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Probability Models for Distributions
of Discrete Variables
1
Randomly select a college student. Determine x, the
number of credit cards the student has.
x = # of cards
x
0
1
2
3
4
5
p(x)
0.20
0.30
0.20
0.15
0.10
0.05
p(x) = probability of x occurring
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
2
Populations / Samples
A population is a collection of all units of interest.
Example: All college students
A sample is a collection of units drawn from the
population.
Example: Any subcollection of college students.
Probabilities go with populations.
Scientific studies randomly sample from the entire
population.
Each unit in the sample is chosen randomly.
The entire sample is random as well.
3
Probability Models and Populations
For discrete data, a population and a sample are
summarized the same way (for instance, as a
table of values and accompanying relative
frequencies).
A probability distribution (or model) for a
discrete variable is a description of values, with
each value accompanied by a probability.
4
Probability Models and Populations
Definitions of Probability
2. the probability of an event is the long term
(technically forever) relative frequency of occurrence of
the event, when the experiment is performed repeatedly
under identical starting conditions.
3. The probability of an event is the relative frequency
of units in the population for which the event applies.
To aggregate these meanings:
The probability associated with an event is its relative
frequency of occurrence over all possible ways the
phenomena can take place.
5
Probability Models
“All models are wrong. Some are useful.”
George Box
-industrial statistician
6
A probability distribution for a discrete variable is
tabulated with a set of values, x and probabilities, p(x).
x
p(x)
0
0.20
1
0.30
2
0.20
3
0.15
4
0.10
Probabilities
Must be
nonnegative.
0.35
0.30
0.25
0.20
0.15
5
0.05
0.10
0.05
0.00
7
0
1
2
3
4
5
A probability distribution for a discrete variable is
tabulated with a set of values, x and probabilities, p(x).
x
p(x)
0
0.20
1
0.30
2
0.20
3
0.15
4
0.10
5
0.05
SUM
1.00
Probabilities
Must be
nonnegative.
Must sum to 1.
Within rounding
error.
8
The mean  of a probability distribution is the
mean value observed for all possible outcomes
of the phenomena.
9
Consider idealized data sets
x
0
1
2
3
4
5
p(x)
0.20
0.30
0.20
0.15
0.10
0.05
20 0s
30 1s
20 2s
15 3s
10 4s
5 5s
10
Idealized data set n = 100
0
0
1
1
2
3
3
5
0
0
1
1
2
3
4
5
0
0
1
1
2
3
4
0
0
1
1
2
3
4
0
0
1
1
2
3
4
0
0
1
1
2
3
4
0
1
1
1
2
3
4
Mean = 1.80
0
1
1
1
2
3
4
0
1
1
2
2
3
4
0
1
1
2
2
3
4
0
1
1
2
2
3
4
0
1
1
2
2
3
5
SD = 1.44
0
1
1
2
2
3
5
0
1
1
2
2
3
5
11
Consider idealized data sets
x
0
1
2
3
4
5
p(x)
0.20
0.30
0.20
0.15
0.10
0.05
200 0s
300 1s
200 2s
150 3s
100 4s
50 5s
12
Idealized data set n = 1000
0 0 0 0
1 1 1 1
2 2 2 2
3 3 3 3
4 4 …4
5…5
0 0 0 … 0
1 1 1 1 1 1…1
2 2 … 2
…3
Mean = 1.80
(200)
(300)
(200)
(150)
(100)
(50)
SD = 1.44
13
Values for the mean and standard deviation don’t
depend on the number of data values; they
depend instead on the relative location of the
data values – they depend on the distribution in
relative frequency terms.
14
The mean  of a probability distribution is the
mean value observed for all possible outcomes
of the phenomena.
