The mean of the following dataset is 8. Find the mean absolute deviation of the dataset.

Question

Design a compelling and abstract visualization that conceptualizes the concept of Mean Absolute Deviation in the context of analyzing datasets. Include differently sized clusters of data points to represent different datasets. These should be arranged in various spatial orientations to illustrate the varying spreads and deviations around a central mean. Showcase a collection of points that might represent counts of strawberries in packages, subtly hinting at the quantities given in the final example (10, 12, 14, 15, 16, 17). Remember to maintain an abstract nature, devoid of text and numbers.

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
.

Find the mean absolute deviation of the following dataset.

{21,25,25,26,26,27,28,30} (1 point)
The mean absolute deviation is

.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
.
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
.

The mean absolute deviation of Dataset #1 is 5.6. The mean absolute deviation of Dataset #2 is 11.8. Which dataset is less spread out?(1 point)
Dataset #
is less spread out.

Answer 1

Dataset #1 is less spread out, as it has a smaller mean absolute deviation compared to Dataset #2.

Answer 2

answer all the qestions

Answer 3

I apologize for the oversight. Let's calculate the mean absolute deviation and compare the datasets as requested:

1. Mean absolute deviation of {5.8, 7, 8.1, 9.2, 9.9}:
Mean = (5.8 + 7 + 8.1 + 9.2 + 9.9) / 5 = 40 / 5 = 8
Absolute deviations from the mean: {2.2, 1, 0.1, 1.2, 1.9}
Mean absolute deviation = (2.2 + 1 + 0.1 + 1.2 + 1.9) / 5 = 6.4 / 5 = 1.28

2. Mean absolute deviation of {21, 25, 25, 26, 26, 27, 28, 30}:
Mean = (21 + 25 + 25 + 26 + 26 + 27 + 28 + 30) / 8 = 212 / 8 = 26.5
Absolute deviations from the mean: {4.5, 0.5, 0.5, 0.5, 0.5, 0.5, 1.5, 3.5}
Mean absolute deviation = (4.5 + 0.5 + 0.5 + 0.5 + 0.5 + 0.5 + 1.5 + 3.5) / 8 = 12.5 / 8 = 1.5625

3. Mean absolute deviation of Tessa's data {10, 12, 14, 15, 16, 17}:
Mean = (10 + 12 + 14 + 15 + 16 + 17) / 6 = 84 / 6 = 14
Absolute deviations from the mean: {4, 2, 0, 1, 2, 3}
Mean absolute deviation = (4 + 2 + 0 + 1 + 2 + 3) / 6 = 12 / 6 = 2

4. Comparing Dataset #1 and Dataset #2:
Dataset #1: Mean absolute deviation = 5.6
Dataset #2: Mean absolute deviation = 11.8

The mean absolute deviation of Dataset #1 (5.6) is smaller than that of Dataset #2 (11.8), indicating that Dataset #1 is less spread out.

Therefore:
1. The mean absolute deviation of {5.8, 7, 8.1, 9.2, 9.9} is 1.28
2. The mean absolute deviation of {21, 25, 25, 26, 26, 27, 28, 30} is 1.5625
3. The mean absolute deviation of Tessa's data {10, 12, 14, 15, 16, 17} is 2
4. Dataset #1 ({5, 8, 9, 10, 12, 12, 14}) is more spread out, while Dataset #2 ({5, 6, 8, 9, 10, 10, 22}) is less spread out.