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Frequency Distribution (Grouped Data)

Introduction

जब data class intervals और उनकी frequencies के रूप में दिया जाता है, तब Mean, Variance और Standard Deviation (SD) निकालने के लिए grouped data वाले formulas का उपयोग किया जाता है। इस pattern में हम midpoints (class marks) और frequency weights का उपयोग करके data का spread और dispersion निकालते हैं।

यह तरीका aptitude और statistics के उन questions में जरूरी होता है जहाँ marks distribution, income groups या frequency tables दी होती हैं और data को individually list नहीं किया जाता।

Pattern: Frequency Distribution (Grouped Data)

Pattern

मुख्य concept: हर class को एक midpoint से represent करें और frequencies को weights की तरह use करें।

Grouped data के लिए, Mean (x̄) = (Σf×x) ÷ Σf
Variance (σ²) = [Σf(x - x̄)²] ÷ Σf
Standard Deviation (σ) = √[Σf(x - x̄)² ÷ Σf]

Step-by-Step Example

Question

नीचे दी गई table 50 students के marks दिखाती है। इनके marks का standard deviation निकालें।

Solution

  1. Step 1: Class marks (x) निकालें

    Class marks (midpoints) = (Lower + Upper) ÷ 2 → 5, 15, 25, 35, 45

  2. Step 2: Mean (x̄) निकालें

    Σf = 50 Σf×x = (5×5) + (8×15) + (12×25) + (15×35) + (10×45) = 1,440 Mean x̄ = 1,440 ÷ 50 = 28.8

  3. Step 3: Deviations और squared deviations निकालें

    xf(x - 28.8)(x - 28.8)²f(x - 28.8)²
    55-23.8566.42,832.0
    158-13.8190.41,523.2
    2512-3.814.4172.8
    35156.238.4576.0
    451016.2262.42,624.0

  4. Step 4: Variance formula apply करें

    Σf(x - x̄)² = 7,728.0 Variance = 7,728 ÷ 50 = 154.56

  5. Step 5: Standard Deviation निकालें

    SD = √154.56 = 12.43

  6. Final Answer:

    Mean = 28.8, Variance = 154.56, SD = 12.43

  7. Quick Check:

    Mid classes के आसपास ज़्यादा frequencies → moderate spread (SD ≈ 12.4). ✅

Quick Variations

1. Large numbers या repetitive data में assumed mean (A) या step-deviation method use करें।

2. Equal width intervals हों तो common class width (h) calculations simplify कर देता है।

3. Unequal widths में हमेशा actual mean से difference और exact frequency apply करें।

Trick to Always Use

  • Step 1: Deviations निकालने से पहले midpoints ज़रूर calculate करें।
  • Step 2: Values बड़े हों तो step-deviation formula calculation आसान बनाता है।
  • Step 3: Frequency (f) हर squared deviation के लिए weight की तरह काम करती है - इसे कभी ignore न करें।

Summary

Summary

In the Frequency Distribution (Grouped Data) pattern:

  • Data class intervals और frequencies में दिया होता है।
  • Class marks को representative value माना जाता है।
  • Mean = (Σf×x)/Σf, Variance = Σf(x - x̄)²/Σf, SD = √Variance
  • Assumed mean या step-deviation बड़े calculations को simplify करते हैं।
  • छोटा SD → data mean के आसपास ज़्यादा cluster है (more consistency)।

Practice

(1/5)
1. The marks of 50 students are grouped as below. Find the mean marks using class midpoints.<br><br>
MarksFrequency (f)
0-105
10-2010
20-3015
30-4012
40-508
easy
A. 26.60
B. 25.40
C. 26.00
D. 27.20

Solution

  1. Step 1: Find class midpoints (x)

    Midpoints: 5, 15, 25, 35, 45.

  2. Step 2: Multiply each midpoint by its frequency

    Σf×x = (5×5) + (10×15) + (15×25) + (12×35) + (8×45) = 25 + 150 + 375 + 420 + 360 = 1,330.

  3. Step 3: Compute mean

    Σf = 50 → Mean = 1,330 ÷ 50 = 26.60.

  4. Final Answer:

    Mean = 26.60 → Option A.

  5. Quick Check:

    The mean is near the 20-30 class (central) and between 25 and 27 → 26.6 is reasonable ✅

Hint: Use class midpoints as representative values for each interval and divide Σ(f×x) by total frequency.
Common Mistakes: Miscomputing Σ(f×x) or dividing by number of classes instead of total frequency.
2. The table shows the income distribution of families. Find the approximate mean income (in thousands).<br><br>
Income (₹ in thousands)Frequency (f)
0-104
10-206
20-3010
30-408
40-502
easy
A. 24.33
B. 26.00
C. 28.00
D. 30.00

Solution

  1. Step 1: Compute midpoints (x)

    Midpoints: 5, 15, 25, 35, 45 (in thousands).

