Math, asked by 9337795477, 9 months ago

The following data points represent the yearly salaries of high school cheerleading coaches in Dakota County (in thousands of dollars). \qquad 41,38, 36 , 57, 4341,38,36,57,4341, comma, 38, comma, 36, comma, 57, comma, 43 Find the mean absolute deviation (MAD) of the data set.

Answers

Answered by amitnrw
22

Given : yearly salaries of high school cheerleading coaches in Dakota County

To find : the mean absolute deviation (MAD) of the data set.

Solution:

Data given

41, 38, 36 , 57, 43

Mean = (41 + 38 + 36 + 57 + 43)/5

= 215/5

= 43

mean absolute deviation (MAD)   = (1/5) ∑ | Mean - xi |      i = 1 to 5

| 43 - 41| = 2

| 43 - 38 | = 5

| 43 - 36| = 7

| 43 - 57} = 14

| 43 - 43 | =0

mean absolute deviation (MAD)   = (1/5) (2 + 5 + 7 + 14 + 0)

=28/5

= 5.6

mean absolute deviation (MAD) of the data set. = 5.6

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Answered by lundse61
0

Answer:The mean is 434343 thousand dollars.

Step-by-step explanation:First, let's add up all of the salaries.

\qquad41+38+36+57+43=\blue{215}41+38+36+57+43=21541, plus, 38, plus, 36, plus, 57, plus, 43, equals, start color #6495ed, 215, end color #6495ed thousand dollars

Now we need to divide the total by the number of coaches.

\qquad \dfrac{\blue{215}}{5} =~ ?

5

215

= ?start fraction, start color #6495ed, 215, end color #6495ed, divided by, 5, end fraction, equals, space, question mark

The mean is 434343 thousand dollars.

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