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Shortcut (Assumed Mean) Method

Introduction

The Assumed Mean (shortcut) method is a clever way to simplify calculations of mean, variance and standard deviation when numbers are large or grouped. Instead of working with original values, you pick a convenient number (assumed mean) and work with deviations from it - this reduces arithmetic and avoids large squares. This pattern is important for speed and accuracy in exam-style questions and when dealing with grouped frequency distributions.

Pattern: Shortcut (Assumed Mean) Method

Pattern

The key concept: Choose an assumed mean A, compute deviations d = x - A (or class mark - A), and use those to find mean and SD using smaller numbers.

For ungrouped data (or class marks):
Let A = assumed mean, d_i = x_i - A, N = total frequency (or number of items).
Mean = A + (Σd_i) ÷ N
Variance = [Σ(d_i)² ÷ N] - [(Σd_i ÷ N)²]
SD = √(Variance)

Step-by-Step Example

Question

Marks (grouped as class-marks) and frequencies are: 45 (f=2), 50 (f=3), 55 (f=5), 60 (f=4). Use assumed mean A = 55 to find the mean and standard deviation using the shortcut method.

Solution

  1. Step 1: List data and choose A

    Class marks x: 45, 50, 55, 60. Frequencies f: 2, 3, 5, 4. Choose assumed mean A = 55.

  2. Step 2: Compute deviations d = x - A and f × d

    d: 45-55 = -10; 50-55 = -5; 55-55 = 0; 60-55 = +5.
    f×d: 2×(-10)=-20; 3×(-5)=-15; 5×0=0; 4×5=20.

  3. Step 3: Compute f×d² (for variance)

    d²: 100, 25, 0, 25.
    f×d²: 2×100=200; 3×25=75; 5×0=0; 4×25=100.

  4. Step 4: Sum columns and compute N

    Σf = 2 + 3 + 5 + 4 = 14 = N.
    Σ(f×d) = -20 -15 + 0 + 20 = -15.
    Σ(f×d²) = 200 + 75 + 0 + 100 = 375.

  5. Step 5: Find Mean using assumed mean formula

    Mean = A + (Σf×d) ÷ N = 55 + (-15) ÷ 14 ≈ 55 - 1.071 = 53.93.

  6. Step 6: Find Variance and SD

    Variance = [Σ(f×d²) ÷ N] - [ (Σ(f×d) ÷ N)² ]
    = (375 ÷ 14) - ( (-15 ÷ 14)² )
    = 26.7857 - (1.1489) ≈ 25.6368.
    SD = √25.6368 ≈ 5.06.

  7. Final Answer:

    Mean ≈ 53.93, Standard Deviation ≈ 5.06.

  8. Quick Check:

    Using A reduced arithmetic (d values small). If you recompute mean via direct weighted mean you get same result - confirms correctness ✅

Quick Variations

1. Use A close to central value (median class mark) for smallest deviations.

2. For continuous class-intervals, use class midpoints as x and treat frequencies similarly.

3. For ungrouped large numbers, pick A as a round number near the data to simplify d calculations.

Trick to Always Use

  • Step 1: Choose A near the center (median or mean estimate) so d values are small.
  • Step 2: Work with f×d and f×d² only - you avoid squaring large x values.
  • Step 3: Use Mean = A + (Σf×d) ÷ N, Variance formula as shown, then square root for SD.

Summary

Summary

In the Assumed Mean (Shortcut) Method pattern:

  • Pick an assumed mean A near the data center to reduce arithmetic.
  • Compute deviations d = x - A (or midpoint - A) and their frequency-weighted sums.
  • Mean = A + (Σf×d) ÷ N; Variance = [Σ(f×d²) ÷ N] - [ (Σ(f×d) ÷ N)² ]; SD = √(Variance).
  • This method is faster and less error-prone for grouped data or large numbers.
  • Always do a quick direct-check (weighted mean or a rough estimate) to confirm results.

Practice

(1/5)
1. The marks of students (class marks) and their frequencies are: 40 (f=2), 50 (f=3), 60 (f=5). Use A = 50. Find the mean using the assumed mean method.
easy
A. 53.00
B. 54.00
C. 55.00
D. 52.00

Solution

  1. Step 1: Compute deviations

    d = x - A → for 40: -10, for 50: 0, for 60: +10.

  2. Step 2: Multiply by frequency

    f×d → (2×-10)=-20; (3×0)=0; (5×10)=50.

  3. Step 3: Sum and apply formula

    Σf×d = (-20 + 0 + 50) = 30; Σf = 2+3+5 = 10.
    Mean = A + (Σf×d ÷ Σf) = 50 + (30 ÷ 10) = 53.00.

  4. Final Answer:

    Mean = 53.00 → Option A.

