Question

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, , 38, , 36, , 57, 43

Find the mean absolute deviation (MAD) of the data set.

_____ thousand dollars

Answer 1

This question is not correct

Correct Question:

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

Answer:

5.6 thousand dollars

Explanation:

Mean Absolute Deviation is mean of or the average of absolute deviation of a set of data from it's central point.

We were given the following sets of data in thousands( the yearly salaries of high school cheer leading coaches)

43, 41, 38, 36, and 57

We would determine the Mean Absolute Deviation in the following steps:

Step 1

The first step is calculate the mean of the salaries

= (43+ 41 + 38+ 36+ 57) ÷ 5

= 215 ÷ 5

= 43

The mean of the salaries is 43 in thousand dollars.

Step 2

The second step would be to find the absolute difference of the values by subtracting the mean of the salaries from each of the salaries of the high school teachers , find the absolute values, after which we would sum up this absolute values together and divide by the number of the observed values

The number of observed values = 5

[|43 - 43| + |41 - 43| + |38- 43| + |36 - 43| + |57- 43| ]÷ 5

= 0 + |- 2|+ |- 5 |+|-7| + |-14| ÷ 5

= (0 + 2+ 5 +7 + 14) ÷ 5

= (28 ÷ 5) in thousand dollars

= 5.6 in thousand dollars

Therefore, the mean absolute deviation (MAD) of the data set in thousand dollars is 5.6.

Answer 2

Explanation:

I think the data is repeated

The given data should be

41, 38, 36, 57, and 43

So, we want to find the absolute mean deviation, absolute here means we will discard the negative sign and make it possible

So, let find the mean of data

X~ = Σx / N

x is the given data

And N is the number of data given

X~ = (41 + 38 + 36 + 57 + 43) / 5

X~ = 215 / 5

X~ = 43

The mean is 43, so we and to find now the data given deviate from the mean.

X....... X-X~.......….|X-X~|

36..... —7............... 7

38..... —5.............. 5

41...... —2............... 2

43....... 0.........., ...0

57...... 14............. 14

So, the sum of the mean deviation is expected to be 0

ΣX-X~ = -7 -5 - 2 + 0 + 14 = 0

As expected

But the sum of the absolute mean is

Σ|X-X~| = 7 + 5 + 2 + 0 + 14 = 28

The mean deviation is given as

M•D = Σ|X-X~| / N

M•D = 28 / 5

The mean absolute deviation is 5.6 thousand dollars