McGraw Hill Glencoe Algebra 1, 2012
MH
McGraw Hill Glencoe Algebra 1, 2012 View details
Standardized Test Practice

Exercise 2 Page 820

How do we find the mean absolute deviation of a data set?

F

Practice makes perfect

We are given a table that shows the number of calories in twelve different snacks.

Number of Calories in Snacks
122 91 149 121
64 138 342 72
179 105 99 114

We are asked to find the mean absolute deviation. Let's recall the formula to find it. MAD = | x-x_1| + | x-x_2| + ... +| x-x_n|/n In this formula n is the number of values, x_n are the data values, and x is the mean of the data. Therefore, we need to find the mean of the data first. To do so, we will add all the values and divide the sum by 12.

x = (122+91+149+121+64+138+342+72+179+105+99+114)/12
x = 1596/12
x = 133

Now we can find the absolute value of the difference between each data value and the mean. Let's do it!

x_i |x - x_i| Simplify
122 |133 - 122| 11
91 |133 - 91| 42
149 |133 - 149| 16
121 |133 - 121| 12
64 |133 - 64| 69
138 |133 - 138| 5
342 |133 - 342| 209
72 |133 - 72| 61
179 |133 - 179| 46
105 |133 - 105| 28
99 |133 - 99| 34
114 |133 - 114| 19

Now let's add all of these values together and divide the sum by 12 to find the mean absolute deviation.

MAD = (11+42+16+12+69+5+209+61+46+28+34+19)/12
MAD = 552/12
MAD = 46

The mean absolute deviation of the calories of each snack is 46 calories. Therefore, the correct answer is F.