Greek letter “myou”
Formula:
SUM symbol
   x  p x 
 is synonymous with “population mean”
15
   x  p x
Multiply each value by its probability
Sum the products
x
0
1
2
3
4
5
p(x)
0.20
0.30
0.20
0.15
0.10
0.05
1.00
x  p(x)
0  0.20 = 0.00
1  0.30 = 0.30
2  0.20 = 0.40
3  0.15 = 0.45
4  0.10 = 0.40
5  0.05 = 0.25
1.80
Mean  = 1.80
16
The standard deviation  of a probability
distribution is the standard deviation of the
values observed for all possible outcomes of the
phenomena.
Formula:

x   
2
 p x 
Greek letter “sigma”
 denotes “population standard deviation”
17
First obtain the variance.
x
0
p(x)
0.20
1
2
3
4
5
0.30
0.20
0.15
0.10
0.05
    x     p x 
2
2
(0 – 1.8)2  0.20 = 0.648
(1 – 1.8)2  0.30 = 0.192
(2 – 1.8)2  0.20 = 0.008
(3 – 1.8)2  0.15 = 0.216
(4 – 1.8)2  0.10 = 0.484
(5 – 1.8)2  0.05 = 0.512
 2 = 2.060
(take square root to obtain)
 = 1.44
18
0
0
1
1
2
3
3
5
0
0
1
1
2
3
4
5
0
0
1
1
2
3
4
0
0
1
1
2
3
4
0
0
1
1
2
3
4
0
0
1
1
2
3
4
0
1
1
1
2
3
4
0
1
1
1
2
3
4
0
1
1
2
2
3
4
0
1
1
2
2
3
4
0
1
1
2
2
3
4
0
1
1
2
2
3
5
0
1
1
2
2
3
5
0
1
1
2
2
3
5
Mean = 1.80
SD = 1.44
Mean – SD = 0.56
Mean + SD = 3.24
65 / 100 = 65%
19
x
0
1
2
3
4
5
p(x)
0.20
0.30
0.20
0.15
0.10
0.05
Mean = 1.80
SD = 1.44
Mean – SD = 0.56
Mean + SD = 3.24
0.30 + 0.20 + 0.15 = 0.65
20
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
0.30
0.25
0.20
Probability
x
1
2
x = # children in randomly selected
college student’s family.
0.15
0.10
0.05
0.00
1
2
3
4
5
6
# of Children
7
8
9
10
21
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
x = # children in randomly selected
college student’s family.
0.2194 = 21.94% of all college
students come from a 1 child family.
22
Guess at mean? Above 2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
(right skew  mean > mode).
0.30
0.25
0.20
Probability
x
1
2
0.15
0.10
0.05
0.00
1
2
3
4
5
6
# of Children
7
8
9
10
23
x
1
2
p(x)
0.2194
0.2806
x  p(x)
1(0.2194) = 0.2194
2(0.2806) = 0.5612
3
4
5
0.2329
0.1442
0.0736
3(0.2329) = 0.6987
: = 0.5768
0.3680
6
7
8
9
0.0317
0.0124
0.0043
0.0005
0.1902
0.0868
0.0344
0.0045
10
55
0.0003
1.0000
0.0030
Mean:  = 2.7430
To determine the
mean, multiply
values by
probabilities,
xp(x)
and sum these.
55/10 = 5.50 is
not the mean
1.000/10 = 0.10
is not the mean
24
x
p(x)
1 0.2194
(x – )2  p(x)
(1 – 2.743)2  0.2194 = 0.6665
2 0.2806
3 0.2329
4 0.1442
(2 – 2.743)2  0.2806 = 0.1549
(3 – 2.743)2  0.2329 = 0.0154
: = 0.2278
5
6
7
0.0736
0.0317
0.0124
0.3749
0.3363
0.2247
8
9
10
0.0043
0.0005
0.0003
0.1188
0.0196
0.0158
55 1.0000
Variance:  2 = 2.1548
To determine the
variance,
multiply squared
deviations from
the mean by
probabilities,
(x – )2p(x)
and sum these.
25
The standard deviation is the square root of the
variance.
