### Chapter 1: Introduction to Statistics

```COURSE: JUST 3900
INTRODUCTORY STATISTICS
FOR CRIMINAL JUSTICE
Chapter 3: Central Tendency
Peer Tutor Slides
Instructor:
Mr. Ethan W. Cooper, Lead Tutor
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Key Terms: Don’t Forget
Notecards
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Central Tendency (p. 73)
Mean (p. 74)
Weighted Mean (p. 77)
Median (p. 83)
Mode (p. 87)
Unimodal (p. 88)
Bimodal (p. 88)
Multimodal (p. 88)
HINT: Review distribution shapes from Ch. 2!
Key Formulas
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Population Mean:  =
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Sample Mean:  =


Overall (Weighted) Mean:  =
1 + 2
1 +2
Mean

Question 1: Find the mean for the sample of n=5 scores:
1, 8, 7, 5, 9
Mean

 M=ΣX
n
 M = 1+8+7+5+9
5
 M = 30
5
 M=6
Mean

Question 2: A sample of n=6 scores has a mean of M=8.
What is the value of Σ X for this sample?
Mean

 M=ΣX
n
 8=ΣX

6
Σ X = 48
Mean

Question 3: One sample has n=5 scores with a mean of
M=4. A second sample has n=3 scores with a mean of
M=10. If the two samples are combined, what is the
mean for the combined sample?
Mean

1 + 2
1 +2
20+30
5+3
50
8
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=
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=
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=
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= 6.25
Remember:
1 is calculated by 1 ∗ 1
2 is calculated by 2 ∗ 2
Mean
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Question 4: A sample of n=6 scores has a mean of M=40.
One new score is added to the sample and the new mean is
found to be M=35. What can you conclude about the value of
the new score?
a)
b)
It must be greater than 40.
It must be less than 40.
Mean

 B) It must be less than 40. A score higher than 40
would have increased the mean.
Mean

Question 5: Find the values for n, Σ X, and M for the
following sample:
X
f
5
1
4
2
3
3
2
5
1
1
Mean

 n = 1+2+3+5+1
 n = 12
 Σ X = 5+4+4+3+3+3+2+2+2+2+2+1
 Σ X = 33
 M = 33
12
 M = 2.75
Mean

Question 6: Adding a new score to a distribution always
changes the mean. True or False?
Mean

 False. If the score is equal to the mean, it does not
change the mean.
Mean

Question 7: A population has a mean of μ = 40.
a)
b)
If 5 points were added to every score, what would be the value
for the new mean?
If every score were multiplied by 3, what would be the value of
the new mean?
Mean

a) The new mean would be 45. When a constant is
the mean.
b) The new mean would be 120. When every score is
multiplied (or divided) by a constant, the mean
changes in the same way.
Mean

Question 8: What is the mean of the following
population?
Mean


μ=7
Mean

Question 9: Using the scores from question 8, fill in the
following table.
Mean

4
Below
1
Below
1
Below
2
4
Above
Above
Median

Question 10: Find the median for each distribution of
scores:
a)
b)
3, 4, 6, 7, 9, 10, 11
8, 10, 11, 12, 14, 15
Median

a)
b)
The median is X = 7
The median is X = 11.5
Median

Question 11:The following is a distribution of
measurements for a continuous variable. Find the
precise median that divides the distribution exactly in
half.
Median

2/3
6
1/3
5
Count 8 boxes
1
2
3
4
5
7
4
2
6
3
1
Median = 6.83
Median

Question 11 Explanation:

To find the precise median, we first observe that the distribution
contains n = 16 scores. The median is the point with exactly 8
boxes on each side. Starting at the left-hand side and moving up
the scale of measurement, we accumulate a total of 7 boxes
when we reach a value of 6.5. We need 1 more box to reach our
goal of 8 boxes (50%), but the next interval contains 3 boxes.
The solution is to take a fraction of each box so that the fractions
combine to give you one box. The fraction is determined by the
number of boxes needed to reach 50% (numerator) and the
number that exists in the interval (denominator).
Median

Question 11 Explanation:

For this example, we needed 1 out of the 3 boxes in the interval,
so the fraction is 1/3. The median is the point located exactly
one-third of the way through the interval. The interval for X = 7
extends from 6.5 to 7.5. The interval width is one point, so onethird of the interval corresponds to approximately 0.33 points.
Starting at the bottom of the interval and moving up 0.33 points
produces a value of 6.50 + 0.33 = 6.83. This is the median, with
exactly 50% of the distribution (8 boxes) on each side.
Mode

Question 12: What is the mode(s) of the following
distribution? Is the distribution unimodal or bimodal?
Mode

The modes are 2 and 8
The distribution is
bimodal.
Note: While this is a bimodal distribution,
both modes have the same frequency.
Thus, there is no “minor” or “major” mode.
Selecting a Measure of Central
Tendency
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Question 13: Which measure of central tendency is most
affected if one extremely large score is added to a
distribution? (mean, median, mode)
Selecting a Measure of Central
Tendency


Mean
Selecting a Measure of Central
Tendency

Question 14: Why is it usually inappropriate to compute
a mean for scores measured on an ordinal scale?
Selecting a Measure of Central
Tendency

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The definition of the mean is based on distances (the mean
balances the distances) and ordinal scales do no measure
distance.
Selecting a Measure of Central
Tendency
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Question 15: In a perfectly symmetrical distribution, the
mean, the median, and the mode will all have the same
value. (True or False)
Selecting a Measure of Central
Tendency


False, if the distribution is bimodal.
Selecting a Measure of Central
Tendency

Question 16: A distribution with a mean of 70 and a
median of 75 is probably positively skewed. (True or
False)
Selecting a Measure of Central
Tendency


False. The mean is displaced toward the tail on the left-hand
side.
Central Tendency and
Distribution Shape

Graphs make life so much easier!
Symmetrical
Distributions
Negatively Skewed
Distribution
Positively Skewed
Distribution
Notice how the means
Note: Median
usually falls
between mean
and mode.
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Interpolation
Real Limits
Median for Continuous Variables
Frequency Distribution
Cumulative Distributions
Weighted Mean
FAQs

How do I find the median for a continuous variable?
FAQs

Step 1: Count the total number of boxes.
50% of 16
is 8.
1


2
3
4
14
16 boxes
10 13
5
7
9 12 16
6
8 11 15
Step 2: How many boxes are necessary to reach 50%?
Step 3: Count the necessary number of boxes starting
from the left (in this case 8).
FAQs
Uh-oh!
What now?
1/3
2/3
7
1
2
3
4
5
6
6.5
7.5
• Step 4: We need one more box to reach 8, but there are
three boxes over the interval spanning 6.5 – 7.5. Thus,
we need 1/3 of each box to reach 50%.
FAQs
Median = 6.83

Step 5: We stopped counting when we reached seven
boxes at the interval X = 6, which has an upper real limit
of 6.5. We want 1/3 of the boxes in the next interval, so
we add 6.5 + (1/3) = 6.83.
FAQs

FAQs

How do I calculate the weighted mean?

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Compute the weighted mean for two sets of scores. The first set
consists of n = 12 scores and has a mean of M = 6. The second
set consists of n = 8 scores and has a mean of M = 7.
Step 1: Find  for both samples.

1 = 12 ∗ 6 = 72

2 = 8 ∗ 7 = 56
Plug numbers into the formula  =
72+56
12+8
128
20
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=
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=
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= 6.4
1 + 2
1 +2
```