EXAMPLE 2 Finding Median Mode and Range Find

EXAMPLE 2 Finding Median, Mode, and Range. Find the median, mode(s), and range of the prices. $7. 20, $13. 25, $14. 94, $16. 56, $18. 74, $19. 99, $29. 49 Median The data set has an even number of prices, so the median is the mean of the two middle values, $16. 56 and $18. 74. Median = $16. 56 + $18. 74 = $35. 30 = $17. 65 2 2 Mode: The price that occurs most often is $19. 99. This is the mode.

EXAMPLE 2 Finding Median, Mode, and Range: Find the difference of the greatest and the least values. Range = $29. 49 – $7. 20 = $22. 29

EXAMPLE 3 Standardized Test Practice SOLUTION Mean = 40 = 4 10 Median = 2 + 2 = 4 2 2 Mode: 2 = 2

EXAMPLE 3 Standardized Test Practice ANSWER Because 20 is an outlier, the mean is greater than all but two ratings. It does not represent the data well. The mode and median best represent the data. The correct answer is B.

GUIDED PRACTICE for Example 2 and 3 Find the median, mode(s), and range of the data. 3. 14, 13, 20, 24, 15, 10, 22, 17, 18 Median = 17 Mode: There is no mode. Range: Find the difference of greatest and the least value. Range = 24 – 10 = 14

GUIDED PRACTICE for Example 2 and 3 Find the median, mode(s), and range of the data. 4. 9, 7, 4, 9, 4, 10, 5, 14, 9, 4 Median =8 Mode: The number that occurs most often are 4 and 9. This is the mode. Range: Find the difference of greatest and the least value. Range = 14 – 4 = 10

GUIDED PRACTICE for Example 2 and 3 5. What If? In Example 3, suppose another group tries the new cereal. Their ratings are: 1, 1, 1, 2, 3, 4, 4, 9, 10. Which average or averages best represent the data? SOLUTION Mean = 35 = 3. 5 10 Median = 3 + 4 = 7 2 2 Mode = 2 = 3. 5

GUIDED PRACTICE for Example 2 and 3 ANSWER Median; Most of the data is around 3. 5