  2.1548  1.468
Examining the data set consisting of # of children in
the family recorded for all students: The mean is
2.743; the standard deviation is 1.468.
26
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
Determine the probability a student is
from a family with more than 5
siblings.
P(x > 5)
27
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
Determine the probability a student is
from a family with more than 5
siblings.
P(x > 5)
28
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
Determine the probability a student is
from a family with more than 5
siblings.
P(x > 5)
29
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
Determine the probability a student is
from a family with more than 5
siblings.
P(x > 5)
30
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
Determine the probability a student is
from a family with more than 5
siblings.
P(x > 5)
31
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
Determine the probability a student is
from a family with more than 5
siblings.
P(x > 5)
= 0.0317
+ 0.0124
+ 0.0043
+ 0.0005
+ 0.0003
32
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
Determine the probability a student is
from a family with more than 5
siblings.
P(x > 5)
= 0.0317
+ 0.0124
+ 0.0043
+ 0.0005
+ 0.0003
= 0.0492
33
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
Determine the probability a student is
from a family with more than 5
siblings.
P(x > 5)
= 0.0492
4.92% of all college students come
from families with more than 5
children (they have 4 or more
brothers and sisters).
34
Determine the probability a student is
from a family with at most 3 siblings.
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
+ 0.2806
6
7
8
0.0317
0.0124
0.0043
= 0.7329
9
10
0.0005
0.0003
P(x  3)
= 0.2194
+ 0.2329
35
x
1
2
p( x)
0.2194
0.2806
3
4
5
0.2329
0.1442
0.0736
6
7
8
0.0317
0.0124
0.0043
9
10
0.0005
0.0003
Determine the probability a student is
from a family with at least 7 siblings.
P(x  7)
= 0.0124
+ 0.0043
+ 0.0005
+ 0.0003
= 0.0175
Good idea: Take the reciprocal of a
small probability…
1/.0175 = 57.1  1 in 57 students
36
Determine the probability a student is
from a family with fewer than 5
siblings.
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
+ 0.2806
7
8
9
10
0.0124
0.0043
0.0005
0.0003
+ 0.1442
P(x < 5)
= 0.2194
+ 0.2329
= 0.8771
37
at most 3
at least 7


less than or equal to 3
greater than or equal to 7


no more than 3
no fewer/less than 7


x3
x7
38
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
7
8
9
10
0.0124
0.0043
0.0005
0.0003
Determine the probability a student’s
number of siblings falls within 1
standard deviation of the mean.
Guess?
0.68
39
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
7
8
9
10
0.0124
0.0043
0.0005
0.0003
Determine the probability a student’s
number of siblings falls within 1
standard deviation of the mean.
Mean
 = 2.743
SD
 = 1.468
1 SD below the mean
2.743 – 1.468 = 1.275
1 SD above the mean
2.743 + 1.468 = 4.211
40
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
7
8
9
10
0.0124
0.0043
0.0005
0.0003
Determine the probability a student’s
number of siblings falls within 1
standard deviation of the mean.
1 SD below the mean = 1.275
1 SD above the mean = 4.211
Values are within 1 SD of the mean if
they are between these.
41
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
7
8
9
10
0.0124
0.0043
0.0005
0.0003
Determine the probability a student’s
number of siblings falls within 1
standard deviation of the mean.
1 SD below the mean = 1.275
1 SD above the mean = 4.211
Values are within 1 SD of the mean if
they are between these.
42
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
7
8
9
10
0.0124
0.0043
0.0005
0.0003
Determine the probability a student’s
number of siblings falls within 1
standard deviation of the mean.
1 SD below the mean = 1.275
1 SD above the mean = 4.211
Values are within 1 SD of the mean if
they are between these.
The probability of being between
these:
0.2806 + 0.2329 + 0.1442 = 0.6577
43
Determine the probability a student’s
number of siblings falls within 2
standard deviations of the mean.
