Big Ideas Math: Modeling Real Life, Grade 6
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Big Ideas Math: Modeling Real Life, Grade 6 View details
5. Mean Absolute Deviation
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Exercise 20 Page 443

The mean absolute deviation is the average of the absolute values of the differences between the mean and each value in the data set. Start by calculating the mean of the given set of numbers.

Mean Absolute Deviation: 320
Interpretation: The data, on average, are 320 units away from the mean of 12.

Practice makes perfect
The mean absolute deviation (MAD) is the average of the absolute values of the differences between the mean and each value in the data set. Let's see the given data set. 1/4, 5/8, 3/8, 3/4, 1/2 To simplify the calculations, we will adjust the data set by rewriting all fractions to have the same denominator of 8. Beginning with 14 and 34, we will multiply both the numerator and denominator of these fractions by 2. 1/4 = 1 * 2/4 * 2 = 2/8 3/4 = 3 * 2/4 * 2 = 6/8Now, let's multiply the numerator and the denominator of 12 by 4. 1/2 = 1 * 4/2 * 4 = 4/8 This gives us new data set that is easier to work with! 2/8, 5/8, 3/8, 6/8, 4/8 We are ready to calculate the MAD! We will continue by calculating the mean of the given set of numbers. We can see that there is 5 values in our data set. Let's calculate the mean!
Mean = Sum of values/Number of values
Mean = 2/8+ 5/8+ 3/8+ 6/8+ 4/8/5
Mean = 2+5+3+6+4/8/5
Mean = 20/8/5
Mean = 20/8* 5
Mean = 20/40
Mean = 20Ă· 20/40Ă· 20
Mean = 1/2
We found that the mean of the given data set is 12. We will write it as a fraction with a denominator of 8.
Mean = 1/2
Mean = 1 * 4/2 * 4
Mean = 4/8
We found that the mean of the given data set is 48. We are ready to calculate the MAD. As previously stated, the MAD of a set of data is the average of the absolute values of the differences between the mean and each value in the data set. | x- x_1|+| x- x_2|+...+| x- x_n|/n In this formula, x_1,...,x_n are the values in the set of data, x is the mean, and n is the number of values. We already know that x= 48 and n= 5. Let's use a table to find the sum of the absolute values of the differences.
x_i x-x_i |x-x_i|
2/8 4/8- 2/8=2/8 |2/8|=2/8
5/8 4/8- 5/8=- 1/8 |- 1/8|=1/8
3/8 4/8- 3/8=1/8 |1/8|=1/8
6/8 4/8- 6/8=- 2/8 |- 2/8|=2/8
4/8 4/8- 4/8=0 |0|=0
Sum of Values 2/8 + 1/8+1/8 + 2/8 + 0 = 6/8
Finally, we need to divide by 5.
MAD = 6/8/5
MAD = 6/8 * 5
MAD = 6/40
MAD = 6 Ă· 2/40 Ă· 2
MAD = 3/20
A MAD of 320 indicates that the data, on average, are 320 of a unit away from the mean of 12.