Exercise 13.1
Q1
A survey was conducted by a group of students regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14 Number of houses 1 2 1 5 6 2 3
| Number of plants | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 |
|---|---|---|---|---|---|---|---|
| Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Solution:
We use the Direct Method as the numerical values of observations and frequencies are small.
| Class Interval | Frequency (\(f_i\)) | Class Mark (\(x_i\)) | \(f_i x_i\) |
|---|---|---|---|
| 0-2 | 1 | 1 | 1 |
| 2-4 | 2 | 3 | 6 |
| 4-6 | 1 | 5 | 5 |
| 6-8 | 5 | 7 | 35 |
| 8-10 | 6 | 9 | 54 |
| 10-12 | 2 | 11 | 22 |
| 12-14 | 3 | 13 | 39 |
| Total | \(\Sigma f_i = 20\) | \(\Sigma f_i x_i = 162\) |
\[ \text{Mean } (\bar{x}) = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{162}{20} = 8.1 \]
Answer: The mean number of plants per house is 8.1.
Q2
Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.
Daily wages (in ₹) 500-520 520-540 540-560 560-580 580-600 Number of workers 12 14 8 6 10
| Daily wages (in ₹) | 500-520 | 520-540 | 540-560 | 560-580 | 580-600 |
|---|---|---|---|---|---|
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Solution:
We use the Step Deviation Method. Let assumed mean \(a = 550\) and class size \(h = 20\).
| Class Interval | Frequency (\(f_i\)) | Class Mark (\(x_i\)) | \(u_i = \frac{x_i - 550}{20}\) | \(f_i u_i\) |
|---|---|---|---|---|
| 500-520 | 12 | 510 | -2 | -24 |
| 520-540 | 14 | 530 | -1 | -14 |
| 540-560 | 8 | 550 | 0 | 0 |
| 560-580 | 6 | 570 | 1 | 6 |
| 580-600 | 10 | 590 | 2 | 20 |
| Total | \(\Sigma f_i = 50\) | \(\Sigma f_i u_i = -12\) |
\[ \text{Mean } (\bar{x}) = a + \left( \frac{\Sigma f_i u_i}{\Sigma f_i} \right) \times h \]
\[ \bar{x} = 550 + \left( \frac{-12}{50} \right) \times 20 = 550 - 4.8 = 545.2 \]
Answer: The mean daily wages of the workers is ₹ 545.20.
Q3
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency \(f\).
Daily allowance (in ₹) 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Number of children 7 6 9 13 \(f\) 5 4
| Daily allowance (in ₹) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
|---|---|---|---|---|---|---|---|
| Number of children | 7 | 6 | 9 | 13 | \(f\) | 5 | 4 |
Solution:
Given Mean (\(\bar{x}\)) = 18.
| Class Interval | Frequency (\(f_i\)) | Class Mark (\(x_i\)) | \(f_i x_i\) |
|---|---|---|---|
| 11-13 | 7 | 12 | 84 |
| 13-15 | 6 | 14 | 84 |
| 15-17 | 9 | 16 | 144 |
| 17-19 | 13 | 18 | 234 |
| 19-21 | \(f\) | 20 | \(20f\) |
| 21-23 | 5 | 22 | 110 |
| 23-25 | 4 | 24 | 96 |
| Total | \(44 + f\) | \(752 + 20f\) |
Using the mean formula:
\[ 18 = \frac{752 + 20f}{44 + f} \]
\[ 18(44 + f) = 752 + 20f \]
\[ 792 + 18f = 752 + 20f \]
\[ 20f - 18f = 792 - 752 \]
\[ 2f = 40 \Rightarrow f = 20 \]
Answer: The missing frequency \(f\) is 20.