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
2 SD below the mean
7
8
9
10
0.0124
0.0043
0.0005
0.0003
2 SD above the mean
Guess?
0.95
1.275 – 1.468 = -0.193
4.211+ 1.468 = 5.679
Between -0.193 and 5.679.
44
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
7
8
9
10
0.0124
0.0043
0.0005
0.0003
Determine the probability a student’s
number of siblings falls within 2
standard deviations of the mean.
Between -0.193 and 5.679.
(Equivalent to 5 or fewer.)
45
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
7
8
9
10
0.0124
0.0043
0.0005
0.0003
Determine the probability a student’s
number of siblings falls within 2
standard deviations of the mean.
Between -0.193 and 5.679.
(Equivalent to 5 or fewer.)
We know an outcome more than 5
has probability 0.0492.
46
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
7
8
9
10
0.0124
0.0043
0.0005
0.0003
Determine the probability a student’s
number of siblings falls within 2
standard deviations of the mean.
Between -0.193 and 5.679.
(Equivalent to 5 or fewer.)
We know an outcome more than 5
has probability 0.0492.
The probability of an outcome at
most 5 is 1 – 0.0492 = 0.9508.
47
x
1
2
p( x)
0.2194
0.2806
3
0.2329
4
5
6
0.1442
0.0736
0.0317
7
8
9
10
0.0124
0.0043
0.0005
0.0003
Determine the probability a student’s
number of siblings falls within 2
standard deviations of the mean.
Between -0.193 and 5.679.
0.9508.
48
Pollutant Particles in
Streamwater
A company monitors pollutants downstream of
discharge into a stream.
Data were collected on 200 days from a point 1
mile downstream of the plant on Stream A.
Data were collected on 100 days from a point 1
miles downstream of the plant on Stream B.
49
70
How do means
compare?
Frequency
60
50
40
30
20
10
0
0
1
2
3
4
5
6
(What are the
means?)
Stream B
How do SDs
compare?
70
Frequency
60
(What are the
SDs?)
50
40
30
20
10
0
0
1
2
3
4
5
6
Stream A
50
70
Similar Means.
Frequency
60
50
40
30
20
10
0
0
1
2
3
4
5
6
Similar
Standard
Deviations.
Stream B
(Similar
everything
except ns.)
70
Frequency
60
50
40
30
20
10
0
0
1
2
3
4
5
6
Stream A
51
35
30
Percent
25
20
15
10
5
0
0
1
2
3
4
5
6
4
5
6
Stream B
35
30
Percent
25
20
15
10
5
0
0
1
2
3
Stream A
52
35
Stream B
30
Percent
25
Mean = 1.775
20
15
10
SD = 1.242
5
0
0
1
2
3
4
5
6
Stream B
35
30
Stream A
Percent
25
20
15
Mean = 1.770
10
5
0
0
1
2
3
4
5
6
SD = 1.340
Stream A
53
Here is the probability distribution for the
number of diners seated at a table in a small café.
x
p(x)
1
0.10
2
0.20
a) Fill in the blank
3 ____
4
0.40
54
Here is the probability distribution for the
number of diners seated at a table in a small café.
x
p(x)
1
0.10
2
0.20
3
0.30
4
0.40
a) Fill in the blank
55
Here is the probability distribution for the
number of diners seated at a table in a small café.
x
p(x)
1
0.10
2
0.20
3
0.30
4
0.40
b) Determine the
mean 
Start by
computing xp(x)
for each row.
56
Here is the probability distribution for the
number of diners seated at a table in a small café.
x
p(x)
1
0.10
2
0.20
3
0.30
4
0.40
xp(x)
b) Determine the
mean 
Start by
computing xp(x)
for each row.
57
Here is the probability distribution for the
number of diners seated at a table in a small café.
x
p(x)
xp(x)
1
0.10 10.10 = 0.10
2
0.20
3
0.30
4
0.40
b) Determine the
mean 
Start by
computing xp(x)
for each row.
