Table of Contents
Exercise 14.2 Solutions
Class X Mathematics
State Council of Educational Research and Training, Telangana, Hyderabad
Problem 1
The following table shows the ages of the patients admitted in a hospital on a particular day:
| Age (in years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 |
|---|---|---|---|---|---|---|
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
Mode Calculation:
The modal class is the class with the highest frequency. Here, the highest frequency is 23, which corresponds to the class 35-45.
Using the mode formula:
Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
Where:
L = lower limit of modal class = 35
f₁ = frequency of modal class = 23
f₀ = frequency of class preceding modal class = 21
f₂ = frequency of class succeeding modal class = 14
h = class width = 10
Mode = 35 + [(23 – 21)/(2×23 – 21 – 14)] × 10
= 35 + [2/(46 – 35)] × 10
= 35 + (2/11) × 10
= 35 + 1.818 ≈ 36.82
Mean Calculation:
| Class Interval | Midpoint (xi) | Frequency (fi) | fixi |
|---|---|---|---|
| 5-15 | 10 | 6 | 60 |
| 15-25 | 20 | 11 | 220 |
| 25-35 | 30 | 21 | 630 |
| 35-45 | 40 | 23 | 920 |
| 45-55 | 50 | 14 | 700 |
| 55-65 | 60 | 5 | 300 |
| Total | 80 | 2830 | |
Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2830}{80} = 35.375\)
Comparison and Interpretation:
Mode = 36.82 years, Mean = 35.38 years
The mode (36.82) is slightly higher than the mean (35.38), indicating that the most common age group of patients is slightly older than the average age of all patients. Both values suggest that most patients admitted are in their mid-30s to mid-40s.
Problem 2
The following data gives the information on the observed life times (in hours) of 225 electrical components:
| Lifetimes (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
|---|---|---|---|---|---|---|
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
Solution:
The modal class is the class with the highest frequency. Here, the highest frequency is 61, which corresponds to the class 60-80.
Using the mode formula:
Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
Where:
L = lower limit of modal class = 60
f₁ = frequency of modal class = 61
f₀ = frequency of class preceding modal class = 52
f₂ = frequency of class succeeding modal class = 38
h = class width = 20
Mode = 60 + [(61 – 52)/(2×61 – 52 – 38)] × 20
= 60 + [9/(122 – 90)] × 20
= 60 + (9/32) × 20
= 60 + 5.625 = 65.625
Modal lifetime of components = 65.63 hours
Problem 3
The following data gives the distribution of total monthly household expenditure of 200 families of Gummadidala village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:
| Expenditure (in rupees) | 1000-1500 | 1500-2000 | 2000-2500 | 2500-3000 | 3000-3500 | 3500-4000 | 4000-4500 | 4500-5000 |
|---|---|---|---|---|---|---|---|---|
| Number of families | 24 | 40 | 33 | 28 | 30 | 22 | 16 | 7 |
Solution:
Mode Calculation:
The modal class is the class with the highest frequency. Here, the highest frequency is 40, which corresponds to the class 1500-2000.
Using the mode formula:
Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
Where:
L = lower limit of modal class = 1500
f₁ = frequency of modal class = 40
f₀ = frequency of class preceding modal class = 24
f₂ = frequency of class succeeding modal class = 33
h = class width = 500
Mode = 1500 + [(40 – 24)/(2×40 – 24 – 33)] × 500
= 1500 + [16/(80 – 57)] × 500
= 1500 + (16/23) × 500
= 1500 + 347.83 ≈ 1847.83
Mean Calculation:
We’ll use the step-deviation method with assumed mean a = 2750 and class width h = 500
| Class Interval | Midpoint (xi) | Frequency (fi) | di = xi – a | ui = di/h | fiui |
|---|---|---|---|---|---|
| 1000-1500 | 1250 | 24 | -1500 | -3 | -72 |
| 1500-2000 | 1750 | 40 | -1000 | -2 | -80 |
| 2000-2500 | 2250 | 33 | -500 | -1 | -33 |
| 2500-3000 | 2750 (a) | 28 | 0 | 0 | 0 |
| 3000-3500 | 3250 | 30 | 500 | 1 | 30 |
| 3500-4000 | 3750 | 22 | 1000 | 2 | 44 |
| 4000-4500 | 4250 | 16 | 1500 | 3 | 48 |
| 4500-5000 | 4750 | 7 | 2000 | 4 | 28 |
| Total | -35 | ||||
Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 2750 + \left(\frac{-35}{200}\right) \times 500 = 2750 – 87.5 = 2662.5\)
Modal monthly expenditure = ₹1847.83
Mean monthly expenditure = ₹2662.50
Problem 4
The following distribution gives the state-wise, teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
| Number of students per teacher | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 |
|---|---|---|---|---|---|---|---|---|
| Number of States | 3 | 8 | 9 | 10 | 3 | 0 | 0 | 2 |
Solution:
Mode Calculation:
The modal class is the class with the highest frequency. Here, the highest frequency is 10, which corresponds to the class 30-35.
