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Shortcut (Assumed Mean) Method

Introduction

Assumed Mean (shortcut) method एक smart तरीका है जिससे बड़े या grouped numbers के लिए mean, variance और standard deviation निकालना आसान हो जाता है। Original values पर directly काम करने की बजाय आप एक convenient number (assumed mean) चुनते हैं और उसी से deviations calculate करते हैं - इससे calculations छोटे numbers में हो जाती हैं और बड़े squares से बचा जा सकता है। यह pattern exam-style questions और grouped frequency distributions में तेज़ और accurate calculation के लिए बहुत useful है।

Pattern: Shortcut (Assumed Mean) Method

Pattern

मुख्य concept: कोई assumed mean A चुनें, फिर deviations d = x - A (या class mark - A) निकालें, और इनके basis पर mean और SD छोटे numbers के साथ निकालें।

Ungrouped data (या class marks) के लिए:
A = assumed mean, d_i = x_i - A, N = total frequency (या 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 (class-marks) और frequencies दिए गए हैं: 45 (f=2), 50 (f=3), 55 (f=5), 60 (f=4)। Assumed mean A = 55 लें और shortcut method से mean और standard deviation निकालें।

Solution

  1. Step 1: Data list करें और A चुनें

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

  2. Step 2: Deviations d = x - A और f × d निकालें

    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: Variance calculation के लिए f×d² निकालें

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

  4. Step 4: Columns का sum और N निकालें

    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: Assumed mean formula से Mean निकालें

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

  6. Step 6: Variance और SD निकालें

    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:

    A चुनने से arithmetic आसान हो गया (d values छोटे रहे)। Direct weighted mean से भी यही result आता है - answer सही है ✅

Quick Variations

1. A को central value (median class mark) के पास रखें ताकि deviations छोटे हों।

2. Continuous class-intervals में class midpoints को x मानकर frequencies इसी तरह use करें।

3. Ungrouped बड़े numbers में A को किसी पास के round number जैसा रखें ताकि d quickly निकले।

Trick to Always Use

  • Step 1: A को हमेशा data center के पास चुनें ताकि d values small रहें।
  • Step 2: बस f×d और f×d² पर काम करें - बड़े x values को square करने की ज़रूरत ही नहीं पड़ेगी।
  • Step 3: Mean = A + (Σf×d) ÷ N और variance formula apply करें, फिर SD के लिए square root लें।

Summary

Summary

In the Assumed Mean (Shortcut) Method pattern:

  • A को data center के पास चुनें ताकि arithmetic कम हो।
  • d = x - A (या midpoint - A) और उनकी frequency-weighted sums calculate करें।
  • Mean = A + (Σf×d) ÷ N; Variance = [Σ(f×d²) ÷ N] - [ (Σ(f×d) ÷ N)² ]; SD = √(Variance)
  • Grouped data और बड़े numbers के लिए यह method तेज़ और कम error वाला है।
  • हमेशा एक quick direct-check (जैसे weighted mean estimate) करके result verify करें।

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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