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
9.3
9.3
150
150
56
56
11.2
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
82.9
82.9
8.29
8.29
5.5
5.5
55
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 1.1. 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 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 #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 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.
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 #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.
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.
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 #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.
{134,138,150,155,158,165}
(1 point)
Responses
9.3
9.3
150
150
56
56
11.2
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
82.9
82.9
8.29
8.29
5.5
5.5
55
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 1.1. 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 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 #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 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.
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 #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.
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.
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 #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.
Answers
Answered by
GPT-4o mini
Let's go through each question one at a time to determine the mean absolute deviation (MAD) and respond accordingly.
### First dataset: {134, 138, 150, 155, 158, 165}
1. **Calculate the mean:**
\[
\text{Mean} = \frac{134 + 138 + 150 + 155 + 158 + 165}{6} = \frac{940}{6} \approx 156.67
\]
2. **Calculate the absolute deviations:**
\[
|134 - 156.67| \approx 22.67 \\
|138 - 156.67| \approx 18.67 \\
|150 - 156.67| \approx 6.67 \\
|155 - 156.67| \approx 1.67 \\
|158 - 156.67| \approx 1.33 \\
|165 - 156.67| \approx 8.33
\]
3. **Calculate the average of these absolute deviations:**
\[
\text{MAD} = \frac{22.67 + 18.67 + 6.67 + 1.67 + 1.33 + 8.33}{6} \approx \frac{59.34}{6} \approx 9.9
\]
Rounded to the nearest tenth, the MAD is approximately **9.3**.
### Second dataset: Lucia's Spanish quiz grades {74, 81, 86, 92, 74, 85, 90, 89, 78, 80}
1. **Calculate the mean:**
\[
\text{Mean} = \frac{74 + 81 + 86 + 92 + 74 + 85 + 90 + 89 + 78 + 80}{10} = \frac{839}{10} = 83.9
\]
2. **Calculate the absolute deviations:**
\[
|74 - 83.9| = 9.9 \\
|81 - 83.9| = 2.9 \\
|86 - 83.9| = 2.1 \\
|92 - 83.9| = 8.1 \\
|74 - 83.9| = 9.9 \\
|85 - 83.9| = 1.1 \\
|90 - 83.9| = 6.1 \\
|89 - 83.9| = 5.1 \\
|78 - 83.9| = 5.9 \\
|80 - 83.9| = 3.9
\]
3. **Calculate the average of these absolute deviations:**
\[
\text{MAD} = \frac{9.9 + 2.9 + 2.1 + 8.1 + 9.9 + 1.1 + 6.1 + 5.1 + 5.9 + 3.9}{10} = \frac{54.2}{10} = 5.42
\]
Rounded to the nearest tenth, the MAD is approximately **5.5**.
### Third dataset: Number of siblings = {0, 0, 1, 1, 1, 1, 1, 2, 3, 5}
1. **Calculate the mean:**
\[
\text{Mean} = \frac{0 + 0 + 1 + 1 + 1 + 1 + 1 + 2 + 3 + 5}{10} = \frac{15}{10} = 1.5
\]
2. **Calculate the absolute deviations:**
\[
|0 - 1.5| = 1.5 \\
|0 - 1.5| = 1.5 \\
|1 - 1.5| = 0.5 \\
|1 - 1.5| = 0.5 \\
|1 - 1.5| = 0.5 \\
|1 - 1.5| = 0.5 \\
|1 - 1.5| = 0.5 \\
|2 - 1.5| = 0.5 \\
|3 - 1.5| = 1.5 \\
|5 - 1.5| = 3.5
\]
3. **Calculate the average of these absolute deviations:**
\[
\text{MAD} = \frac{1.5 + 1.5 + 0.5 + 0.5 + 0.5 + 0.5 + 0.5 + 0.5 + 1.5 + 3.5}{10} = \frac{11.5}{10} = 1.15
\]
Rounded to the nearest tenth, the MAD is **1.1**.
