The following are the scores of ten students in Sections A, B and C. Compute for the range, variance and standard deviation of each section. Which section has the most consistent score? 

Section A - 25, 24, 24, 20, 20, 20, 16, 12, 10, 7
Section B - 25, 24, 24, 20, 18, 18, 17, 10, 9, 7
Section C - 25, 25, 25, 22, 21, 21, 21, 18, 18, 18

R= HV-LV
s²=Σ(x-x̄)²/n-
s=√s²


Let's compute the **range**, **variance**, and **standard deviation** for each section, following these steps:

### **Step 1: Compute the Range (R)**
The range is the difference between the highest value (HV) and the lowest value (LV).

\[
R = HV - LV
\]

### **Step 2: Compute the Variance (s²)**
The formula for variance is:

\[
s^2 = \frac{\sum(x - \bar{x})^2}{n-1}
\]

Where:
- \( x \) represents each score,
- \( \bar{x} \) is the mean (average) score,
- \( n \) is the number of scores.

### **Step 3: Compute the Standard Deviation (s)**
The standard deviation is the square root of the variance:

\[
s = \sqrt{s^2}
\]

### **Section A: 25, 24, 24, 20, 20, 20, 16, 12, 10, 7**
- **Range (R):**
  \[
  R_A = 25 - 7 = 18
  \]
- **Mean ( \( \bar{x} \) ):**
  \[
  \bar{x}_A = \frac{25 + 24 + 24 + 20 + 20 + 20 + 16 + 12 + 10 + 7}{10} = \frac{178}{10} = 17.8
  \]
- **Variance (s²):**
  \[
  s_A^2 = \frac{(25 - 17.8)^2 + (24 - 17.8)^2 + \cdots + (7 - 17.8)^2}{9} = \frac{(52.84 + 37.44 + \cdots + 118.81)}{9} = 43.51
  \]
- **Standard Deviation (s):**
  \[
  s_A = \sqrt{43.51} \approx 6.6
  \]

### **Section B: 25, 24, 24, 20, 18, 18, 17, 10, 9, 7**
- **Range (R):**
  \[
  R_B = 25 - 7 = 18
  \]
- **Mean ( \( \bar{x} \) ):**
  \[
  \bar{x}_B = \frac{25 + 24 + 24 + 20 + 18 + 18 + 17 + 10 + 9 + 7}{10} = \frac{172}{10} = 17.2
  \]
- **Variance (s²):**
  \[
  s_B^2 = \frac{(25 - 17.2)^2 + (24 - 17.2)^2 + \cdots + (7 - 17.2)^2}{9} = \frac{(60.84 + 46.24 + \cdots + 104.04)}{9} = 40.71
  \]
- **Standard Deviation (s):**
  \[
  s_B = \sqrt{40.71} \approx 6.38
  \]

### **Section C: 25, 25, 25, 22, 21, 21, 21, 18, 18, 18**
- **Range (R):**
  \[
  R_C = 25 - 18 = 7
  \]
- **Mean ( \( \bar{x} \) ):**
  \[
  \bar{x}_C = \frac{25 + 25 + 25 + 22 + 21 + 21 + 21 + 18 + 18 + 18}{10} = \frac{214}{10} = 21.4
  \]
- **Variance (s²):**
  \[
  s_C^2 = \frac{(25 - 21.4)^2 + (25 - 21.4)^2 + \cdots + (18 - 21.4)^2}{9} = \frac{(12.96 + 12.96 + \cdots + 11.56)}{9} = 3.60
  \]
- **Standard Deviation (s):**
  \[
  s_C = \sqrt{3.60} \approx 1.90
  \]

### **Summary of Results**
| Section   | Range (R) | Variance (s²) | Standard Deviation (s) |
|-----------|------------|---------------|------------------------|
| Section A | 18         | 43.51         | 6.60                   |
| Section B | 18         | 40.71         | 6.38                   |
| Section C | 7          | 3.60          | 1.90                   |

### **Conclusion:**
Section C has the **most consistent scores**, as it has the smallest range, variance, and standard deviation.

---

Would you like further explanation on any part of the calculations? Here are some related questions:

1. How does the range affect the consistency of scores?
2. Why is the standard deviation important for comparing consistency?
3. What would happen if the sample size was larger?
4. How can outliers influence the standard deviation?
5. Can we apply these methods to non-numerical data?

**Tip:** A smaller standard deviation indicates that the data points are closer to the mean, showing more consistency.

The following are the scores of ten students in Sections A, B and C. Compute for the range, variance and standard deviation of each section. Which section has the most consistent score?

Section A - 25, 24, 24, 20, 20, 20, 16, 12, 10, 7
Section B - 25, 24, 24, 20, 18, 18, 17, 10, 9, 7
Section C - 25, 25, 25, 22, 21, 21, 21, 18, 18, 18

R= HV-LV
s²=Σ(x-x̄)²/n-
s=√s²

Calculate the range, variance, and standard deviation of student scores in Sections A, B, and C. This example uses core statistical concepts to assess consistency, ideal for students in Grades 9-12 learning about measures of dispersion.

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