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John LohmanMichael Mattocks

Aubrey Urwick

Chapter 3: Central Tendency

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!

More Key Terms: Think Notecards  

Mean Question 1: Find the mean for the sample of n=5 scores:

1, 8, 7, 5, 9 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 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 1 Answer:

M = Σ X n

M = 1+8+7+5+9 5

M = 30 5

M = 6

Question 2 Answer: M = Σ X

n 8 = Σ X

6 Σ X = 48

Mean

Mean  

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.

Question 5: Find the values for n, Σ X, and M for the following sample:

Mean

X f5 14 23 32 51 1

Mean Question 4 Answer:

B) It must be less than 40. A score higher than 40 would have increased the 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? 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 6 Answer:

False. If the score is equal to the mean, it does not change the 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 9: Using the scores from question 8, fill in the

following table.

Mean Question 8 Answer:

μ = 7 Question 9 Answer:

4

4

1

1

2

Below

BelowBelow

AboveAbove

Median Question 10: Find the median for each distribution of

scores:a) 3, 4, 6, 7, 9, 10, 11b) 8, 10, 11, 12, 14, 15

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 10 Answers:

a) The median is X = 7b) The median is X = 11.5

Question 11 Answer:

1/3

2/3

Median = 6.83

1 2 3 4 5 6

7Count 8 boxes

12

34

5

6

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)

Question 14: Why is it usually inappropriate to compute a mean for scores measured on an ordinal scale?

Question 15: In a perfectly symmetrical distribution, the mean, the median, and the mode will all have the same value. (True or False)

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 13 Answer: Mean

Question 14 Answer: The definition of the mean is based on distances (the mean

balances the distances) and ordinal scales do no measure distance.

Question 15 Answer: False, if the distribution is bimodal.

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

1112

1314

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