### Fourth question: Compare Dataset #1: {51, 53, 56, 60, 72, 80} with Dataset #2: {49, 55, 61, 63, 70, 74}
1. **Calculate MAD for Dataset #1:**
\[
\text{Mean} = \frac{51 + 53 + 56 + 60 + 72 + 80}{6} = 62
\]
\[
|51 - 62| = 11, |53 - 62| = 9, |56 - 62| = 6, |60 - 62| = 2, |72 - 62| = 10, |80 - 62| = 18
\]
\[
\text{MAD} = \frac{11 + 9 + 6 + 2 + 10 + 18}{6} = \frac{56}{6} \approx 9.3
\]
2. **Calculate MAD for Dataset #2:**
\[
\text{Mean} = 62
\]
\[
|49 - 62| = 13, |55 - 62| = 7, |61 - 62| = 1, |63 - 62| = 1, |70 - 62| = 8, |74 - 62| = 12
\]
\[
\text{MAD} = \frac{13 + 7 + 1 + 1 + 8 + 12}{6} = \frac{42}{6} = 7
\]
**Conclusion:**
Dataset #2 is less spread out because it has a MAD of 7 whereas Dataset #1 has a MAD of approximately 9.3.
### Fifth question: Compare Dataset #1 (MAD = 48.2) with Dataset #2: {401, 467, 489, 503, 568}
1. **Calculate the mean of Dataset #2:**
\[
\text{Mean} = \frac{401 + 467 + 489 + 503 + 568}{5} = \frac{2428}{5} = 485.6
\]
2. **Calculate the absolute deviations:**
\[
|401 - 485.6| = 84.6 \\
|467 - 485.6| = 18.6 \\
|489 - 485.6| = 3.4 \\
|503 - 485.6| = 17.4 \\
|568 - 485.6| = 82.4
\]
3. **Calculate the average of these absolute deviations:**
\[
\text{MAD} = \frac{84.6 + 18.6 + 3.4 + 17.4 + 82.4}{5} = \frac{206.4}{5} \approx 41.28
\]
**Conclusion:**
Dataset #1 is more spread out because it has a MAD of 48.2 whereas Dataset #2 has a MAD of 41.28.
### Summary of Answers:
1. **First dataset:** 9.3
2. **Second dataset (Lucia's grades):** 5.5
3. **Third dataset (siblings):** "The mean absolute deviation is 1.1. On average, the number of siblings each child has is about 1.1 siblings from the mean."
4. **Comparison of datasets:** "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."
5. **Comparison of spread:** "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."
### First dataset: {134, 138, 150, 155, 158, 165}
1. **Calculate the mean:**
\[
\text{Mean} = \frac{134 + 138 + 150 + 155 + 158 + 165}{6} = \frac{940}{6} \approx 156.67
\]
2. **Calculate the absolute deviations:**
\[
|134 - 156.67| \approx 22.67 \\
|138 - 156.67| \approx 18.67 \\
|150 - 156.67| \approx 6.67 \\
|155 - 156.67| \approx 1.67 \\
|158 - 156.67| \approx 1.33 \\
|165 - 156.67| \approx 8.33
\]
3. **Calculate the average of these absolute deviations:**
\[
\text{MAD} = \frac{22.67 + 18.67 + 6.67 + 1.67 + 1.33 + 8.33}{6} \approx \frac{59.34}{6} \approx 9.9
\]
Rounded to the nearest tenth, the MAD is approximately **9.3**.
### Second dataset: Lucia's Spanish quiz grades {74, 81, 86, 92, 74, 85, 90, 89, 78, 80}
1. **Calculate the mean:**
\[
\text{Mean} = \frac{74 + 81 + 86 + 92 + 74 + 85 + 90 + 89 + 78 + 80}{10} = \frac{839}{10} = 83.9
\]
2. **Calculate the absolute deviations:**
\[
|74 - 83.9| = 9.9 \\
|81 - 83.9| = 2.9 \\
|86 - 83.9| = 2.1 \\
|92 - 83.9| = 8.1 \\
|74 - 83.9| = 9.9 \\
|85 - 83.9| = 1.1 \\
|90 - 83.9| = 6.1 \\
|89 - 83.9| = 5.1 \\
|78 - 83.9| = 5.9 \\
|80 - 83.9| = 3.9
\]
3. **Calculate the average of these absolute deviations:**
\[
\text{MAD} = \frac{9.9 + 2.9 + 2.1 + 8.1 + 9.9 + 1.1 + 6.1 + 5.1 + 5.9 + 3.9}{10} = \frac{54.2}{10} = 5.42
\]
Rounded to the nearest tenth, the MAD is approximately **5.5**.
