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Frequency Distribution (Grouped Data)

Introduction

When data are given in class intervals with frequencies, calculating the Mean, Variance, and Standard Deviation (SD) requires working with grouped data formulas. In this pattern, we use midpoints (class marks) and frequency weights to compute spread and dispersion.

This method is essential in aptitude and statistics questions involving marks distribution, income groups, and frequency tables where data are summarized rather than listed individually.

Pattern: Frequency Distribution (Grouped Data)

Pattern

The key concept: Use midpoints as representative values of each class and apply weighted formulas using frequencies.

For grouped data, 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

The following table shows the marks of 50 students. Find the standard deviation of their marks.

Solution

  1. Step 1: Find class marks (x)

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

  2. Step 2: Compute the mean (x̄)

    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: Compute deviations and squared deviations

    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: Apply the variance formula

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

  5. Step 5: Find the Standard Deviation

    SD = √154.56 = 12.43

  6. Final Answer:

    Mean = 28.8, Variance = 154.56, SD = 12.43

  7. Quick Check:

    Higher frequencies around mid classes → moderate spread (SD ≈ 12.4). ✅

Quick Variations

1. Use assumed mean (A) or step-deviation method when data values are large or repetitive.

2. For class intervals with equal widths, you can simplify calculations using a common class width (h).

3. For unequal class widths, always multiply by the exact frequency and difference from the actual mean.

Trick to Always Use

  • Step 1: Always calculate midpoints before finding deviations.
  • Step 2: Use the step-deviation formula if the mean or class values are large.
  • Step 3: Remember that frequency (f) acts as a weight for each squared deviation.

Summary

Summary

In the Frequency Distribution (Grouped Data) pattern:

  • Data are organized into class intervals and frequencies.
  • Use class marks as representative values.
  • Mean = (Σf×x)/Σf, Variance = Σf(x - x̄)²/Σf, SD = √Variance.
  • The assumed mean or step-deviation method simplifies large calculations.
  • Smaller SD → tighter clustering of data around the mean (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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