Q4
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
Heartbeats/min 65-68 68-71 71-74 74-77 77-80 80-83 83-86 Number of women 2 4 3 8 7 4 2
| Heartbeats/min | 65-68 | 68-71 | 71-74 | 74-77 | 77-80 | 80-83 | 83-86 |
|---|---|---|---|---|---|---|---|
| Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Solution:
Using Step Deviation Method. Let \(a = 75.5\) and \(h = 3\).
| Class Interval | \(f_i\) | \(x_i\) | \(u_i = \frac{x_i - 75.5}{3}\) | \(f_i u_i\) |
|---|---|---|---|---|
| 65-68 | 2 | 66.5 | -3 | -6 |
| 68-71 | 4 | 69.5 | -2 | -8 |
| 71-74 | 3 | 72.5 | -1 | -3 |
| 74-77 | 8 | 75.5 | 0 | 0 |
| 77-80 | 7 | 78.5 | 1 | 7 |
| 80-83 | 4 | 81.5 | 2 | 8 |
| 83-86 | 2 | 84.5 | 3 | 6 |
| Total | 30 | 4 |
\[ \bar{x} = 75.5 + \left( \frac{4}{30} \right) \times 3 = 75.5 + 0.4 = 75.9 \]
Answer: 75.9 beats per minute.
Q5
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box.
Number of mangoes 50-52 53-55 56-58 59-61 62-64 Number of boxes 15 110 135 115 25
| Number of mangoes | 50-52 | 53-55 | 56-58 | 59-61 | 62-64 |
|---|---|---|---|---|---|
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
Solution:
The class intervals are not continuous (50-52, 53-55). However, the class mark \(x_i\) (midpoint) remains consistent with a gap of 3. We use the Step Deviation Method. Let \(a = 57\) and \(h = 3\).
| Class Interval | \(f_i\) | \(x_i\) | \(u_i = \frac{x_i - 57}{3}\) | \(f_i u_i\) |
|---|---|---|---|---|
| 50-52 | 15 | 51 | -2 | -30 |
| 53-55 | 110 | 54 | -1 | -110 |
| 56-58 | 135 | 57 | 0 | 0 |
| 59-61 | 115 | 60 | 1 | 115 |
| 62-64 | 25 | 63 | 2 | 50 |
| Total | 400 | 25 |
\[ \bar{x} = 57 + \left( \frac{25}{400} \right) \times 3 = 57 + \frac{3}{16} = 57 + 0.1875 \approx 57.19 \]
Answer: 57.19 mangoes.
Q6
The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.
Daily expenditure (in ₹) 100-150 150-200 200-250 250-300 300-350 Number of households 4 5 12 2 2
| Daily expenditure (in ₹) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
|---|---|---|---|---|---|
| Number of households | 4 | 5 | 12 | 2 | 2 |
Solution:
Using Step Deviation Method. Let \(a = 225\) and \(h = 50\).
| Class Interval | \(f_i\) | \(x_i\) | \(u_i = \frac{x_i - 225}{50}\) | \(f_i u_i\) |
|---|---|---|---|---|
| 100-150 | 4 | 125 | -2 | -8 |
| 150-200 | 5 | 175 | -1 | -5 |
| 200-250 | 12 | 225 | 0 | 0 |
| 250-300 | 2 | 275 | 1 | 2 |
| 300-350 | 2 | 325 | 2 | 4 |
| Total | 25 | -7 |
\[ \bar{x} = 225 + \left( \frac{-7}{25} \right) \times 50 = 225 - 14 = 211 \]
Answer: ₹ 211.
Q7
To find out the concentration of \(SO_2\) in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below. Find the mean concentration of \(SO_2\) in the air.
Conc. of \(SO_2\) (in ppm) 0.00-0.04 0.04-0.08 0.08-0.12 0.12-0.16 0.16-0.20 0.20-0.24 Frequency 4 9 9 2 4 2
| Conc. of \(SO_2\) (in ppm) | 0.00-0.04 | 0.04-0.08 | 0.08-0.12 | 0.12-0.16 | 0.16-0.20 | 0.20-0.24 |
|---|---|---|---|---|---|---|
| Frequency | 4 | 9 | 9 | 2 | 4 | 2 |
Solution:
Using Direct Method.
| Class Interval | \(f_i\) | \(x_i\) | \(f_i x_i\) |
|---|---|---|---|
| 0.00 - 0.04 | 4 | 0.02 | 0.08 |
| 0.04 - 0.08 | 9 | 0.06 | 0.54 |
| 0.08 - 0.12 | 9 | 0.10 | 0.90 |
| 0.12 - 0.16 | 2 | 0.14 | 0.28 |
| 0.16 - 0.20 | 4 | 0.18 | 0.72 |
| 0.20 - 0.24 | 2 | 0.22 | 0.44 |
| Total | 30 | 2.96 |
\[ \bar{x} = \frac{2.96}{30} = 0.0986... \approx 0.099 \]
Answer: 0.099 ppm.