58
Here is the probability distribution for the
number of diners seated at a table in a small café.
x
p(x)
xp(x)
1
0.10 10.10 = 0.10
2
0.20 20.20 = 0.40
3
0.30
4
0.40
b) Determine the
mean 
Start by
computing xp(x)
for each row.
59
Here is the probability distribution for the
number of diners seated at a table in a small café.
x
p(x)
xp(x)
1
0.10 10.10 = 0.10
2
0.20 20.20 = 0.40
3
0.30 30.30 = 0.90
4
0.40 40.40 = 1.60
b) Determine the
mean 
Start by
computing xp(x)
for each row.
60
Here is the probability distribution for the
number of diners seated at a table in a small café.
x
p(x)
xp(x)
1
0.10 10.10 = 0.10
2
0.20 20.20 = 0.40
3
0.30 30.30 = 0.90
4
0.40 40.40 = 1.60
b) Determine the
mean 
Sum these.
61
Here is the probability distribution for the
number of diners seated at a table in a small café.
x
p(x)
xp(x)
1
0.10
10.10 = 0.10
2
0.20
20.20 = 0.40
3
0.30
30.30 = 0.90
4
0.40
40.40 = 1.60
b) Determine the
mean 
Sum these.
 = 3.00
62
Here is the probability distribution for the
number of diners seated at a table in a small café.
b) Determine the
standard
deviation 
x
p(x)
1
0.10
2
0.20
3
0.30
Start by
computing
0.40
( x –  ) 2 p(x)
4
for each row.
63
Here is the probability distribution for the
number of diners seated at a table in a small café.
b) Determine the
standard
deviation 
x
p(x)
1
0.10
2
0.20
3
0.30
Start by
computing
0.40
( x –  )2 p(x)
4
for each row.
=3
64
Here is the probability distribution for the
number of diners seated at a table in a small café.
(x–
3)2
x
p(x)
1
0.10
2
0.20
3
0.30
Start by
computing
0.40
( x – 3)2 p(x)
4
p(x)
b) Determine the
standard
deviation 
for each row.
=3
65
Here is the probability distribution for the
number of diners seated at a table in a small café.
(x–
3)2
x
p(x)
1
0.10 (1 – 3)20.10 = 0.40
2
0.20
3
0.30
Start by
computing
0.40
( x – 3 ) 2 p(x)
4
p(x)
b) Determine the
standard
deviation 
for each row.
=3
66
Here is the probability distribution for the
number of diners seated at a table in a small café.
(x–
3)2
x
p(x)
1
0.10 (1 – 3)20.10 = 0.40
2
0.20 (2 – 3)20.20 = 0.20
3
0.30
Start by
computing
0.40
( x – 3 ) 2 p(x)
4
p(x)
b) Determine the
standard
deviation 
for each row.
=3
67
Here is the probability distribution for the
number of diners seated at a table in a small café.
(x–
3)2
x
p(x)
1
0.10 (1 – 3)20.10 = 0.40
2
0.20 (2 – 3)20.20 = 0.20
3
0.30 (3 – 3)20.20 = 0.00
4
0.40 (4 –
3)20.20
p(x)
= 0.40
b) Determine the
standard
deviation 
Start by
computing
(x – 3 ) 2 p(x)
for each row.
=3
68
Here is the probability distribution for the
number of diners seated at a table in a small café.
(x–
3)2
x
p(x)
p(x)
1
0.10 (1 – 3)20.10 = 0.40
2
0.20 (2 – 3)20.20 = 0.20
3
0.30 (3 – 3)20.20 = 0.00
4
0.40 (4 – 3)20.20 = 0.40
b) Determine the
standard
deviation 
Sum these
69
Here is the probability distribution for the
number of diners seated at a table in a small café.
