 steps are used to construct a box-and-whisker plot.
Back to kitten weights
1. Find the median = 57
Q1  47
IQR  20
Q3  67
2. Find the points that are 1.5  IQR below Q1 , and 1.5  IQR above Q3 .
These are called the lower inner fence and the upper inner fence respectively.
For the kittens 1.5  20  30
Lower Inner Fence ( f L )  Q1  30  47  30  17
Upper Inner Fence ( f U )  Q3  30  67  30  97
kittens – 37
kittens – 91
3. Determine:the smallest value within the two inner fences
The largest value within the two inner fences
4. Draw a horizontal line and mark observation values on it so that all data in the set are covered.
5. Draw a box that has Q1 as one side and Q3 as the other. Draw a vertical line at Q2.
whisker
whisker
This shows the kitten data is skewed to the right.
(the right whisker is longer)
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35
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40
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45
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50
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55
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60
|
65
|
70
|
75
|
80
|
85
|
90
Make a mark at the lowest value inside the lower inner fence. Make a mark at the largest value inside the
upper inner fence. Connect these to the box. These are the whiskers
Any values outside the two inner fences would be indicated by asterisks. These would be outliers.
For the kittens there are none outside the fences. So no outliers for the kittens.
Another example:
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95
Problem 30 page 114
This is the number of runners left on bases by each of 30 Major league teams.
15.5
8th
3
4
4
5
5
5
5
5
6
6
6
6
6
6
6
7
7
8
8
8
8
8th
8
Place of Q2 
8
9
9
10
10
11
13
18
30  1
 15.5 Remember this is the place of the value.
2
Q2 
67
 6.5
2
Place of Q1 
15  1
8
2
Therefore Q2  5
Place of Q1 
15  1
8
2
Therefore Q2  5 , Q3  8
IQR  Q3  Q1  8  5  3
1.5  IQR  1.5  3  4.5
Lower inner fence:
f L  5  4.5  .5
Upper inner fence:
fU  8  4.5  12.5
Notice, there are two values not within the fences: 13 and 18. These are outliers and we represent
them with asterisks
*
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0
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2
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4
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6
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8
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10
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12
*
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14
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16
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18
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20
We can see from the box and whisker plot that the data is skewed right with two outliers.