COMPARING DIFFERENCES IN CENTERS TO VARIABILITY

Recall that to find the mean absolute deviation (MAD) of a data set, first find the mean of the data. Next, take the absolute value of the difference between the mean and each data point. Finally, find the mean of those absolute values.

Example :

The tables show the number of minutes per day students in a class spend exercising and playing video games. What is the difference of the means as a multiple of the mean absolute deviations ?

Solution : 

Step 1 :

Calculate the mean number of minutes per day exercising.

0 + 7 + 7 + 18 + 20 + 38 + 33 + 24 + 22 + 18 + 11 + 6  =  204

Divide the sum by the number of students.

204 ÷ 12  =  17

Step 2 : 

Calculate the mean absolute deviation for the number of minutes exercising.

|0-17|  =  17

|7-17|  =  10

|7-17|  =  10

|18-17|  =  1

|20-17|  =  3

|38-17|  =  21

|33-17|  =  16

|24-17|  =  7

|22-17|  =  5

|18-17|  =  1

|11-17|  =  6

|6-17|  =  11

Find the mean of the absolute values.

17 + 10 + 10 + 1 + 3 + 21+ 16 + 7 + 5 + 1 + 6 + 11  =  108

Divide the sum by the number of students.

108 ÷ 12  =  9

Step 3 :

Calculate the mean number of minutes per day playing video games. Round to the nearest tenth.

13+18+19+30+32+46+50+34+36+30+23+19  =  350

Divide the sum by the number of students.

350 ÷ 12  ≈  29.2

Step 4 : 

Calculate the mean absolute deviation for the numbers of minutes playing video games.

|13-29.2|  =  16.2

|18-29.2|  =  11.2

|19-29.2|  =  10.2

|30-29.2|  =  0.8

|32-29.2|  =  2.8

|46-29.2|  =  16.8

|50-29.2|  =  20.8

|34-29.2|  =  4.8

|36-29.2|  =  6.8

|30-29.2|  =  0.8

|23-29.2|  =  6.2

|19-29.2|  =  10.2

Find the mean of the absolute values. Round to the nearest tenth.

16.2 + 11.2 + 10.2 + 0.8 + 2.8 + 16.8 + 20.8 + 4.8 + 6.8 + 0.8 + 6.2 + 10.2  =  107.6

Divide the sum by the number of students.

107.6 ÷ 12  ≈  9

Step 5 :

Find the difference in the means.

Subtract the lesser mean from the greater mean.

29.2 - 17  =  12.2

Step 6 :

Write the difference of the means as a multiple of the mean absolute deviations, which are similar but not identical.

Divide the difference of the means by the MAD.

12.2 ÷ 9  ≈  1.36

The means of the two data sets differ by about 1.4 times the variability of the two data sets.

Reflect

The high jumps in inches of the students on two intramural track and field teams are shown below. What is the difference of the means as a multiple of the mean absolute deviations ?

Solution :

About 1.1 times the MAD.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 199)

    Jul 02, 25 07:06 AM

    digitalsatmath268.png
    Digital SAT Math Problems and Solutions (Part - 199)

    Read More

  2. Logarithm Questions and Answers Class 11

    Jul 01, 25 10:27 AM

    Logarithm Questions and Answers Class 11

    Read More

  3. Digital SAT Math Problems and Solutions (Part -198)

    Jul 01, 25 07:31 AM

    digitalsatmath267.png
    Digital SAT Math Problems and Solutions (Part -198)

    Read More