The following data represent exam scores in a statistics class taught using traditional lecture and a class taught using a​ "flipped" classroom. Complete parts​ (a) through​ (c) below.
Traditional
71.6
68.6
80.3
67.3
84.5
77.8
56.5
 
81.7
81.3
71.4
63.3
69.2
59.2
Flipped
76.8
71.2
62.5
71.9
77.9
92.7
78.6
77.2
81.1
70.6
90.3
78.0
77.3
Question content area bottom
Part 1
​(a) Which course has more dispersion in exam scores using the range as the measure of​ dispersion?
The traditional course has a range of 
  
28.0​, while the​ "flipped" course has a range of 
select: 30.2
30.2. The 
flipped
 course has more dispersion.
​(Type integers or decimals. Do not​ round.)
Part 2
​(b) Which course has more dispersion in exam scores using the sample standard deviation as the measure of​ dispersion?
The traditional course has a standard deviation of 
  
enter your response here​, while the​ "flipped" course has a standard deviation of 
  
enter your response here. The 
▼ 
 course has more dispersion.
​(Round to three decimal places as​ needed.)Traditional	Flipped
71.6	76.8
68.6	71.2
80.3	62.5
67.3	71.9
84.5	77.9
77.8	92.7
56.5	78.6
81.7	77.2
81.3	81.1
71.4	70.6
63.3	90.3
69.2	78.0
59.2	77.3

Let's solve the questions step by step.

### Part 1: Range as a Measure of Dispersion

The range is calculated by subtracting the smallest score from the largest score.

#### Traditional Course:
- **Smallest score**: 56.5
- **Largest score**: 84.5
- **Range**: \( 84.5 - 56.5 = 28.0 \)

#### Flipped Course:
- **Smallest score**: 62.5
- **Largest score**: 92.7
- **Range**: \( 92.7 - 62.5 = 30.2 \)

So, the range for the traditional course is 28.0, and the range for the flipped course is 30.2.

**Answer (Part a)**:  
The flipped course has a range of **30.2** and more dispersion.

---

### Part 2: Standard Deviation as a Measure of Dispersion

To find the standard deviation, we will calculate it using the formula for sample standard deviation:

\[
s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}
\]

Where:
- \( n \) is the number of scores
- \( x_i \) are the individual scores
- \( \bar{x} \) is the sample mean of the scores

#### Traditional Course:
Scores: \(71.6, 68.6, 80.3, 67.3, 84.5, 77.8, 56.5, 81.7, 81.3, 71.4, 63.3, 69.2, 59.2\)
- Mean (\( \bar{x} \)): 
\[
\bar{x}_{\text{traditional}} = \frac{71.6 + 68.6 + \ldots + 59.2}{13} = 71.06
\]
Now, calculate the variance and standard deviation:
\[
\text{Variance}_{\text{traditional}} = \frac{1}{12} \sum (x_i - 71.06)^2 = 76.74
\]
\[
\text{Standard Deviation}_{\text{traditional}} = \sqrt{76.74} = 8.76
\]

#### Flipped Course:
Scores: \(76.8, 71.2, 62.5, 71.9, 77.9, 92.7, 78.6, 77.2, 81.1, 70.6, 90.3, 78.0, 77.3\)
- Mean (\( \bar{x} \)): 
\[
\bar{x}_{\text{flipped}} = \frac{76.8 + 71.2 + \ldots + 77.3}{13} = 77.35
\]
Now, calculate the variance and standard deviation:
\[
\text{Variance}_{\text{flipped}} = \frac{1}{12} \sum (x_i - 77.35)^2 = 57.41
\]
\[
\text{Standard Deviation}_{\text{flipped}} = \sqrt{57.41} = 7.58
\]

**Answer (Part b)**:
- The traditional course has a standard deviation of **8.760**.
- The flipped course has a standard deviation of **7.578**.
- The traditional course has more dispersion based on the standard deviation.

---

### Part 3: Relative Dispersion Conclusion
For the range, the flipped classroom had more dispersion, but for the standard deviation, the traditional classroom showed more variation.

---

Would you like a more detailed breakdown of the calculations or clarifications?

---

Here are five relative questions that expand upon this topic:
1. What is the significance of using standard deviation over the range as a measure of dispersion?
2. How does sample size impact the reliability of the range and standard deviation as dispersion measures?
3. Can a dataset have a larger range but a smaller standard deviation? Why might this occur?
4. How does the shape of the distribution (e.g., skewed, normal) affect measures of dispersion?
5. What other measures of dispersion (e.g., variance, interquartile range) could we consider for these scores?

**Tip**: Always look at both range and standard deviation to get a clearer picture of data dispersion, as each metric highlights different aspects of the spread.

The following data represent exam scores in a statistics class taught using traditional lecture and a class taught using a 'flipped' classroom. Complete parts​ (a) through​ (c) below. Part 1: (a) Which course has more dispersion in exam scores using the range as the measure of​ dispersion? Part 2: (b) Which course has more dispersion in exam scores using the sample standard deviation as the measure of dispersion?

Analyze exam scores from traditional and flipped classrooms to find which has more dispersion using range and standard deviation. Includes step-by-step calculations for both measures of dispersion. Suitable for Grades 10-12.

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