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Combined Standard Deviation (Two Data Sets)

Introduction

Sometimes two separate groups of observations are merged and you must find the combined standard deviation. The combined SD uses each group's size, mean and variance - it accounts for within-group spread and the shift between group means.

This pattern is important for problems that combine class scores, batch measurements, or sample results from two subgroups.

Pattern: Combined Standard Deviation (Two Data Sets)

Pattern

The key concept: Combined variance = weighted sum of (each group's variance + squared difference between group mean and combined mean), divided by total size. SD = square root of combined variance.

Let group 1 have size n₁, mean x̄₁, SD σ₁.
Let group 2 have size n₂, mean x̄₂, SD σ₂.

Combined mean:
x̄ = (n₁ × x̄₁ + n₂ × x̄₂) ÷ (n₁ + n₂)

Combined variance formula:
σ² = [ n₁ × (σ₁² + (x̄₁ - x̄)²) + n₂ × (σ₂² + (x̄₂ - x̄)²) ] ÷ (n₁ + n₂)
Combined standard deviation:
σ = √σ²

Step-by-Step Example

Question

Class A: n₁ = 10 students, mean = 50, SD = 4.
Class B: n₂ = 15 students, mean = 55, SD = 5.
Find the combined mean and combined standard deviation for all 25 students.

Solution

  1. Step 1: Compute combined mean x̄

    x̄ = (n₁ × x̄₁ + n₂ × x̄₂) ÷ (n₁ + n₂)
    = (10 × 50 + 15 × 55) ÷ 25
    = (500 + 825) ÷ 25 = 1,325 ÷ 25 = 53.

  2. Step 2: Compute squared deviations of group means from combined mean

    (x̄₁ - x̄) = 50 - 53 = -3 → (x̄₁ - x̄)² = 9.
    (x̄₂ - x̄) = 55 - 53 = 2 → (x̄₂ - x̄)² = 4.

  3. Step 3: Form n × (σ² + (x̄ - x̄)²) for each group

    For group A: n₁ × (σ₁² + (x̄₁ - x̄)²) = 10 × (4² + 9) = 10 × (16 + 9) = 10 × 25 = 250.
    For group B: n₂ × (σ₂² + (x̄₂ - x̄)²) = 15 × (5² + 4) = 15 × (25 + 4) = 15 × 29 = 435.

  4. Step 4: Sum and divide by total size to get combined variance

    Total = 250 + 435 = 685.
    Combined variance σ² = 685 ÷ (10 + 15) = 685 ÷ 25 = 27.4.

  5. Step 5: Take square root for combined SD

    Combined SD σ = √27.4 ≈ 5.24.

  6. Final Answer:

    Combined Mean = 53
    Combined SD ≈ 5.24

  7. Quick Check:

    Combined mean (53) lies between 50 and 55 - closer to Class B (more students).
    Combined SD (≈5.24) lies between 4 and 5 - slightly higher due to mean difference ✅

Quick Variations

1. If group sizes are equal (n₁ = n₂), combined mean = simple average of means.

2. If group means are equal (x̄₁ = x̄₂), combined variance simplifies to [n₁σ₁² + n₂σ₂²] ÷ (n₁ + n₂).

3. For more than two groups, extend the formula by summing nᵢ(σᵢ² + (x̄ᵢ - x̄)²) for all groups and dividing by total N.

Trick to Always Use

  • Step 1: Always calculate the combined mean first - it’s needed in the formula.
  • Step 2: Remember each bracket includes two parts: within-group variance and between-group difference.
  • Step 3: Larger groups (higher n) influence the combined SD more.

Summary

Summary

In the Combined Standard Deviation (Two Data Sets) pattern:

  • Combined Mean: (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂)
  • Combined Variance: [ n₁(σ₁² + (x̄₁ - x̄)²) + n₂(σ₂² + (x̄₂ - x̄)²) ] ÷ (n₁ + n₂)
  • Combined SD = √(Combined Variance)
  • Formula accounts for both within-group spread and between-group mean differences.

Practice

(1/5)
1. Class A has 20 students with Mean = 60 and SD = 4. Class B has 30 students with Mean = 65 and SD = 5. Find the combined Mean of both classes.
easy
A. 63.0
B. 62.5
C. 64.0
D. 61.5

Solution

  1. Step 1: Identify values

    n₁ = 20, x̄₁ = 60; n₂ = 30, x̄₂ = 65.

  2. Step 2: Apply combined mean formula

    x̄ = (n₁×x̄₁ + n₂×x̄₂) ÷ (n₁ + n₂) = (20×60 + 30×65) ÷ 50 = (1,200 + 1,950) ÷ 50 = 3,150 ÷ 50 = 63.0.

  3. Final Answer:

    Combined Mean = 63.0 → Option A.