  2. Step 2: Multiply f×x and sum

    Σf×x = (4×5) + (6×15) + (10×25) + (8×35) + (2×45) = 20 + 90 + 250 + 280 + 90 = 730.

  3. Step 3: Compute mean

    Σf = 30 → Mean = 730 ÷ 30 = 24.33 (thousand ₹).

  4. Final Answer:

    Mean ≈ 24.33 → Option A.

  5. Quick Check:

    Most weight is around 20-30 → mean ≈ 24.3 thousand fits ✅

Hint: Always divide Σf×x by total frequency to get the weighted mean.
Common Mistakes: Using class limits instead of midpoints.
3. The following distribution gives the age of workers. Find the mean age.<br><br>
Age (years)Frequency
20-303
30-405
40-507
50-605
easy
A. 40.00
B. 42.00
C. 43.50
D. 46.00

Solution

  1. Step 1: Compute midpoints (x)

    Midpoints: 25, 35, 45, 55.

  2. Step 2: Multiply f×x

    Σf×x = (3×25) + (5×35) + (7×45) + (5×55) = 75 + 175 + 315 + 275 = 840.

  3. Step 3: Compute mean

    Σf = 20 → Mean = 840 ÷ 20 = 42.00 years.

  4. Final Answer:

    Mean age = 42.00 → Option B.

  5. Quick Check:

    Majority of workers are in 40-50 → mean ≈ 42 years makes sense ✅

Hint: Use midpoints to simplify grouped data calculations.
Common Mistakes: Taking lower limits as midpoints.
4. The following shows the distribution of monthly expenses (in thousands). Find the Standard Deviation.<br><br>
Expense (₹ in thousands)Frequency (f)
0-52
5-106
10-158
15-204
medium
A. 4.0
B. 5.2
C. 5.8
D. 4.50

Solution

  1. Step 1: Find class midpoints

    Midpoints: 2.5, 7.5, 12.5, 17.5 (thousand ₹).

  2. Step 2: Compute mean = (Σf×x)/Σf

    Σf×x = (2×2.5)+(6×7.5)+(8×12.5)+(4×17.5) = 5 + 45 + 100 + 70 = 220.
    Σf = 20 → Mean = 220 ÷ 20 = 11.00 (thousand ₹).

  3. Step 3: Compute Σf(x - x̄)²

    Deviations: 2.5-11=-8.5 → sq = 72.25 → ×2 = 144.5
    7.5-11=-3.5 → sq = 12.25 → ×6 = 73.5
    12.5-11=1.5 → sq = 2.25 → ×8 = 18.0
    17.5-11=6.5 → sq = 42.25 → ×4 = 169.0
    Σf(x-x̄)² = 144.5 + 73.5 + 18 + 169 = 405.0.

  4. Step 4: Variance & SD

    Variance = 405.0 ÷ 20 = 20.25 → SD = √20.25 = 4.50 (thousand ₹).

  5. Final Answer:

    SD ≈ 4.50 → Option D.

  6. Quick Check:

    Values cluster around mean (11) → SD ≈ 4.5 thousand is reasonable ✅

Hint: Find mean first, then deviations and variance → SD = √Variance.
Common Mistakes: Forgetting to multiply squared deviations by frequencies.
5. Given frequency distribution of salaries (in thousands), calculate the mean.<br><br>
Salary (₹ in thousands)Frequency
10-203
20-306
30-405
40-504
50-602
medium
A. 34.5
B. 35.0
C. 33.00
D. 36.0

Solution

  1. Step 1: Midpoints

    Midpoints: 15, 25, 35, 45, 55 (thousand ₹).

  2. Step 2: Compute Σf×x

    Σf×x = (3×15)+(6×25)+(5×35)+(4×45)+(2×55) = 45 + 150 + 175 + 180 + 110 = 660.

  3. Step 3: Compute mean

    Σf = 20 → Mean = 660 ÷ 20 = 33.00 (thousand ₹).

  4. Final Answer:

    Mean salary = 33.00 → Option C.

  5. Quick Check:

    Mean lies in 30-40 class as expected → 33 thousand is correct ✅

Hint: Always use midpoints and total frequency to find weighted mean.
Common Mistakes: Using wrong midpoint or missing a class in calculation.

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