Exercises: Informally assess the degree of visual overlap of two numerical data distributions - Common Core Math - Grade 7

A

Recall / Warm-Up

1.

The data set is 4, 6, 7, 9, 14. What is the mean?

A.

$7$

B.

$8$

C.

$9$

D.

$10$

2.

Evaluate $|8.5 - 12|$ .

A.

$-3.5$

B.

$3.5$

C.

$20.5$

D.

$4.0$

3.

Compute the mean of the data set: 5, 7, 7, 9, 10, 10.

B

Fluency Practice

1.

The data set is 3, 5, 6, 7, 9, and the mean is 6. Compute the Mean Absolute Deviation (MAD).

2.

Group A has a mean of 24. Group B has a mean of 18. The average MAD for both groups is 3. Compute the MAD-multiple (difference in means divided by average MAD).

3.

Two data distributions have a MAD-multiple of 2.1. Which statement is most accurate?

A.

The distributions completely separate — there is no overlap between them.

B.

The distributions are noticeably separated — most values in one group exceed most values in the other, though some overlap may still exist near the boundary.

C.

The distributions have about the same center — the difference is not meaningful.

D.

The distributions substantially overlap — most values fall in the same region.

4.

The data set is 8, 10, 12, 14, 16, 18, 20, 22. Compute the MAD.

5.

A MAD-multiple of 0.5 most likely indicates which of these?

A.

Noticeable separation — most values in the two groups are distinctly different.

B.

Substantial overlap — the two distributions mostly occupy the same region of the number line.

C.

No variability within either group.

D.

The two groups have identical means.

C

Varied Practice

D

Word Problems

1.

A teacher surveys two reading groups about the number of books each student read over the summer. Class A (6 students): 5, 6, 7, 7, 8, 9. Class B: mean = 10 books, MAD = 1.0 book.

1.

Find the mean and MAD for Class A.

2.

Using Class B's mean (10 books) and MAD (1.0 book), compute the MAD-multiple and write one complete sentence describing the separation between the two classes.

2.

A basketball team has a mean height of 182 cm and a soccer team has a mean height of 170 cm. The average MAD for both teams is 4 cm. Compute the MAD-multiple.

3.

Group P (reaction time): mean = 5.0 sec, MAD = 1.5 sec. Group Q (reaction time): mean = 8.0 sec, MAD = 1.5 sec. (a) Compute the MAD-multiple using the average MAD. (b) Write a complete comparison statement that includes specific numbers and a description of the separation.

4.

A statistical analysis finds that basketball players' mean height is 12 cm greater than soccer players' mean height, with a MAD-multiple of 3.0. A student writes: 'Basketball players are better athletes than soccer players because they are taller.' What is wrong with this statement?

A.

Nothing — a higher mean always indicates better performance.

B.

The MAD-multiple of 3.0 is too high to support any conclusion.

C.

Statistics describes differences in distributions — it does not make value judgments about which group is better or worse.

D.

The student computed the MAD-multiple incorrectly.

E

Error Analysis

$Two-column contrast card: left shows raw deviations summing to 0 (wrong), right shows absolute deviations summing to 12 giving MAD 2.4 (correct).$

1.

Jordan finds the mean of the data set {8, 10, 12, 14, 16} to be 12. He then computes deviations from the mean:

$8 - 12 = -4$
$10 - 12 = -2$
$12 - 12 = 0$
$14 - 12 = +2$
$16 - 12 = +4$

He averages the deviations: $(-4 + (-2) + 0 + 2 + 4) \div 5 = 0 \div 5 = 0$ .
He concludes: "MAD = 0, so there is no variability in this data set."

What error did Jordan make?

A.

Jordan computed the mean incorrectly.

B.

Jordan forgot to take the absolute value of each deviation before averaging — without absolute values, positive and negative deviations cancel, always giving a sum near zero.

C.

Jordan should have subtracted the mean from each value in a different order.

D.

Jordan should have used the range instead of this procedure.

2.

Two data distributions have these statistics:

Group A: mean = 180 cm, range = 20 cm, MAD = 4 cm
Group B: mean = 170 cm, range = 18 cm, MAD = 4 cm

Nina computes the MAD-multiple as:
$(180 - 170) \div \text{average range} = 10 \div 19 \approx 0.53$

She concludes: "With a ratio of only 0.53, the separation is barely noticeable."

What error did Nina make, and what is the correct answer?

A.

Nina subtracted the means incorrectly.

B.

Nina used range as the denominator instead of MAD, making the ratio too small and leading to the wrong interpretation.

C.

Nina should have used the median instead of the mean.

D.

Nina's calculation is correct — the separation is barely noticeable.

F

Challenge

1.

Here are test scores for two classes. Class M: 65, 68, 72, 75, 78, 80, 82, 84. Class N: 78, 80, 83, 85, 88, 90, 92, 96. (a) Compute the mean and MAD for each class. (b) Compute the MAD-multiple using the average MAD. (c) Write a complete interpretation statement describing the separation.

2.