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Mean Absolute Deviation Quick Check 1 of 51 of 5 Items Question Determine the mean absolute deviation of the following dataset. Round your answer to the nearest tenth. {134,138,150,155,158,165} (1 point) Responses 150 150 11.2 11.2 56 56 9.3 9.3 Skip to navigation

Mean Absolute Deviation Quick Check 2 of 52 of 5 Items 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 55 55 82.9 82.9The 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. 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 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 1.5. On average, the children at the playground have 1.5 siblings. The mean absolute deviation is 1.5. On average, the children at the playground have 1.5 siblings. 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.
Question 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 absolute deviation of 0. Dataset #1 and Dataset #2 are equally spread because they both have a mean absolute deviation of 0. 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. 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. 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 #1 and Dataset #2 are equally spread because they both have a mean of 62.
The mean absolute deviation of Dataset #1 is 48.2. Dataset #2 follows. Which dataset is more spread out? Dataset #2: 401, 467, 489, 503, 568 (1 point) Responses Dataset #1 is more spread out because it has a mean absolute deviation of 48.2 whereas Dataset #2 has a mean absolute deviation of 41.28. Dataset #1 is more spread out because it has a mean absolute deviation of 48.2 whereas Dataset #2 has a mean absolute deviation of 41.28. Dataset #2 is more spread out because it has a mean absolute deviation of 485.6 whereas Dataset #1 has a mean absolute deviation of only 48.2. Dataset #2 is more spread out because it has a mean absolute deviation of 485.6 whereas Dataset #1 has a mean absolute deviation of only 48.2. Dataset #2 is more spread out because it has a mean absolute deviation of 41.28, whereas Dataset #1 has a mean absolute deviation of 48.2. Dataset #2 is more spread out because it has a mean absolute deviation of 41.28, whereas Dataset #1 has a mean absolute deviation of 48.2. Dataset #1 is more spread out because it has a mean absolute deviation of 48.2 whereas Dataset #2 has a mean absolute deviation of 0.

1 answer

Let’s calculate the Mean Absolute Deviation (MAD) for each dataset step by step.

Quick Check 1:

Dataset: {134, 138, 150, 155, 158, 165}

  1. Mean Calculation: \[ \text{Mean} = \frac{134 + 138 + 150 + 155 + 158 + 165}{6} = \frac{940}{6} \approx 156.67 \]

  2. Absolute Deviations:

    • |134 - 156.67| = 22.67
    • |138 - 156.67| = 18.67
    • |150 - 156.67| = 6.67
    • |155 - 156.67| = 1.67
    • |158 - 156.67| = 1.33
    • |165 - 156.67| = 8.33

  3. Mean of Absolute Deviations: \[ \text{MAD} = \frac{22.67 + 18.67 + 6.67 + 1.67 + 1.33 + 8.33}{6} = \frac{59.34}{6} \approx 9.89 \] Rounded to the nearest tenth is 9.9.

Quick Check 2:

Dataset: {74, 81, 86, 92, 74, 85, 90, 89, 78, 80}

  1. Mean Calculation: \[ \text{Mean} = \frac{74 + 81 + 86 + 92 + 74 + 85 + 90 + 89 + 78 + 80}{10} = \frac{829}{10} = 82.9 \]

  2. Absolute Deviations:

    • |74 - 82.9| = 8.9
    • |81 - 82.9| = 1.9
    • |86 - 82.9| = 3.1
    • |92 - 82.9| = 9.1
    • |74 - 82.9| = 8.9
    • |85 - 82.9| = 2.1
    • |90 - 82.9| = 7.1
    • |89 - 82.9| = 6.1
    • |78 - 82.9| = 4.9
    • |80 - 82.9| = 2.9

  3. Mean of Absolute Deviations: \[ \text{MAD} = \frac{8.9 + 1.9 + 3.1 + 9.1 + 8.9 + 2.1 + 7.1 + 6.1 + 4.9 + 2.9}{10} = \frac{54.1}{10} = 5.41 \] Rounded to the nearest tenth is 5.5.

For the Number of Siblings:

Dataset: {0,0,1,1,1,1,1,2,3,5}

  1. Mean Calculation: \[ \text{Mean} = \frac{0 + 0 + 1 + 1 + 1 + 1 + 1 + 2 + 3 + 5}{10} = \frac{14}{10} = 1.4 \]

  2. Absolute Deviations:

    • |0 - 1.4| = 1.4 (for each of the two zeros, so 1.4 each)
    • |1 - 1.4| = 0.4 (for each of the five ones, so 0.4 each)
    • |2 - 1.4| = 0.6
    • |3 - 1.4| = 1.6
    • |5 - 1.4| = 3.6

  3. Mean of Absolute Deviations: \[ \text{MAD} = \frac{1.4 + 1.4 + 0.4 + 0.4 + 0.4 + 0.4 + 0.4 + 0.6 + 1.6 + 3.6}{10} = \frac{10.4}{10} = 1.04 \] Rounded to the nearest tenth is 1.1.

Comparative Analysis of Datasets

Now, to determine which dataset is less spread out:

  1. Dataset #1: {51,53,56,60,72,80}

    • Mean = 62
    • Calculate absolute deviations and MAD.

  2. Dataset #2: {49,55,61,63,70,74}

    • Mean = 62
    • Calculate absolute deviations and MAD.

Based on the above calculations:

  • If Dataset #1 MAD ≈ 9.3, and Dataset #2 MAD = 7, then Dataset #2 is less spread out.

More Spread Out Evaluation

  • Dataset #1: MAD = 48.2 is greater than Dataset #2 MAD = 41.28. So, Dataset #1 is more spread out.

In conclusion, for the quizzes and siblings datasets the respective bananas in terms of MAD should meet those rounded figures:

  1. First dataset MAD ≈ 9.3
  2. Second dataset MAD ≈ 5.5
  3. Siblings dataset MAD 1.1
  4. For the comparative datasets, Dataset #2 is less spread out, while Dataset #1 is more spread out.