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Vikentia [17]
3 years ago
8

28, 45, 12, 34, 36, 45, 19, 20

Mathematics
1 answer:
Alborosie3 years ago
7 0

1) Mean of the set of data: 29.88

2) Mean absolute deviation: 10.13

3) See explanation

Step-by-step explanation:

1)

The mean of a set of data it is calculated as

\bar x = \frac{1}{N}\sum x_i

where

N is the number of data in the set

x_i is the value of each point in the  dataset

For the set of data in this problem, we have:

x_i =[28, 45, 12, 34, 36, 45, 19, 20]

And the number of values is

N = 8

Therefore, we can calculate the mean:

\bar x = \frac{1}{8}(28+ 45+ 12+ 34+ 36+ 45+ 19+ 20)=\frac{239}{8}=29.88

2)

The mean absolute deviation of a set of data is given by

\delta = \frac{1}{N}\sum |x_i-\bar x|

where

N is the number of values in the dataset

x_i are the single values

\bar x is the mean of the dataset

The dataset here is

x_i =[28, 45, 12, 34, 36, 45, 19, 20]

The mean, calculated in part 1), is

\bar x = 29.88

And

N = 8

Therefore the mean absolute deviation is

\delta = \frac{1}{8}(|28-29.88|+|45-29.88|+|12-29.88|+|34-29.88|+|36-29.88|+|45-29.88|+|19-29.88|+|20-29.88|)=\frac{81}{8}=10.13

3)

The mean of a dataset is the sum of the single values of the dataset divided by the number of values. The mean represents the value \bar x for which, if the dataset would have N values all equal to \bar x, the sum of the values of the dataset would be the same as the sum of the actual values.

The mean absolute deviation for a set of data represents the average of the absolute deviations of the single points from the mean of the dataset. This quantity gives a measure of the "dispersion" of the points around the mean: in fact, the larger the mean absolute deviation is, the more the points are "spread" around the mean of the dataset. Instead, if the mean absolute deviation is small, it means that the points are closer to the mean value.

Learn more about mean and spread of a distribution:

brainly.com/question/6073431

brainly.com/question/8799684

brainly.com/question/4625002

#LearnwithBrainly

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