Table of Contents
Exercise 14.3 Solutions
Class X Mathematics – State Council of Educational Research and Training, Telangana, Hyderabad
1. Electricity Consumption Problem
| Monthly consumption | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 | 185-205 |
|---|---|---|---|---|---|---|---|
| Number of consumers | 4 | 5 | 13 | 20 | 14 | 8 | 4 |
Solution:
Median:
First, we calculate cumulative frequencies:
| Class | Frequency | Cumulative Frequency |
|---|---|---|
| 65-85 | 4 | 4 |
| 85-105 | 5 | 9 |
| 105-125 | 13 | 22 |
| 125-145 | 20 | 42 |
| 145-165 | 14 | 56 |
| 165-185 | 8 | 64 |
| 185-205 | 4 | 68 |
Median class is where cumulative frequency ≥ n/2 = 34 → 125-145
Using median formula:
Median = L + [(n/2 – CF)/f] × h
Where L = 125, n = 68, CF = 22, f = 20, h = 20
Median = 125 + [(34 – 22)/20] × 20 = 125 + 12 = 137
Mean:
Using assumed mean method with A = 135:
| Class | Midpoint (x) | Frequency (f) | d = (x – A)/h | f × d |
|---|---|---|---|---|
| 65-85 | 75 | 4 | -3 | -12 |
| 85-105 | 95 | 5 | -2 | -10 |
| 105-125 | 115 | 13 | -1 | -13 |
| 125-145 | 135 | 20 | 0 | 0 |
| 145-165 | 155 | 14 | 1 | 14 |
| 165-185 | 175 | 8 | 2 | 16 |
| 185-205 | 195 | 4 | 3 | 12 |
| Total | 7 | |||
Mean = A + (Σfd/Σf) × h = 135 + (7/68) × 20 ≈ 137.06
Mode:
Modal class is 125-145 (highest frequency = 20)
Using mode formula:
Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
Where L = 125, f₁ = 20, f₀ = 13, f₂ = 14, h = 20
Mode = 125 + [(20 – 13)/(40 – 13 – 14)] × 20 ≈ 125 + (7/13) × 20 ≈ 135.77
Comparison: Mean (137.06) > Median (137) > Mode (135.77)
2. Median Problem with Missing Frequencies
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
|---|---|---|---|---|---|---|
| Frequency | 5 | x | 20 | 15 | y | 5 |
Given median = 28.5, n = 60
Solution:
Total observations: 5 + x + 20 + 15 + y + 5 = 60 ⇒ x + y = 15
Median class is where cumulative frequency ≥ n/2 = 30 → 20-30
Using median formula:
28.5 = 20 + [(30 – (5 + x))/20] × 10
8.5 = (25 – x)/2 ⇒ 17 = 25 – x ⇒ x = 8
Since x + y = 15 ⇒ y = 7
Solution: x = 8, y = 7
3. Insurance Policy Holders Age Distribution
| Age (in years) | Below 20 | Below 25 | Below 30 | Below 35 | Below 40 | Below 45 | Below 50 | Below 55 | Below 60 |
|---|---|---|---|---|---|---|---|---|---|
| Number of policy holders | 2 | 6 | 24 | 45 | 78 | 89 | 92 | 98 | 100 |
Solution:
First convert to frequency distribution:
| Class | Frequency | Cumulative Frequency |
|---|---|---|
| 18-20 | 2 | 2 |
| 20-25 | 4 | 6 |
| 25-30 | 18 | 24 |
| 30-35 | 21 | 45 |
| 35-40 | 33 | 78 |
| 40-45 | 11 | 89 |
| 45-50 | 3 | 92 |
| 50-55 | 6 | 98 |
| 55-60 | 2 | 100 |
Median class is where cumulative frequency ≥ n/2 = 50 → 35-40
Using median formula:
Median = 35 + [(50 – 45)/33] × 5 ≈ 35 + 0.76 ≈ 35.76 years
4. Leaves Length Measurement
| Length (in mm) | 118-126 | 127-135 | 136-144 | 145-153 | 154-162 | 163-171 | 172-180 |