Using the mode formula:
Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
Where:
L = lower limit of modal class = 30
f₁ = frequency of modal class = 10
f₀ = frequency of class preceding modal class = 9
f₂ = frequency of class succeeding modal class = 3
h = class width = 5
Mode = 30 + [(10 – 9)/(2×10 – 9 – 3)] × 5
= 30 + [1/(20 – 12)] × 5
= 30 + (1/8) × 5
= 30 + 0.625 = 30.625
Mean Calculation:
| Class Interval | Midpoint (xi) | Frequency (fi) | fixi |
|---|---|---|---|
| 15-20 | 17.5 | 3 | 52.5 |
| 20-25 | 22.5 | 8 | 180 |
| 25-30 | 27.5 | 9 | 247.5 |
| 30-35 | 32.5 | 10 | 325 |
| 35-40 | 37.5 | 3 | 112.5 |
| 40-45 | 42.5 | 0 | 0 |
| 45-50 | 47.5 | 0 | 0 |
| 50-55 | 52.5 | 2 | 105 |
| Total | 35 | 1022.5 | |
Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{1022.5}{35} \approx 29.21\)
Interpretation:
Mode = 30.62 students per teacher
Mean = 29.21 students per teacher
The mode (30.62) is slightly higher than the mean (29.21), indicating that the most common student-teacher ratio is slightly higher than the average ratio across all states. This suggests that while most states have about 30-31 students per teacher, some states with lower ratios bring the average down slightly.
Problem 5
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
| Runs | 3000-4000 | 4000-5000 | 5000-6000 | 6000-7000 | 7000-8000 | 8000-9000 | 9000-10000 | 10000-11000 |
|---|---|---|---|---|---|---|---|---|
| Number of batsmen | 4 | 18 | 9 | 7 | 6 | 3 | 1 | 1 |
Find the mode of the data.
Solution:
The modal class is the class with the highest frequency. Here, the highest frequency is 18, which corresponds to the class 4000-5000.
Using the mode formula:
Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
Where:
L = lower limit of modal class = 4000
f₁ = frequency of modal class = 18
f₀ = frequency of class preceding modal class = 4
f₂ = frequency of class succeeding modal class = 9
h = class width = 1000
Mode = 4000 + [(18 – 4)/(2×18 – 4 – 9)] × 1000
= 4000 + [14/(36 – 13)] × 1000
= 4000 + (14/23) × 1000
= 4000 + 608.70 ≈ 4608.70
Modal runs scored by batsmen = 4608.70 runs
Problem 6
A student noted the number of cars passing through a spot on a road for 100 periods, each of 3 minutes, and summarised this in the table given below.
| 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 of the data.
Solution:
The modal class is the class with the highest frequency. Here, the highest frequency is 20, which corresponds to the class 40-50.
Using the mode formula:
Mode = L + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
Where:
L = lower limit of modal class = 40
f₁ = frequency of modal class = 20
f₀ = frequency of class preceding modal class = 12
f₂ = frequency of class succeeding modal class = 11
h = class width = 10
Mode = 40 + [(20 – 12)/(2×20 – 12 – 11)] × 10
= 40 + [8/(40 – 23)] × 10
= 40 + (8/17) × 10
= 40 + 4.71 ≈ 44.71
Mode number of cars passing = 44.71 cars per 3-minute period
