The mean of the following dataset is 8. Find the mean absolute deviation of the dataset. {5.8,7,8.1,9.2,9.9} (1 point) The mean absolute deviation is .
1.12
Wrong
I apologize for the error in my previous response. Let's calculate the mean absolute deviation correctly:
Step 1: Find the absolute deviation for each data point from the mean.
Absolute deviation for each data point:
- 5.8: |5.8 - 8| = 2.2
- 7: |7 - 8| = 1
- 8.1: |8.1 - 8| = 0.1
- 9.2: |9.2 - 8| = 1.2
- 9.9: |9.9 - 8| = 1.9
Step 2: Find the mean of the absolute deviations.
Mean absolute deviation = (2.2 + 1 + 0.1 + 1.2 + 1.9) / 5
Mean absolute deviation = 6.4 / 5
Mean absolute deviation = 1.28
Therefore, the mean absolute deviation of the dataset {5.8, 7, 8.1, 9.2, 9.9} is 1.28.
Find the mean absolute deviation of the following dataset. {21,25,25,26,26,27,28,30} (1 point) The mean absolute deviation is .
To find the mean absolute deviation (MAD) of the given dataset {21, 25, 25, 26, 26, 27, 28, 30}, follow these steps:
Step 1: Calculate the mean of the dataset.
Mean = (21 + 25 + 25 + 26 + 26 + 27 + 28 + 30) / 8
Mean = 208 / 8
Mean = 26
Step 2: Find the absolute deviation for each data point from the mean.
Absolute deviation for each data point:
- 21: |21 - 26| = 5
- 25: |25 - 26| = 1
- 25: |25 - 26| = 1
- 26: |26 - 26| = 0
- 26: |26 - 26| = 0
- 27: |27 - 26| = 1
- 28: |28 - 26| = 2
- 30: |30 - 26| = 4
Step 3: Find the mean of the absolute deviations.
Mean absolute deviation = (5 + 1 + 1 + 0 + 0 + 1 + 2 + 4) / 8
Mean absolute deviation = 14 / 8
Mean absolute deviation = 1.75
Therefore, the mean absolute deviation of the dataset {21, 25, 25, 26, 26, 27, 28, 30} is 1.75.
Tessa is investigating the number of strawberries in a 1-pound package. She has 6 packages of strawberries and counts the number of strawberries in each package. Her results follow. {10,12,14,15,16,17} What is the mean absolute deviation of her data? (1 point) The mean absolute deviation is .
To find the mean absolute deviation (MAD) of the given data {10, 12, 14, 15, 16, 17} representing the number of strawberries in each package, follow these steps:
Step 1: Calculate the mean of the dataset.
Mean = (10 + 12 + 14 + 15 + 16 + 17) / 6
Mean = 84 / 6
Mean = 14
Step 2: Find the absolute deviation for each data point from the mean.
Absolute deviation for each data point:
- 10: |10 - 14| = 4
- 12: |12 - 14| = 2
- 14: |14 - 14| = 0
- 15: |15 - 14| = 1
- 16: |16 - 14| = 2
- 17: |17 - 14| = 3
Step 3: Find the mean of the absolute deviations.
Mean absolute deviation = (4 + 2 + 0 + 1 + 2 + 3) / 6
Mean absolute deviation = 12 / 6
Mean absolute deviation = 2
Therefore, the mean absolute deviation of Tessa's data {10, 12, 14, 15, 16, 17} representing the number of strawberries in each package is 2.
The mean of each dataset that follows is 10. Which dataset is more spread out? Justify your answer using the mean absolute deviation of each dataset. Round your answers to the nearest tenth. Dataset #1: {5,8,9,10,12,12,14} Dataset #2: {5,6,8,9,10,10,22} (2 points) Dataset # is more spread out because it has a mean absolute deviation of while Dataset # has a mean absolute deviation of .
To determine which of the two datasets {5, 8, 9, 10, 12, 12, 14} and {5, 6, 8, 9, 10, 10, 22} is more spread out based on their mean absolute deviation (MAD), we will calculate the MAD for each dataset:
For Dataset #1: {5, 8, 9, 10, 12, 12, 14}
- Mean = 10
- Absolute deviation for each data point:
- 5: |5 - 10| = 5
- 8: |8 - 10| = 2
- 9: |9 - 10| = 1
- 10: |10 - 10| = 0
- 12: |12 - 10| = 2
- 12: |12 - 10| = 2
- 14: |14 - 10| = 4
- MAD = (5 + 2 + 1 + 0 + 2 + 2 + 4) / 7
- MAD = 16 / 7
- MAD = 2.3 (rounded to the nearest tenth)
For Dataset #2: {5, 6, 8, 9, 10, 10, 22}
- Mean = 10
- Absolute deviation for each data point:
- 5: |5 - 10| = 5
- 6: |6 - 10| = 4
- 8: |8 - 10| = 2
- 9: |9 - 10| = 1
- 10: |10 - 10| = 0
- 10: |10 - 10| = 0
- 22: |22 - 10| = 12
- MAD = (5 + 4 + 2 + 1 + 0 + 0 + 12) / 7
- MAD = 24 / 7
- MAD = 3.4 (rounded to the nearest tenth)
Therefore, Dataset #2 {5, 6, 8, 9, 10, 10, 22} is more spread out because it has a mean absolute deviation of 3.4, while Dataset #1 {5, 8, 9, 10, 12, 12, 14} has a mean absolute deviation of 2.3.