(x–
3)2
p(x)
b) Determine the
standard
deviation 
x
p(x)
1
0.10 (1 – 3)20.10 = 0.40
2
0.20 (2 – 3)20.20 = 0.20
Sum these
3
0.30 (3 – 3)20.30 = 0.00
Variance = 1.00
4
0.40 (4 – 3)20.40 = 0.40
SD:  = 1.00
70
[Optional] Application
This framework makes it possible to obtain fairly
good approximations to means and standard
deviations from a histogram of continuous data.
71
Example
Here are waiting times between student arrivals
in a class. There are 21 students (20 waits).
Approximate the mean
and median. How do they
compare?
10
Frequency
8
6
4
2
0
0
10
20
30
Waiting Time
40
50
72
Example: Mean
For each class, determine its frequency and
corresponding midpoint.
Frequency = 10
10
Frequency
8
Midpoint = 5
6
4
2
0
0
10
20
30
Waiting Time
40
50
73
Example: Mean
Tabulate frequencies and midpoints.
Midpoint Frequency
5
10
74
Example: Mean
Tabulate frequencies and midpoints.
Midpoint Frequency
5
10
15
5
25
3
35
1
45
1
Total
20
75
Example: Mean
Obtain relative frequencies.
Midpoint Frequency
5
10
15
5
25
3
35
1
45
1
Total
20
Relative Frequency
10/20
= 0.50
76
Example: Mean
Obtain relative frequencies.
Midpoint Frequency
Relative Frequency
5
10
10/20
= 0.50
15
5
5/20
= 0.25
25
3
3/20
= 0.15
35
1
1/20
= 0.05
45
1
1/20
= 0.05
Total
20
1.00
77
Example: Mean
Proceed with the formula
Mean   x  p x
Midpoint
Rel Freq
Product
5
0.50
5(0.50) = 2.50
15
0.25
25
0.15
35
0.05
45
0.05
Total
20
78
Example: Mean
Proceed as a discrete population distribution.
Midpoint
Rel Freq
Product
5
0.50
5(0.50) = 2.50
15
0.25
15(0.25) = 3.75
25
0.15
25(0.15) = 3.75
35
0.05
35(0.05) = 1.75
45
0.05
45(0.05) = 2.25
Total
20
Mean
79
Example: Mean
Proceed as a discrete population distribution.
Midpoint
Rel Freq
Product
5
0.50
5(0.50) = 2.50
15
0.25
15(0.25) = 3.75
25
0.15
25(0.15) = 3.75
35
0.05
35(0.05) = 1.75
45
0.05
45(0.05) = 2.25
Total
20
14.00
Mean
 14.00
80
Example: Median
Find the value with 50% below and 50% above.
10
Frequency
8
6
4
2
0
0
10
20
30
Waiting Time
40
50
81
Example: Median
Obtain relative frequencies.
Midpoint
Rel Freq
5
0.50
15
0.20
25
0.15
35
0.05
45
0.05
Total
1.00
82
Example: Median
Find the value with 50% below and 50% above.
10 of 20 = 50% below 10
Median  10.00
10
Frequency
8
Mean  14.00
6
Range  44
4
S.D.  11
2
0
0
10
20
30
Waiting Time
40
50
83
Example: Data / Exact Values
1.3
3.6
11.2
33.6
1.9
3.7
13.5
43.5
1.9
5.9
15.9
Approximations:
2.5
9.7
21.4
2.6
10.4
27.5
3.0
10.6
29.8
Actual Values:
Median  10.0.05
Median =
Mean  14.0
Mean =
Range  44
Range =
SD  11
SD =
84
Example: Data / Exact Values
1.3
3.6
11.2
33.6
1.9
3.7
13.5
43.5
1.9
5.9
15.9
Approximations:
2.5
9.7
21.4
2.6
10.4
27.5
3.0
10.6
29.8
Actual Values:
Median  10.0.05
Median = 10.05
Mean  14.0
Mean = 12.68
Range  44
Range = 42.2
SD  11
SD = 12.31
85