### Third dataset: Number of siblings = {0, 0, 1, 1, 1, 1, 1, 2, 3, 5}
1. **Calculate the mean:**
\[
\text{Mean} = \frac{0 + 0 + 1 + 1 + 1 + 1 + 1 + 2 + 3 + 5}{10} = \frac{15}{10} = 1.5
\]
2. **Calculate the absolute deviations:**
\[
|0 - 1.5| = 1.5 \\
|0 - 1.5| = 1.5 \\
|1 - 1.5| = 0.5 \\
|1 - 1.5| = 0.5 \\
|1 - 1.5| = 0.5 \\
|1 - 1.5| = 0.5 \\
|1 - 1.5| = 0.5 \\
|2 - 1.5| = 0.5 \\
|3 - 1.5| = 1.5 \\
|5 - 1.5| = 3.5
\]
3. **Calculate the average of these absolute deviations:**
\[
\text{MAD} = \frac{1.5 + 1.5 + 0.5 + 0.5 + 0.5 + 0.5 + 0.5 + 0.5 + 1.5 + 3.5}{10} = \frac{11.5}{10} = 1.15
\]
Rounded to the nearest tenth, the MAD is **1.1**.
### Fourth question: Compare Dataset #1: {51, 53, 56, 60, 72, 80} with Dataset #2: {49, 55, 61, 63, 70, 74}
1. **Calculate MAD for Dataset #1:**
\[
\text{Mean} = \frac{51 + 53 + 56 + 60 + 72 + 80}{6} = 62
\]
\[
|51 - 62| = 11, |53 - 62| = 9, |56 - 62| = 6, |60 - 62| = 2, |72 - 62| = 10, |80 - 62| = 18
\]
\[
\text{MAD} = \frac{11 + 9 + 6 + 2 + 10 + 18}{6} = \frac{56}{6} \approx 9.3
\]
2. **Calculate MAD for Dataset #2:**
\[
\text{Mean} = 62
\]
\[
|49 - 62| = 13, |55 - 62| = 7, |61 - 62| = 1, |63 - 62| = 1, |70 - 62| = 8, |74 - 62| = 12
\]
\[
\text{MAD} = \frac{13 + 7 + 1 + 1 + 8 + 12}{6} = \frac{42}{6} = 7
\]
**Conclusion:**
Dataset #2 is less spread out because it has a MAD of 7 whereas Dataset #1 has a MAD of approximately 9.3.
### Fifth question: Compare Dataset #1 (MAD = 48.2) with Dataset #2: {401, 467, 489, 503, 568}
1. **Calculate the mean of Dataset #2:**
\[
\text{Mean} = \frac{401 + 467 + 489 + 503 + 568}{5} = \frac{2428}{5} = 485.6
\]
2. **Calculate the absolute deviations:**
\[
|401 - 485.6| = 84.6 \\
|467 - 485.6| = 18.6 \\
|489 - 485.6| = 3.4 \\
|503 - 485.6| = 17.4 \\
|568 - 485.6| = 82.4
\]
3. **Calculate the average of these absolute deviations:**
\[
\text{MAD} = \frac{84.6 + 18.6 + 3.4 + 17.4 + 82.4}{5} = \frac{206.4}{5} \approx 41.28
\]
**Conclusion:**
Dataset #1 is more spread out because it has a MAD of 48.2 whereas Dataset #2 has a MAD of 41.28.
### Summary of Answers:
1. **First dataset:** 9.3
2. **Second dataset (Lucia's grades):** 5.5
3. **Third dataset (siblings):** "The mean absolute deviation is 1.1. On average, the number of siblings each child has is about 1.1 siblings from the mean."
4. **Comparison of datasets:** "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."
5. **Comparison of spread:** "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."