1 / 8

Statistics: Measures of Variation

Statistics: Measures of Variation. The central tendencies (mean, median, and mode) are used to describe the data set. Measures of variation are used to describe the distribution of the data. Vocabulary. Range the difference between the greatest and least values in the data Quartiles

chaney
Download Presentation

Statistics: Measures of Variation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Statistics: Measures of Variation The central tendencies (mean, median, and mode) are used to describe the data set. Measures of variation are used to describe the distribution of the data.

  2. Vocabulary • Range • the difference between the greatest and least values in the data • Quartiles • the values that divide the data into four equal parts • The median separates the data into 2 equal parts. • Lower quartile (LQ) • the median of the lower half of the set of data • Upper quartile (UQ) • the median of the upper half of the set of data

  3. Vocabulary • Interquartile range • the range of the middle half of the data • the difference between the upper quartile and the lower quartile • Outlier • a value that is much greater or much less than the median • Data that are more than 1.5 times the value of the interquartile range beyond the quartiles

  4. Ex 1) Find a) the range of the data set b) the median c) the upper and lower quartiles d) the interquartile range Data set: 80, 100, 110, 115, 120, 120, 170 Median (middle) • Range = 170 – 80 = 90 Lower Quartile (median of lower half of data) Upper Quartile (median of upper half of data) b) Median = 115 c) Lower quartile = 100 Upper quartile = 120 d) Interquartile range = UQ – LQ = 120 – 100 = 20

  5. Ex 2) Data set: 27, 37, 21, 58, 46, 35, 19, 54 Find a) the range of the data set b) the median c) the upper and lower quartiles d) the interquartile range Data set in order from least to greatest: 19, 21, 27, 35, 37, 46, 54, 58 • Range = 58 – 19 = 39 Median (mean of 2 middle values) b) Median Lower Quartile (median of lower half of data) Upper Quartile (median of upper half of data) • Lower quartile • Upper quartile • Interquartile range = UQ – LQ • = 50 – 24 • = 26

  6. Ex 3) The chart shows the average number of hours that adults spend online. Find the upper and lower quartile of the data set. First put the data in order from least to greatest. 12.9, 12.9, 13.1, 13.3, 13.4, 14.2, 14.4, 14.9, 14.9, 15.8 Lower Quartile Upper Quartile Median Locate the median to separate the data into the halves. To find the upper and lower quartiles, find the median in the lower half and in the upper half of the data. The lower quartile is 13.1 and the upper quartile is 14.9.

  7. Ex 4) Find any outliers for the data in the table. • Locate the median, upper quartile, and lower quartile. UpperQuartile b) Interquartile Range = 6.5 - 2.1 = 4.4 c) Multiply the interquartile range by 1.5. 4.4(1.5) = 6.6 d) Find the limits for the outliers. -Subtract 6.6 from lower quartile. 2.1 – 6.6 = -4.5 lower limit -Add 6.6 to the upper quartile. 6.5 + 6.6 = 13.1 upper limit Median Lower Quartile e) The only outlier is 16.6 since that value is above the upper limit.

  8. Practice Problem • Find the median, upper and lower quartiles, interquartile range and any outliers for the following data set: 81, 79, 88, 67, 89, 87, 85, 83, 83 Put them in order from least to greatest: 67, 79, 81, 83, 83, 85, 87, 88, 89 Upper half: 85, 87, 88, 89 Upper Quartile (87+88)÷2 = 87.5 Lower half: 67, 79, 81, 83 Lower Quartile (79+81)÷2 = 80 83 is the median Outliers 7.5(1.5) = 11.25 Lower limit 80-11.25 = 68.75 Upper limit 87.5+11.25 = 98.75 Interquartile range Upper Quartile – Lower Quartile 87.5 – 80 = 7.5 67 is an outlier because it is less than 68.75

More Related