Part (a) - Calculating the Range for Both Courses

To calculate the range, we use the formula:

$\text{Range} = \text{Maximum Value} - \text{Minimum Value}$

For the Traditional Course:

Scores:
71.2, 69.6, 80.1, 67.9, 86.4, 78.2, 56.4, 80.6, 80.7, 70.4, 63.5, 70.7, 60.2

• Maximum value: 86.4
• Minimum value: 56.4

$\text{Range}_{\text{Traditional}} = 86.4 - 56.4 = 30$

For the Flipped Course:

Scores:
77.0, 72.3, 64.2, 71.4, 78.9, 90.7, 79.4, 76.3, 81.5, 70.5, 91.1, 77.3, 77.4

• Maximum value: 91.1
• Minimum value: 64.2

$\text{Range}_{\text{Flipped}} = 91.1 - 64.2 = 26.9$

Conclusion:

• The Traditional course has a range of 30.
• The Flipped course has a range of 26.9.

Since $30 > 26.9$, the Traditional course has more dispersion in exam scores based on the range.

Would you like details on the next parts or have any questions?

Related Questions:

1. How would the standard deviation compare between these two sets of scores?
2. How does range differ from other measures of dispersion like variance or interquartile range?
3. What are some limitations of using the range to describe data variability?
4. Can we use box plots to visualize the dispersion of these two courses? How would they look?
5. How would changing one of the extreme values affect the range for each course?

Tip: The range is very sensitive to extreme values (outliers), so other measures like variance or standard deviation might give a better sense of dispersion when extreme scores are present.