COURSE: JUST 3900INTRODUCTORY STATISTICS

FOR CRIMINAL JUSTICEChapter 3: Central Tendency

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Key Terms: Don’t Forget Notecards

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

Population Mean: Sample Mean: Overall (Weighted) Mean:

Mean

Question 1: Find the mean for the sample of n=5 scores: 1, 8, 7, 5, 9

Mean

Question 1 Answer: M = Σ X

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?

Question 2 Answer: M = Σ X

n 8 = Σ X

6 Σ X = 48

Mean

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

Question 3 Answer:

Remember: is calculated by is calculated by

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) It must be greater than 40.

b) It must be less than 40.

Mean

Mean

Question 4 Answer: 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

Question 5 Answer: 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

Question 6 Answer: 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) If 5 points were added to every score, what would be the value

for the new mean?

b) If every score were multiplied by 3, what would be the value of the new mean?

Mean

Question 7 Answer:

a) The new mean would be 45. When a constant is added to every score, the same constant is added to 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

Question 8 Answer: μ = 7

Mean

Question 9: Using the scores from question 8, fill in the following table.

Mean

Question 9 Answer:

4

4

1

1

2

Below

Below

Below

Above

Above

Median

Question 10: Find the median for each distribution of scores:

a) 3, 4, 6, 7, 9, 10, 11

b) 8, 10, 11, 12, 14, 15

Median

Question 10 Answers:a) The median is X = 7

b) 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

Question 11 Answer:

Count 8 boxes

Median = 6.83

1

1/3

2/3

6

5

2 3 4

7

5 6

4 2

3 1

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 one-third 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

Question 12 Answers:

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

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

Question 13 Answer: 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

Question 14 Answer: 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

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

Question 15 Answer: 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

Question 16 Answer: 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

Positively Skewed Distribution

Negatively Skewed Distribution

Notice how the means follow the outliers

Note: Median usually falls between meanand mode.

Frequently Asked Questions

Interpolation Real Limits Median for Continuous Variables Frequency Distribution Cumulative Distributions Weighted Mean

Frequently Asked Question FAQs

How do I find the median for a continuous variable?

Frequently Asked Questions FAQs

Step 1: Count the total number of boxes.

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).

1 2 3 4 5 6

7

8

9

10

11

12

13

14

16

15

16 boxes50% of 16is 8.

Frequently Asked Questions FAQs

1 2 3 4 5 6

7

Uh-oh!What now?

• 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%.

1/3 2/3

7.56.5

Frequently Asked Questions FAQs

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.

Median = 6.83

Frequently Asked Questions FAQs

 

Frequently Asked Questions FAQs

How do I calculate the weighted mean? 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.

Plug numbers into the formula