Q8
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40 Number of students 11 10 7 4 4 3 1
| Number of days | 0-6 | 6-10 | 10-14 | 14-20 | 20-28 | 28-38 | 38-40 |
|---|---|---|---|---|---|---|---|
| Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Solution:
The class widths are unequal, so we use the Assumed Mean Method. Let \(a = 17\).
| Class Interval | \(f_i\) | \(x_i\) | \(d_i = x_i - 17\) | \(f_i d_i\) |
|---|---|---|---|---|
| 0-6 | 11 | 3 | -14 | -154 |
| 6-10 | 10 | 8 | -9 | -90 |
| 10-14 | 7 | 12 | -5 | -35 |
| 14-20 | 4 | 17 | 0 | 0 |
| 20-28 | 4 | 24 | 7 | 28 |
| 28-38 | 3 | 33 | 16 | 48 |
| 38-40 | 1 | 39 | 22 | 22 |
| Total | 40 | -181 |
\[ \bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i} = 17 + \frac{-181}{40} = 17 - 4.525 = 12.475 \]
Answer: 12.48 days (approx).
Q9
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Literacy rate (in %) 45-55 55-65 65-75 75-85 85-95 Number of cities 3 10 11 8 3
| Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 |
|---|---|---|---|---|---|
| Number of cities | 3 | 10 | 11 | 8 | 3 |
Solution:
Using Step Deviation Method. Let \(a = 70\) and \(h = 10\).
| Class Interval | \(f_i\) | \(x_i\) | \(u_i = \frac{x_i - 70}{10}\) | \(f_i u_i\) |
|---|---|---|---|---|
| 45-55 | 3 | 50 | -2 | -6 |
| 55-65 | 10 | 60 | -1 | -10 |
| 65-75 | 11 | 70 | 0 | 0 |
| 75-85 | 8 | 80 | 1 | 8 |
| 85-95 | 3 | 90 | 2 | 6 |
| Total | 35 | -2 |
\[ \bar{x} = 70 + \left( \frac{-2}{35} \right) \times 10 = 70 - \frac{20}{35} = 70 - 0.57 = 69.43 \]
Answer: 69.43%
Exercise 13.2
Q1
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 below. 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
| Age (in years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 |
|---|---|---|---|---|---|---|
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Solution:
(i) Mode:
The maximum frequency is 23. Thus, the modal class is 35-45.
\(l = 35\) (lower limit), \(h = 10\) (class size).
\(f_1 = 23\) (frequency of modal class).
\(f_0 = 21\) (frequency of preceding class).
\(f_2 = 14\) (frequency of succeeding class).
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]
\[ \text{Mode} = 35 + \left( \frac{23 - 21}{46 - 21 - 14} \right) \times 10 \]
\[ = 35 + \left( \frac{2}{11} \right) \times 10 = 35 + 1.818... \approx 36.8 \]
(ii) Mean:
Using Assumed Mean Method. Let assumed mean \(a = 30\) and class size \(h = 10\).
| Class Interval | Frequency (\(f_i\)) | Class Mark (\(x_i\)) | \(d_i = x_i - 30\) | \(f_i d_i\) |
|---|---|---|---|---|
| 5-15 | 6 | 10 | -20 | -120 |
| 15-25 | 11 | 20 | -10 | -110 |
| 25-35 | 21 | 30 | 0 | 0 |
| 35-45 | 23 | 40 | 10 | 230 |
| 45-55 | 14 | 50 | 20 | 280 |
| 55-65 | 5 | 60 | 30 | 150 |
| Total | 80 | 430 |
\[ \text{Mean } (\bar{x}) = a + \frac{\Sigma f_i d_i}{\Sigma f_i} \]
\[ \bar{x} = 30 + \frac{430}{80} = 30 + 5.375 = 35.375 \approx 35.37 \]
Interpretation: The maximum number of patients admitted in the hospital are of the age 36.8 years (Mode), while on average the age of a patient admitted to the hospital is 35.37 years (Mean).
