Math Problem Statement

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

Solution

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=HVLVR = HV - LV

Step 2: Compute the Variance (s²)

The formula for variance is:

s2=(xxˉ)2n1s^2 = \frac{\sum(x - \bar{x})^2}{n-1}

Where:

  • xx represents each score,
  • xˉ\bar{x} is the mean (average) score,
  • nn is the number of scores.

Step 3: Compute the Standard Deviation (s)

The standard deviation is the square root of the variance:

s=s2s = \sqrt{s^2}

Section A: 25, 24, 24, 20, 20, 20, 16, 12, 10, 7

  • Range (R): RA=257=18R_A = 25 - 7 = 18
  • Mean ( xˉ\bar{x} ): xˉA=25+24+24+20+20+20+16+12+10+710=17810=17.8\bar{x}_A = \frac{25 + 24 + 24 + 20 + 20 + 20 + 16 + 12 + 10 + 7}{10} = \frac{178}{10} = 17.8
  • Variance (s²): sA2=(2517.8)2+(2417.8)2++(717.8)29=(52.84+37.44++118.81)9=43.51s_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): sA=43.516.6s_A = \sqrt{43.51} \approx 6.6

Section B: 25, 24, 24, 20, 18, 18, 17, 10, 9, 7

  • Range (R): RB=257=18R_B = 25 - 7 = 18
  • Mean ( xˉ\bar{x} ): xˉB=25+24+24+20+18+18+17+10+9+710=17210=17.2\bar{x}_B = \frac{25 + 24 + 24 + 20 + 18 + 18 + 17 + 10 + 9 + 7}{10} = \frac{172}{10} = 17.2
  • Variance (s²): sB2=(2517.2)2+(2417.2)2++(717.2)29=(60.84+46.24++104.04)9=40.71s_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): sB=40.716.38s_B = \sqrt{40.71} \approx 6.38

Section C: 25, 25, 25, 22, 21, 21, 21, 18, 18, 18

  • Range (R): RC=2518=7R_C = 25 - 18 = 7
  • Mean ( xˉ\bar{x} ): xˉC=25+25+25+22+21+21+21+18+18+1810=21410=21.4\bar{x}_C = \frac{25 + 25 + 25 + 22 + 21 + 21 + 21 + 18 + 18 + 18}{10} = \frac{214}{10} = 21.4
  • Variance (s²): sC2=(2521.4)2+(2521.4)2++(1821.4)29=(12.96+12.96++11.56)9=3.60s_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): sC=3.601.90s_C = \sqrt{3.60} \approx 1.90

Summary of Results

SectionRange (R)Variance (s²)Standard Deviation (s)
Section A1843.516.60
Section B1840.716.38
Section C73.601.90

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.

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Math Problem Analysis

Mathematical Concepts

Statistics
Descriptive Statistics
Measures of Dispersion

Formulas

Range: R = HV - LV
Variance: s² = Σ(x - x̄)² / (n - 1)
Standard Deviation: s = √s²

Theorems

Measure of Dispersion (Variance and Standard Deviation)

Suitable Grade Level

Grades 9-12