Exercise 13.2 Solutions
Statistics
1. The following table shows the ages of the patients admitted in a hospital during a year: Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
| Age (in years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 |
|---|---|---|---|---|---|---|
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Calculation of Mode
The maximum frequency is 23, which corresponds to the class interval 35 – 45. So, the modal class is 35 – 45.
Lower limit of modal class (l) = 35
Frequency of modal class (f₁) = 23
Frequency of preceding class (f₀) = 21
Frequency of succeeding class (f₂) = 14
Class size (h) = 10
Mode = l + [ (f₁ - f₀) / (2f₁ - f₀ - f₂) ] * h
= 35 + [ (23 – 21) / (2*23 – 21 – 14) ] * 10
= 35 + [ 2 / (46 – 35) ] * 10 = 35 + [ 2 / 11 ] * 10
= 35 + 20/11 ≈ 35 + 1.82 = 36.82 years.
Calculation of Mean
We use the Assumed Mean Method. Let Assumed Mean (A) = 40.
| Class Interval | Frequency (fᵢ) | Class mark (xᵢ) | dᵢ = xᵢ – A | fᵢdᵢ |
|---|---|---|---|---|
| 5-15 | 6 | 10 | -30 | -180 |
| 15-25 | 11 | 20 | -20 | -220 |
| 25-35 | 21 | 30 | -10 | -210 |
| 35-45 | 23 | 40 | 0 | 0 |
| 45-55 | 14 | 50 | 10 | 140 |
| 55-65 | 5 | 60 | 20 | 100 |
| Total | Σfᵢ = 80 | Σfᵢdᵢ = -370 |
Mean (x̄) = A + (Σfᵢdᵢ / Σfᵢ) = 40 + (-370 / 80) = 40 – 4.625 = 35.375 years.
Interpretation
The mode (36.8 years) indicates that the maximum number of patients admitted to the hospital are in the age group of approximately 37 years. The mean (35.4 years) represents that the average age of a patient admitted to the hospital is approximately 35.4 years.
2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components: Determine the modal lifetimes of the components.
| Lifetimes (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
|---|---|---|---|---|---|---|
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
The maximum frequency is 61, so the modal class is 60 – 80.
l = 60, h = 20, f₁ = 61, f₀ = 52, f₂ = 38.
Mode = l + [ (f₁ - f₀) / (2f₁ - f₀ - f₂) ] * h
= 60 + [ (61 – 52) / (2*61 – 52 – 38) ] * 20
= 60 + [ 9 / (122 – 90) ] * 20 = 60 + [ 9 / 32 ] * 20
= 60 + 180/32 = 60 + 5.625 = 65.625 hours.
Answer: The modal lifetime of the components is 65.625 hours.
3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.
| Expenditure (in ₹) | Number of families |
|---|---|
| 1000 – 1500 | 24 |
| 1500 – 2000 | 40 |
| 2000 – 2500 | 33 |
| 2500 – 3000 | 28 |
| 3000 – 3500 | 30 |
| 3500 – 4000 | 22 |
| 4000 – 4500 | 16 |
| 4500 – 5000 | 7 |
Calculation of Mode
The maximum frequency is 40, so the modal class is 1500 – 2000.
l = 1500, h = 500, f₁ = 40, f₀ = 24, f₂ = 33.
Mode = 1500 + [ (40 – 24) / (2*40 – 24 – 33) ] * 500
= 1500 + [ 16 / (80 – 57) ] * 500 = 1500 + [ 16 / 23 ] * 500
= 1500 + 8000/23 ≈ 1500 + 347.83 = ₹1847.83.
Calculation of Mean
Using Step-Deviation Method. Let Assumed Mean (A) = 2750. h = 500.
| Class Interval | Freq (fᵢ) | xᵢ | uᵢ=(xᵢ-A)/h | fᵢuᵢ |
|---|---|---|---|---|
| 1000-1500 | 24 | 1250 | -3 | -72 |
| 1500-2000 | 40 | 1750 | -2 | -80 |
| 2000-2500 | 33 | 2250 | -1 | -33 |
| 2500-3000 | 28 | 2750 | 0 | 0 |
| 3000-3500 | 30 | 3250 | 1 | 30 |
| 3500-4000 | 22 | 3750 | 2 | 44 |
| 4000-4500 | 16 | 4250 | 3 | 48 |
| 4500-5000 | 7 | 4750 | 4 | 28 |
| Total | Σfᵢ = 200 | Σfᵢuᵢ = -35 |
Mean (x̄) = A + ( (Σfᵢuᵢ / Σfᵢ) × h )
= 2750 + ( (-35 / 200) × 500 ) = 2750 – (35/2) × 5 = 2750 – 87.5 = ₹2662.50.
