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7.130 Mean absolute deviation Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Grade 7 Math worksheet provides structured practice on 7.130 Mean absolute deviation. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

7.130 Mean absolute deviation - In multiple-choice items, students select the mean absolute deviation for a data set and choose which data set is more variable based on MAD. (CCSS 7.SP.B.4)

Whether you are an educator preparing classroom handouts, a tutor reinforcing weekly topics, or a parent supporting homeschool learning, this printable resource gives students the deliberate practice required to achieve subject fluency and master state standards.

Grade

Grade 7

Subject

Math

Domain

Math Curriculum

Formats

PDF & DOCX

Sample Practice Problems

Preview the types of problems and questions covered in this worksheet

Below are sample questions representative of the exercises included on this worksheet and related 7.130 Mean absolute deviation skill assessments:

Problem 1Difficulty: Mastery

Fatima is analyzing a 6-value data set. Consider the 6-value data set: 9, 11, 13, 15, x, and 24-x. The mean of this data set is 12. The MAD of the data set is 1.67. If x is the greater of the two possible values that make this MAD, what is x?

A14

B13

C11

D10

Solution & Explanation:Because x and 24-x are symmetric around 12, the mean stays 12. MAD = (3+1+1+3+absolute value of (x−12)+absolute value of ((24-x)−12))/6 = (8+2 × absolute value of (x−12))/6. Set (8+2 × absolute value of (x−12))/6 = 1.67. Then 8+2 × absolute value of (x−12) = 6·1.67 = 10, so absolute value of (x−12) = 1. The two solutions are x = 12±1; the greater is 13.

Problem 2Difficulty: Mastery

Tariq is analyzing a 6-value data set. Consider the 6-value data set: 9, 11, 13, 15, x, and 24-x. The mean of this data set is 12. The MAD of the data set is 3.33. If x is the greater of the two possible values that make this MAD, what is x?

A18

Solution & Explanation:Because x and 24-x are symmetric around 12, the mean stays 12. MAD = (3+1+1+3+absolute value of (x−12)+absolute value of ((24-x)−12))/6 = (8+2 × absolute value of (x−12))/6. Set (8+2 × absolute value of (x−12))/6 = 3.33. Then 8+2 × absolute value of (x−12) = 6·3.33 = 20, so absolute value of (x−12) = 6. The two solutions are x = 12±6; the greater is 18.

Problem 3Difficulty: Mastery

Uma is analyzing a 6-value data set. Consider the 6-value data set: 10, 12, 14, 16, x, and 26-x. The mean of this data set is 13. The MAD of the data set is 2.67. If x is the greater of the two possible values that make this MAD, what is x?

A19

D17

Solution & Explanation:Because x and 26-x are symmetric around 13, the mean stays 13. MAD = (3+1+1+3+absolute value of (x−13)+absolute value of ((26-x)−13))/6 = (8+2 × absolute value of (x−13))/6. Set (8+2 × absolute value of (x−13))/6 = 2.67. Then 8+2 × absolute value of (x−13) = 6·2.67 = 16, so absolute value of (x−13) = 4. The two solutions are x = 13±4; the greater is 17.

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Grade 7 Math students. It aligns with educational standard 7.130 (Practice and review Grade 7 math skill 7.130: Mean Absolute Deviation.).