  4. Quick Check:

    63 lies between 60 and 65 and is closer to 65 (larger group) - consistent ✅

Hint: Use weighted average: larger group pulls the combined mean toward its mean.
Common Mistakes: Taking simple average of the two means instead of weighting by group sizes.
2. Two groups have means 40 and 50 with equal sizes of 10 each. Find the combined mean.
easy
A. 44
B. 45
C. 48
D. 46

Solution

  1. Step 1: Note equal sizes

    When group sizes are equal, combined mean = (mean₁ + mean₂) ÷ 2.

  2. Step 2: Compute

    (40 + 50) ÷ 2 = 45.

  3. Final Answer:

    Combined Mean = 45 → Option B.

  4. Quick Check:

    45 is midway between 40 and 50 ✅

Hint: If group sizes equal, take simple average of means.
Common Mistakes: Applying weights when sizes are equal (unnecessary).
3. Group A: n₁ = 10, Mean = 70, SD = 3; Group B: n₂ = 20, Mean = 75, SD = 4. Find the combined Mean.
easy
A. 72.5
B. 74.0
C. 73.33
D. 76.0

Solution

  1. Step 1: Apply weighted mean formula

    x̄ = (n₁×x̄₁ + n₂×x̄₂) ÷ (n₁ + n₂) = (10×70 + 20×75) ÷ 30.

  2. Step 2: Compute

    (700 + 1,500) ÷ 30 = 2,200 ÷ 30 = 73.33 (rounded to 2 decimals).

  3. Final Answer:

    Combined Mean ≈ 73.33 → Option C.

  4. Quick Check:

    Value lies between 70 and 75 and is closer to 75 (larger group) ✅

Hint: Larger group’s mean has greater influence on combined mean.
Common Mistakes: Using simple average instead of weighted average.
4. Class X: n₁ = 12, Mean = 50, SD = 3. Class Y: n₂ = 18, Mean = 55, SD = 4. Find the combined Standard Deviation (rounded to 2 decimals).
medium
A. 3.80
B. 4.00
C. 4.50
D. 4.38

Solution

  1. Step 1: Compute combined mean

    x̄ = (12×50 + 18×55) ÷ 30 = (600 + 990) ÷ 30 = 1,590 ÷ 30 = 53.

  2. Step 2: Compute group contributions

    For X: σ₁² = 3² = 9; (x̄₁ - x̄)² = (50 - 53)² = 9 → term = 12×(9 + 9) = 12×18 = 216.
    For Y: σ₂² = 4² = 16; (x̄₂ - x̄)² = (55 - 53)² = 4 → term = 18×(16 + 4) = 18×20 = 360.

  3. Step 3: Combined variance and SD

    Sum = 216 + 360 = 576. Combined variance = 576 ÷ 30 = 19.2. Combined SD = √19.2 ≈ 4.38.

  4. Final Answer:

    Combined SD ≈ 4.38 → Option D.

  5. Quick Check:

    Combined SD ≈ 4.38 lies between 4 and 5 and is increased by the between-group difference - reasonable ✅

Hint: Include both within-group variance and squared mean differences when combining.
Common Mistakes: Ignoring the between-group term (x̄ᵢ - x̄)².
5. Two departments have the following data: Dept A: n₁ = 25, Mean = 80, SD = 5; Dept B: n₂ = 15, Mean = 70, SD = 4. Find the combined SD (rounded to 2 decimals).
medium
A. 6.72
B. 5.25
C. 6.00
D. 5.50

Solution

  1. Step 1: Combined mean

    x̄ = (25×80 + 15×70) ÷ 40 = (2,000 + 1,050) ÷ 40 = 3,050 ÷ 40 = 76.25.

  2. Step 2: Compute group contributions

    Dept A: σ₁² = 25; (x̄₁ - x̄)² = (80 - 76.25)² = 3.75² = 14.0625 → term = 25×(25 + 14.0625) = 25×39.0625 = 976.5625.
    Dept B: σ₂² = 16; (x̄₂ - x̄)² = (70 - 76.25)² = (-6.25)² = 39.0625 → term = 15×(16 + 39.0625) = 15×55.0625 = 825.9375.

  3. Step 3: Combined variance and SD

    Sum = 976.5625 + 825.9375 = 1,802.5. Combined variance = 1,802.5 ÷ 40 = 45.0625. Combined SD = √45.0625 ≈ 6.72.

  4. Final Answer:

    Combined SD ≈ 6.72 → Option A.

  5. Quick Check:

    Result is closer to the larger group’s SD but increased due to mean difference - makes sense ✅

Hint: Compute weighted sum of (variance + squared mean-difference) then divide by total N and square-root.
Common Mistakes: Forgetting to multiply each group's term by its size before summing.

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