4. The following distribution gives the state-wise teacher-student ratio… Find the mode and mean of this data. Interpret the two measures.
| Number of students per teacher | Number of states/U.T. |
|---|---|
| 15-20 | 3 |
| 20-25 | 8 |
| 25-30 | 9 |
| 30-35 | 10 |
| 35-40 | 3 |
| 40-45 | 0 |
| 45-50 | 0 |
| 50-55 | 2 |
Calculation of Mode
Maximum frequency is 10, so modal class is 30 – 35.
l = 30, h = 5, f₁ = 10, f₀ = 9, f₂ = 3.
Mode = 30 + [ (10 – 9) / (2*10 – 9 – 3) ] * 5
= 30 + [ 1 / (20 – 12) ] * 5 = 30 + 5/8 = 30 + 0.625 = 30.625.
Calculation of Mean
Using Assumed Mean Method. Let A = 32.5.
| Class Interval | Freq (fᵢ) | xᵢ | dᵢ=xᵢ-A | fᵢdᵢ |
|---|---|---|---|---|
| 15-20 | 3 | 17.5 | -15 | -45 |
| 20-25 | 8 | 22.5 | -10 | -80 |
| 25-30 | 9 | 27.5 | -5 | -45 |
| 30-35 | 10 | 32.5 | 0 | 0 |
| 35-40 | 3 | 37.5 | 5 | 15 |
| 40-45 | 0 | 42.5 | 10 | 0 |
| 45-50 | 0 | 47.5 | 15 | 0 |
| 50-55 | 2 | 52.5 | 20 | 40 |
| Total | Σfᵢ = 35 | Σfᵢdᵢ = -115 |
Mean (x̄) = A + (Σfᵢdᵢ / Σfᵢ) = 32.5 + (-115 / 35) ≈ 32.5 – 3.286 = 29.214.
Interpretation
The mode (30.6) indicates that most states/U.T.s have a teacher-student ratio of around 30.6. The mean (29.2) indicates that the average teacher-student ratio across all states/U.T.s is 29.2.
5. The given distribution shows the number of runs scored by some top batsmen… Find the mode of the data.
| Runs scored | Number of batsmen |
|---|---|
| 3000-4000 | 4 |
| 4000-5000 | 18 |
| 5000-6000 | 9 |
| 6000-7000 | 7 |
| 7000-8000 | 6 |
| 8000-9000 | 3 |
| 9000-10000 | 1 |
| 10000-11000 | 1 |
Maximum frequency is 18, so modal class is 4000 – 5000.
l = 4000, h = 1000, f₁ = 18, f₀ = 4, f₂ = 9.
Mode = 4000 + [ (18 – 4) / (2*18 – 4 – 9) ] * 1000
= 4000 + [ 14 / (36 – 13) ] * 1000 = 4000 + [ 14 / 23 ] * 1000
= 4000 + 14000/23 ≈ 4000 + 608.7 = 4608.7 runs.
Answer: The mode of the data is 4608.7 runs.
6. A student noted the number of cars passing through a spot on a road for 100 periods of 3 minutes… Find the mode of the data.
| 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 |
Maximum frequency is 20, so modal class is 40 – 50.
l = 40, h = 10, f₁ = 20, f₀ = 12, f₂ = 11.
Mode = 40 + [ (20 – 12) / (2*20 – 12 – 11) ] * 10
= 40 + [ 8 / (40 – 23) ] * 10 = 40 + [ 8 / 17 ] * 10
= 40 + 80/17 ≈ 40 + 4.7 = 44.7 cars.
Answer: The mode of the data is 44.7 cars.
