Determine the mean absolute deviation of the following dataset. Round your answer to the nearest tenth. {134,138,150,155,158,165} (1 poin) Responses 56 56 11.2 11.2 150 150 9.3
To find the mean absolute deviation, we first need to find the mean of the dataset.
Mean = (134 + 138 + 150 + 155 + 158 + 165) / 6 = 900 / 6 = 150
Next, we find the absolute deviation of each number from the mean, and then find the mean of those absolute deviations.
|134 - 150| = 16
|138 - 150| = 12
|150 - 150| = 0
|155 - 150| = 5
|158 - 150| = 8
|165 - 150| = 15
Mean absolute deviation = (16 + 12 + 0 + 5 + 8 + 15) / 6 = 56 / 6 = 9.3
Therefore, the mean absolute deviation of the dataset is 9.3.
Question Lucia has been recording her grades on her Spanish quizzes. The following is a list of her grades: 74, 81, 86, 92, 74, 85, 90, 89, 78, 80 Determine the mean absolute deviation of Lucia’s Spanish quiz grades. (1 point) Responses 8.29 8.29 5.5 5.5 82.9 82.9 55
To find the mean absolute deviation, we first need to find the mean of Lucia's Spanish quiz grades.
Mean = (74 + 81 + 86 + 92 + 74 + 85 + 90 + 89 + 78 + 80) / 10 = 849 / 10 = 84.9
Next, we find the absolute deviation of each grade from the mean, and then find the mean of those absolute deviations.
|74 - 84.9| = 10.9
|81 - 84.9| = 3.9
|86 - 84.9| = 1.1
|92 - 84.9| = 7.1
|74 - 84.9| = 10.9
|85 - 84.9| = 0.1
|90 - 84.9| = 5.1
|89 - 84.9| = 4.1
|78 - 84.9| = 6.9
|80 - 84.9| = 4.9
Mean absolute deviation = (10.9 + 3.9 + 1.1 + 7.1 + 10.9 + 0.1 + 5.1 + 4.1 + 6.9 + 4.9) / 10 = 55.9 / 10 = 5.59
Therefore, the mean absolute deviation of Lucia’s Spanish quiz grades is 5.59. Rounding to the nearest hundredth, we get 5.5.
The following dataset is a list of the number of siblings for 10 children at a playground. Describe the spread of the dataset using the mean absolute deviation. Number of siblings = {0,0,1,1,1,1,1,2,3,5} (1 point) Responses The mean absolute deviation is 1.1. The number of siblings each child has is within 1.1 siblings from the mean. The mean absolute deviation is 1.1. The number of siblings each child has is within 1.1 siblings from the mean. The mean absolute deviation is 1.5. On average, the number of siblings each child has is about 1.5 siblings from the mean. The mean absolute deviation is 1.5. On average, the number of siblings each child has is about 1.5 siblings from the mean. The mean absolute deviation is 1.1. On average, the number of siblings each child has is about 1.1 siblings from the mean. The mean absolute deviation is 1.1. On average, the number of siblings each child has is about 1.1 siblings from the mean. The mean absolute deviation is 1.5. On average, the children at the playground have 1.5 siblings.
To find the mean absolute deviation, first, we need to find the mean of the number of siblings for the 10 children:
Mean = (0 + 0 + 1 + 1 + 1 + 1 + 1 + 2 + 3 + 5) / 10 = 15 / 10 = 1.5
Next, we find the absolute deviation of each number of siblings from the mean:
|0 - 1.5| = 1.5
|0 - 1.5| = 1.5
|1 - 1.5| = 0.5
|1 - 1.5| = 0.5
|1 - 1.5| = 0.5
|1 - 1.5| = 0.5
|1 - 1.5| = 0.5
|2 - 1.5| = 0.5
|3 - 1.5| = 1.5
|5 - 1.5| = 3.5
Mean absolute deviation = (1.5 + 1.5 + 0.5 + 0.5 + 0.5 + 0.5 + 0.5 + 0.5 + 1.5 + 3.5) / 10 = 10.5 / 10 = 1.05
Therefore, the mean absolute deviation of the number of siblings for the 10 children is 1.1. The interpretation is "The mean absolute deviation is 1.1. The number of siblings each child has is within 1.1 siblings from the mean."
The mean of each dataset that follows is 62. Which dataset is less spread out? Justify your answer using the mean absolute deviation of each dataset. Round your answers to the nearest tenth. Dataset #1: {51,53,56,60,72,80} Dataset #2: {49,55,61,63,70,74} (1 point) Responses Dataset #1 and Dataset #2 are equally spread because they both have a mean of 62. Dataset #1 and Dataset #2 are equally spread because they both have a mean of 62. Dataset #1 and Dataset #2 are equally spread because they both have a mean absolute deviation of 0. Dataset #1 and Dataset #2 are equally spread because they both have a mean absolute deviation of 0. Dataset #1 is less spread out because it has a mean absolute deviation of 9.3 whereas Dataset #2 has a mean absolute deviation of 7. Dataset #1 is less spread out because it has a mean absolute deviation of 9.3 whereas Dataset #2 has a mean absolute deviation of 7. Dataset #2 is less spread out because it has a mean absolute deviation of 7 whereas Dataset #1 has a mean absolute deviation of approximately 9.3.
To determine which dataset is less spread out, we can compare the mean absolute deviation of each dataset.
For Dataset #1:
Mean = 62
Absolute deviation from 62 for each number in Dataset #1:
|51 - 62| = 11
|53 - 62| = 9
|56 - 62| = 6
|60 - 62| = 2
|72 - 62| = 10
|80 - 62| = 18
Mean absolute deviation for Dataset #1:
(11 + 9 + 6 + 2 + 10 + 18) / 6 = 56 / 6 = 9.3
For Dataset #2:
Mean = 62
Absolute deviation from 62 for each number in Dataset #2:
|49 - 62| = 13
|55 - 62| = 7
|61 - 62| = 1
|63 - 62| = 1
|70 - 62| = 8
|74 - 62| = 12
Mean absolute deviation for Dataset #2:
(13 + 7 + 1 + 1 + 8 + 12) / 6 = 42 / 6 = 7
Comparing the mean absolute deviations, we see that Dataset #2 has a mean absolute deviation of 7, while Dataset #1 has a mean absolute deviation of 9.3. Therefore, Dataset #2 is less spread out than Dataset #1.
The correct response is: Dataset #2 is less spread out because it has a mean absolute deviation of 7, whereas Dataset #1 has a mean absolute deviation of 9.3.