|---|---|---|---|---|---|---|---|
| Number of leaves | 3 | 5 | 9 | 12 | 5 | 4 | 2 |
Solution:
Convert to continuous classes as suggested:
| Class | Frequency | Cumulative Frequency |
|---|---|---|
| 117.5-126.5 | 3 | 3 |
| 126.5-135.5 | 5 | 8 |
| 135.5-144.5 | 9 | 17 |
| 144.5-153.5 | 12 | 29 |
| 153.5-162.5 | 5 | 34 |
| 162.5-171.5 | 4 | 38 |
| 171.5-180.5 | 2 | 40 |
Median class is where cumulative frequency ≥ n/2 = 20 → 144.5-153.5
Using median formula:
Median = 144.5 + [(20 – 17)/12] × 9 = 144.5 + 2.25 = 146.75 mm
5. Neon Lamps Life Time
| Life time (in hours) | 1500-2000 | 2000-2500 | 2500-3000 | 3000-3500 | 3500-4000 | 4000-4500 | 4500-5000 |
|---|---|---|---|---|---|---|---|
| Number of lamps | 14 | 56 | 60 | 86 | 74 | 62 | 48 |
Solution:
Calculate cumulative frequencies:
| Class | Frequency | Cumulative Frequency |
|---|---|---|
| 1500-2000 | 14 | 14 |
| 2000-2500 | 56 | 70 |
| 2500-3000 | 60 | 130 |
| 3000-3500 | 86 | 216 |
| 3500-4000 | 74 | 290 |
| 4000-4500 | 62 | 352 |
| 4500-5000 | 48 | 400 |
Median class is where cumulative frequency ≥ n/2 = 200 → 3000-3500
Using median formula:
Median = 3000 + [(200 – 130)/86] × 500 ≈ 3000 + 406.98 ≈ 3406.98 hours
6. Surnames Letters Distribution
| Number of letters | 1-4 | 4-7 | 7-10 | 10-13 | 13-16 | 16-19 |
|---|---|---|---|---|---|---|
| Number of surnames | 6 | 30 | 40 | 16 | 4 | 4 |
Solution:
Median:
First make classes continuous and calculate cumulative frequencies:
| Class | Frequency | Cumulative Frequency |
|---|---|---|
| 0.5-4.5 | 6 | 6 |
| 4.5-7.5 | 30 | 36 |
| 7.5-10.5 | 40 | 76 |
| 10.5-13.5 | 16 | 92 |
| 13.5-16.5 | 4 | 96 |
| 16.5-19.5 | 4 | 100 |
Median class is where cumulative frequency ≥ n/2 = 50 → 7.5-10.5
Median = 7.5 + [(50 – 36)/40] × 3 = 7.5 + 1.05 = 8.55
Mean:
Using midpoint method:
| Class | Midpoint (x) | Frequency (f) | f × x |
|---|---|---|---|
| 1-4 | 2.5 | 6 | 15 |
| 4-7 | 5.5 | 30 | 165 |
| 7-10 | 8.5 | 40 | 340 |
| 10-13 | 11.5 | 16 | 184 |
| 13-16 | 14.5 | 4 | 58 |
| 16-19 | 17.5 | 4 | 70 |
| Total | 100 | 832 | |
Mean = Σfx/Σf = 832/100 = 8.32
Mode:
Modal class is 7-10 (highest frequency = 40)
Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
Where L = 7, f₁ = 40, f₀ = 30, f₂ = 16, h = 3
Mode = 7 + [(40 – 30)/(80 – 30 – 16)] × 3 ≈ 7 + (10/34) × 3 ≈ 7.88
7. Students Weight Distribution
| Weight (in kg) | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |
|---|---|---|---|---|---|---|---|
| Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |
Solution:
Calculate cumulative frequencies:
| Class | Frequency | Cumulative Frequency |
|---|---|---|
| 40-45 | 2 | 2 |
| 45-50 | 3 | 5 |
| 50-55 | 8 | 13 |
| 55-60 | 6 | 19 |
| 60-65 | 6 | 25 |
| 65-70 | 3 | 28 |
| 70-75 | 2 | 30 |
Median class is where cumulative frequency ≥ n/2 = 15 → 55-60
Using median formula:
Median = 55 + [(15 – 13)/6] × 5 ≈ 55 + 1.67 ≈ 56.67 kg