  5. Quick Check:

    Weighted average is closer to higher-frequency classes (50 & 60) → 53 is reasonable ✅

Hint: Pick A near center; Mean = A + (Σf×d)/Σf.
Common Mistakes: Using x instead of d when summing deviations.
2. Given x: 10, 20, 30 with frequencies f: 1, 3, 2. Take A = 20. Find the mean using the assumed mean method.
easy
A. 21.80
B. 21.67
C. 23.34
D. 24.50

Solution

  1. Step 1: Compute deviations

    d = x - A → (10-20)=-10, (20-20)=0, (30-20)=+10.

  2. Step 2: Multiply by frequency

    1×(-10)=-10, 3×0=0, 2×10=20 → Σf×d = (-10 + 0 + 20) = 10; Σf = 6.

  3. Step 3: Apply formula

    Mean = A + (Σf×d ÷ Σf) = 20 + (10 ÷ 6) = 20 + 1.6667 = 21.67.

  4. Final Answer:

    Mean ≈ 21.67 → Option B.

  5. Quick Check:

    More weight on 20 (freq 3) pulls mean slightly above 20 → 21.67 fits ✅

Hint: Use fractional result for final rounding instead of rounding early.
Common Mistakes: Dividing Σf×d by wrong total frequency.
3. Class marks: 30, 40, 50, 60; frequencies: 3, 5, 4, 2. Take A = 45. Find the mean using the assumed mean method.
easy
A. 43.57
B. 45.00
C. 46.57
D. 47.00

Solution

  1. Step 1: Compute d = x - A

    30-45=-15; 40-45=-5; 50-45=+5; 60-45=+15.

  2. Step 2: Multiply by frequency

    3×(-15)=-45; 5×(-5)=-25; 4×5=20; 2×15=30 → Σf×d = -45-25+20+30 = -20. Σf = 14.

  3. Step 3: Apply formula

    Mean = A + (Σf×d ÷ Σf) = 45 + (-20 ÷ 14) = 45 - 1.4286 = 43.57.

  4. Final Answer:

    Mean ≈ 43.57 → Option A.

  5. Quick Check:

    More weight on lower classes (30 & 40) pulls mean below 45 → 43.57 logical ✅

Hint: Keep negative signs for lower classes when summing f×d.
Common Mistakes: Dropping negative sign in Σf×d computation.
4. Given data x: 5, 10, 15, 20 with f: 4, 6, 8, 2. Using A = 12.5, find the standard deviation (SD) using the shortcut method.
medium
A. 4.74
B. 5.10
C. 4.58
D. 6.12

Solution

  1. Step 1: Compute d = x - A

    d = -7.5, -2.5, +2.5, +7.5.

  2. Step 2: Compute f×d and f×d²

    f×d sum = (4×-7.5)+(6×-2.5)+(8×2.5)+(2×7.5) = -30 -15 +20 +15 = -10.
    f×d² sum = (4×56.25)+(6×6.25)+(8×6.25)+(2×56.25) = 225 + 37.5 + 50 + 112.5 = 425.

  3. Step 3: Variance & SD

    Σf = 20.
    Variance = (Σf×d² ÷ Σf) - (Σf×d ÷ Σf)² = (425 ÷ 20) - (-10 ÷ 20)² = 21.25 - 0.25 = 21.00.
    SD = √21.00 ≈ 4.58.

  4. Final Answer:

    SD ≈ 4.58 → Option C.

  5. Quick Check:

    Variance 21 and SD ≈4.58 matches the spread of d values ✅

Hint: Compute Σf×d and Σf×d² carefully; then apply variance formula.
Common Mistakes: Squaring Σf×d instead of (Σf×d ÷ Σf) when computing variance.
5. The wages (₹) and frequencies are: 10 (3), 20 (5), 30 (7), 40 (5). Take A = 25. Find the SD using the shortcut method.
medium
A. 10.20
B. 9.30
C. 8.70
D. 10.05

Solution

  1. Step 1: Compute d = x - A

    d = -15, -5, +5, +15.

  2. Step 2: Compute f×d and f×d²

    f×d = 3(-15)+5(-5)+7(5)+5(15) = -45 -25 +35 +75 = 40.
    Σf×d² = 3(225)+5(25)+7(25)+5(225) = 675 +125 +175 +1125 = 2100.

  3. Step 3: Variance & SD

    Σf = 20.
    Variance = (2100 ÷ 20) - (40 ÷ 20)² = 105 - 4 = 101.
    SD = √101 ≈ 10.05.

  4. Final Answer:

    SD ≈ 10.05 → Option D.

  5. Quick Check:

    Large spread around A = 25 gives SD ≈ 10 - consistent ✅

Hint: Use A near centre and compute Σf×d and Σf×d² to get variance quickly.
Common Mistakes: Using wrong Σf when dividing squared-sum term.

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