2.3

Section 2.3 Measures of Central Tendency Larson/Farber 4th ed.

Measures of Central Tendency ,[object Object],[object Object],[object Object],[object Object],[object Object]

Symbols that are commonly used

[object Object],[object Object],Measure of Central Tendency: Mean

Finding the Mean ,[object Object],[object Object],[object Object],[object Object]

Example: Finding a Sample Mean ,[object Object],[object Object],Larson/Farber 4th ed.

Solution: Finding a Sample Mean ,[object Object],Larson/Farber 4th ed. ,[object Object],[object Object],[object Object],The mean price of the flights is about $527.90.

Measures of Central Tendency: Median ,[object Object],[object Object],[object Object],[object Object]

Finding the Median ,[object Object],[object Object],[object Object],[object Object]

Example: Finding the Median ,[object Object],[object Object],Larson/Farber 4th ed.

Solution: Finding the Median ,[object Object],Larson/Farber 4th ed. ,[object Object],[object Object],[object Object],The median price of the flights is $427.

Solution: Finding the Median ,[object Object],Larson/Farber 4th ed. ,[object Object],[object Object],[object Object],The median price of the flights is $412.

Measures of Central Tendency: Mode ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

Example: Finding the Mode ,[object Object],[object Object],Larson/Farber 4th ed.

Solution: Finding the Mode ,[object Object],Larson/Farber 4th ed. ,[object Object],[object Object],[object Object],The mode of the flight prices is $397.

Example: Finding the Mode ,[object Object],Larson/Farber 4th ed. Political Party Frequency, f Democrat 34 Republican 56 Other 21 Did not respond 9

Solution: Finding the Mode Larson/Farber 4th ed. The mode is Republican (the response occurring with the greatest frequency). In this sample there were more Republicans than people of any other single affiliation. Political Party Frequency, f Democrat 34 Republican 56 Other 21 Did not respond 9

Comparing the Mean, Median, and Mode ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],Larson/Farber 4th ed.

Example: Comparing the Mean, Median, and Mode ,[object Object],Larson/Farber 4th ed. Ages in a class 20 20 20 20 20 20 21 21 21 21 22 22 22 23 23 23 23 24 24 65

Solution: Comparing the Mean, Median, and Mode Larson/Farber 4th ed. Mean: Median: 20 years (the entry occurring with the greatest frequency) Mode: Ages in a class 20 20 20 20 20 20 21 21 21 21 22 22 22 23 23 23 23 24 24 65

Solution: Comparing the Mean, Median, and Mode Larson/Farber 4th ed. Mean ≈ 23.8 years Median = 21.5 years Mode = 20 years ,[object Object],[object Object],[object Object]

Solution: Comparing the Mean, Median, and Mode Larson/Farber 4th ed. Sometimes a graphical comparison can help you decide which measure of central tendency best represents a data set. In this case, it appears that the median best describes the data set.

Weighted Mean ,[object Object],[object Object]

To find the weighted mean: ,[object Object],[object Object],[object Object],[object Object],where w represents the weight of each entry

Example: Finding a Weighted Mean ,[object Object],Larson/Farber 4th ed.

Solution: Finding a Weighted Mean Larson/Farber 4th ed. Your weighted mean for the course is 88.6. You did not get an A. Source Score, x Weight, w x∙w Test Mean 86 0.50 86(0.50)= 43.0 Midterm 96 0.15 96(0.15) = 14.4 Final Exam 82 0.20 82(0.20) = 16.4 Computer Lab 98 0.10 98(0.10) = 9.8 Homework 100 0.05 100(0.05) = 5.0 Σ w = 1 Σ ( x∙w ) = 88.6

Mean of Grouped Data ,[object Object],[object Object]

Finding the Mean of a Frequency Distribution ,[object Object],[object Object],[object Object],[object Object]

Example: Find the Mean of a Frequency Distribution ,[object Object],Larson/Farber 4th ed. Class Midpoint Frequency, f 7 – 18 12.5 6 19 – 30 24.5 10 31 – 42 36.5 13 43 – 54 48.5 8 55 – 66 60.5 5 67 – 78 72.5 6 79 – 90 84.5 2

Solution: Find the Mean of a Frequency Distribution Larson/Farber 4th ed. Class Midpoint, x Frequency, f ( x∙f ) 7 – 18 12.5 6 12.5∙6 = 75.0 19 – 30 24.5 10 24.5∙10 = 245.0 31 – 42 36.5 13 36.5∙13 = 474.5 43 – 54 48.5 8 48.5∙8 = 388.0 55 – 66 60.5 5 60.5∙5 = 302.5 67 – 78 72.5 6 72.5∙6 = 435.0 79 – 90 84.5 2 84.5∙2 = 169.0 n = 50 Σ ( x∙f ) = 2089.0

The Shape of Distribution ,[object Object],[object Object]

The Shape of Distributions Larson/Farber 4th ed. ,[object Object],[object Object],[object Object]

The Shape of Distributions Larson/Farber 4th ed. ,[object Object],[object Object],[object Object],[object Object]

The Shape of Distributions ,[object Object],[object Object]

2.3

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2.3