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MAD Practice #1-6

MAD Practice #1-6. Mean Absolute Deviation Practice. 1. Daily Visitors to a Web Site:. 1. Daily Visitors to a Web Site: The MAD is 18.48 indicating the average difference of visits to a web site from the mean is 18.48 times. The data is spread out in relation to the mean. .

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MAD Practice #1-6

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  1. MAD Practice #1-6

  2. Mean Absolute Deviation Practice 1. Daily Visitors to a Web Site: 1. Daily Visitors to a Web Site: The MAD is 18.48 indicating the average difference of visits to a web site from the mean is 18.48 times. The data is spread out in relation to the mean. Step 2 (cont.): Use the absolute value of each distance Step 2: Find the distance each element is from the mean Step 1: Find the Mean Step 3: Calculate the mean of these distances

  3. Mean Absolute Deviation Practice 2. Zoo Admission Prices: 2. Zoo Admission Prices: The MAD is 0.5 indicating the average difference of prices from the mean is $0.50. The data is less spread out and closer to the mean. Step 1: Find the Mean Step 2 (cont.): Use the absolute value of each distance Step 2: Find the distance each element is from the mean Step 3: Calculate the mean of these distances

  4. Mean Absolute Deviation Practice 3. Height of Waterslides: Splash Lagoon vs. Wild Water Bay Step 2: Find the distance each element is from the mean and use the absolute value of each distance Step 2: Find the distance each element is from the mean and use the absolute value of each distance Step 1: Find the Mean of Splash Lagoon Step 1: Find the Mean of Wild Water Bay Step 3: Calculate the mean of these distances Step 3: Calculate the mean of these distances

  5. Mean Absolute Deviation Practice 3. Height of Waterslides: Splash Lagoon vs. Wild Water Bay

  6. Mean Absolute Deviation Practice 3. Height of Waterslides: Splash Lagoon vs. Wild Water Bay The MAD for Splash Lagoon is 10.32. The MAD for Wild Water Bay is 12.48. A lower MAD indicates the data is less spread out and closer to the mean. Therefore, the height of the waterslides at Splash Lagoon are closer to the mean height demonstrating less variation in the height of each slide in comparison to the slides at Wild Water Bay.

  7. Mean Absolute Deviation Practice 4. Known Moons of Planets: The MAD is 17.875 indicating the average difference of the number of moons per planet from the mean is 17.875. The data is spread out in relation to the mean. 4. Known Moons of Planets: Step 2: Find the distance each element is from the mean Step 2 (cont.): Use the absolute value of each distance Step 1: Find the Mean Step 3: Calculate the mean of these distances

  8. Mean Absolute Deviation Practice 5. Hard Drive (gigabytes): 5. Hard Drive (gigabytes): The MAD is 158.75 indicating the average difference of the number of gigabytes per hard drive from the mean is 158.75. The data is significantly spread out in relation to the mean. Step 2: Find the distance each element is from the mean Step 2 (cont.): Use the absolute value of each distance Step 1: Find the Mean Step 3: Calculate the mean of these distances

  9. Mean Absolute Deviation Practice 6. Digital Camera Prices: The MAD is 26.76 indicating the average difference of the prices from the mean is $26.76. The data is spread out in relation to the mean. 6. Digital Camera Prices: Step 2: Find the distance each element is from the mean Step 2 (cont.): Use the absolute value of each distance Step 1: Find the Mean Step 3: Calculate the mean of these distances

  10. Mean Absolute Deviation Practice Homework: Find the MAD for problem 7, Grand Slam Singles Titles Won (10 different tennis players) Step 2: Find the distance each element is from the mean Step 2 (cont.): Use the absolute value of each distance Step 1: Find the Mean Step 3: Calculate the mean of these distances

  11. Homework: Find the mean absolute deviation (MAD) for problem 7, Grand Slam Singles Titles Won (10 different tennis players) 7. Grand Slam Singles Titles Won: 4. Known Moons of Planets: The MAD is 2.02 indicating the average difference of the number of titles won from the mean is 2.02. The data is less spread out and closer to the mean. Step 2: Find the distance each element is from the mean Step 2 (cont.): Use the absolute value of each distance Step 1: Find the Mean Step 3: Calculate the mean of these distances

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