Question

4 from unit 1, lesson 11
the mean of data set a is 43.5 and the mad is 3.7. the mean of data set b is 12.8 and the mad is 4.1.

a. which data set shows greater variability?
explain your reasoning.

b. what differences would you expect to see when comparing the dot plots of the two data sets?

5 from unit 1, lesson 10
select all the distribution shapes for which the mean and median must be about the same.

a bell - shaped

b bimodal

c skewed

d symmetric

e uniform

6 from unit 1, lesson 11
what is the iqr?

a 5 branches

b 7 branches

c 10 branches

d 12 branches

box plot with label number of branches on 10 - year - old trees and axis from 4 to 18: 4, 6, 8, 10, 12, 14, 16, 18

Explanation:

Question 4a

Step1: Recall MAD meaning

MAD (Mean Absolute Deviation) measures variability. Larger MAD means greater variability.

Step2: Compare MAD values

Data set A: MAD = 3.7; Data set B: MAD = 4.1. Since \( 4.1 > 3.7 \), data set B has greater MAD.

The dot plot of data set B (with higher MAD) should have dots more spread out from its mean (12.8) compared to data set A’s dot plot around its mean (43.5). Also, the centers (means) are different: A’s center is ~43.5, B’s is ~12.8.

Step1: Recall mean - median relationship

• Bell - shaped (normal): Symmetric, mean ≈ median.
• Bimodal: Can be symmetric (e.g., two equal peaks symmetrically placed), mean ≈ median.
• Skewed: Mean pulled toward tail, median ≠ mean.
• Symmetric: If symmetric (including bell - shaped, uniform, symmetric bimodal), mean ≈ median.
• Uniform: Symmetric (all values equally likely), mean ≈ median.

Step2: Select valid options

Bell - shaped (A), bimodal (if symmetric, B), symmetric (D), uniform (E) have mean ≈ median. Skewed (C) does not.

Answer:

Data set B shows greater variability because its Mean Absolute Deviation (MAD) of 4.1 is larger than the MAD of data set A (3.7), and MAD measures the variability of data around the mean.

Question 4b