Question 1
The following table shows the ages of the patients admitted in a hospital during a year:
| Age (years) | 5–15 | 15–25 | 25–35 | 35–45 | 45–55 | 55–65 |
|---|---|---|---|---|---|---|
| No. of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Find the mean and mode of the data. Compare the two measures.
Solution:
Total frequency (Σf) = 6 + 11 + 21 + 23 + 14 + 5 = 80
Class marks (x): 10, 20, 30, 40, 50, 60
f × x = 60 + 220 + 630 + 920 + 700 + 300 = 2830
Mean = Σ(fx) / Σf = 2830 / 80 = 35.38 years
Modal class = 35–45
l = 35, h = 10, f₁ = 23, f₀ = 21, f₂ = 14
Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
= 35 + [(23 − 21) / (46 − 21 − 14)] × 10
= 35 + (2 / 11) × 10 = 36.82 years
Interpretation:
Mean = 35.38 years, Mode = 36.82 years.
Both are close, showing most patients are around 35–37 years old.
Question 2
Lifetimes (in hours) of 225 electrical components:
| Lifetime (hours) | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 | 100–120 |
|---|---|---|---|---|---|---|
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Find the modal lifetime.
Solution:
Modal class = 60–80
l = 60, h = 20, f₁ = 61, f₀ = 52, f₂ = 38
Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
= 60 + [(61 − 52) / (122 − 52 − 38)] × 20
= 60 + (9 / 32) × 20
= 60 + 5.625 = 65.63 hours
Question 3
Monthly household expenditure of 200 families:
| Expenditure (₹) | 1000–1500 | 1500–2000 | 2000–2500 | 2500–3000 | 3000–3500 | 3500–4000 | 4000–4500 | 4500–5000 |
|---|---|---|---|---|---|---|---|---|
| No. of families | 24 | 40 | 33 | 28 | 30 | 22 | 16 | 7 |
Find the mode and mean monthly expenditure.
Solution:
Modal class = 1500–2000
l = 1500, h = 500, f₁ = 40, f₀ = 24, f₂ = 33
Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h
= 1500 + [(40 − 24) / (80 − 24 − 33)] × 500
= 1500 + (16 / 23) × 500 = 1500 + 347.83 = ₹1847.83
Class marks (x): 1250, 1750, 2250, 2750, 3250, 3750, 4250, 4750
Σ(fx) = 532500, Σf = 200
Mean = Σ(fx)/Σf = 532500 / 200 = ₹2662.50
Interpretation:
Mean (₹2662.50) > Mode (₹1847.83), showing the data is right-skewed (some families spend much more).
Question 4
Teacher–student ratio (students per teacher):
| Students per teacher | 15–20 | 20–25 | 25–30 | 30–35 | 35–40 | 40–45 | 45–50 | 50–55 |
|---|---|---|---|---|---|---|---|---|
| No. of states/U.T. | 3 | 8 | 9 | 10 | 3 | 0 | 0 | 2 |
Find the mode and mean of this data.
Solution:
Modal class = 30–35
l = 30, h = 5, f₁ = 10, f₀ = 9, f₂ = 3
Mode = 30 + [(10 − 9) / (20 − 9 − 3)] × 5
= 30 + (1 / 8) × 5 = 30.63
Class marks (x): 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5
Σ(fx) = 1022.5, Σf = 35
Mean = 1022.5 / 35 = 29.21
Interpretation:
Mean = 29.21, Mode = 30.63 → Both indicate roughly 30 students per teacher.
Question 5
Runs scored by top batsmen in one-day internationals:
| Runs scored | 3000–4000 | 4000–5000 | 5000–6000 | 6000–7000 | 7000–8000 | 8000–9000 | 9000–10000 | 10000–11000 |
|---|---|---|---|---|---|---|---|---|
| No. of batsmen | 4 | 18 | 9 | 7 | 6 | 3 | 1 | 1 |
Find the mode of the data.
Solution:
Modal class = 4000–5000
l = 4000, h = 1000, f₁ = 18, f₀ = 4, f₂ = 9
Mode = 4000 + [(18 − 4) / (36 − 4 − 9)] × 1000
= 4000 + (14 / 23) × 1000
= 4000 + 608.7 = 4608.7 runs
Question 6
Number of cars passing a spot during 100 three-minute intervals:
| Number of cars | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 | 70–80 |
|---|---|---|---|---|---|---|---|---|
| Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Find the mode.
Solution:
Modal class = 40–50
l = 40, h = 10, f₁ = 20, f₀ = 12, f₂ = 11
Mode = 40 + [(20 − 12) / (40 − 12 − 11)] × 10
= 40 + (8 / 17) × 10
= 40 + 4.71 = 44.71 cars
Final Answers Summary
| Question | Mean | Mode |
|---|---|---|
| 1 | 35.38 | 36.82 |
| 2 | — | 65.63 |
| 3 | 2662.50 | 1847.83 |
| 4 | 29.21 | 30.63 |
| 5 | — | 4608.7 |
| 6 | — | 44.71 |