Answer: Mode = 36.8 years, Mean = 35.37 years.
Q2
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
| Lifetimes (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
|---|---|---|---|---|---|---|
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Solution:
The maximum frequency is 61. So, the modal class is 60-80.
\(l = 60, h = 20, f_1 = 61, f_0 = 52, f_2 = 38\).
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \]
\[ \text{Mode} = 60 + \left( \frac{61 - 52}{122 - 52 - 38} \right) \times 20 \]
\[ = 60 + \left( \frac{9}{32} \right) \times 20 = 60 + \frac{180}{32} = 60 + 5.625 = 65.625 \]
Answer: The modal lifetime is 65.625 hours.
Q3
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 (₹) 1000-1500 1500-2000 2000-2500 2500-3000 3000-3500 3500-4000 4000-4500 4500-5000 Families 24 40 33 28 30 22 16 7
| Expenditure (₹) | 1000-1500 | 1500-2000 | 2000-2500 | 2500-3000 | 3000-3500 | 3500-4000 | 4000-4500 | 4500-5000 |
|---|---|---|---|---|---|---|---|---|
| Families | 24 | 40 | 33 | 28 | 30 | 22 | 16 | 7 |
Solution:
(i) Mode:
Maximum frequency is 40. Modal Class is 1500-2000.
\(l = 1500, h = 500, f_1 = 40, f_0 = 24, f_2 = 33\).
\[ \text{Mode} = 1500 + \left( \frac{40 - 24}{80 - 24 - 33} \right) \times 500 \]
\[ = 1500 + \left( \frac{16}{23} \right) \times 500 = 1500 + 347.83 = 1847.83 \]
(ii) Mean:
Using Step Deviation Method. \(a = 2750, h = 500\).
| Class Interval | \(f_i\) | \(x_i\) | \(u_i = \frac{x_i - 2750}{500}\) | \(f_i u_i\) |
|---|---|---|---|---|
| 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 | 200 | -35 |
\[ \bar{x} = 2750 + \left( \frac{-35}{200} \right) \times 500 = 2750 - 87.5 = 2662.5 \]
Answer: Mode = ₹ 1847.83, Mean = ₹ 2662.50.
Q4
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.
Students per teacher 15-20 20-25 25-30 30-35 35-40 40-45 45-50 50-55 States/U.T. 3 8 9 10 3 0 0 2
| Students per teacher | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | 40-45 | 45-50 | 50-55 |
|---|---|---|---|---|---|---|---|---|
| States/U.T. | 3 | 8 | 9 | 10 | 3 | 0 | 0 | 2 |
Solution:
(i) Mode:
Maximum frequency is 10. Modal Class is 30-35.
\(l = 30, h = 5, f_1 = 10, f_0 = 9, f_2 = 3\).
\[ \text{Mode} = 30 + \left( \frac{10 - 9}{20 - 9 - 3} \right) \times 5 = 30 + \frac{5}{8} = 30 + 0.625 = 30.625 \]
(ii) Mean:
Using Step Deviation Method. \(a = 32.5, h = 5\).
| Class Interval | \(f_i\) | \(x_i\) | \(u_i = \frac{x_i - 32.5}{5}\) | \(f_i u_i\) |
|---|---|---|---|---|
| 15-20 | 3 | 17.5 | -3 | -9 |
| 20-25 | 8 | 22.5 | -2 | -16 |
| 25-30 | 9 | 27.5 | -1 | -9 |
| 30-35 | 10 | 32.5 | 0 | 0 |
| 35-40 | 3 | 37.5 | 1 | 3 |
| 40-45 | 0 | 42.5 | 2 | 0 |
| 45-50 | 0 | 47.5 | 3 | 0 |
| 50-55 | 2 | 52.5 | 4 | 8 |
| Total | 35 | -23 |
\[ \bar{x} = 32.5 + \left( \frac{-23}{35} \right) \times 5 = 32.5 - \frac{23}{7} = 32.5 - 3.28 = 29.22 \]
Interpretation: Most states have a student-teacher ratio of 30.6, while the average ratio is 29.2.
Answer: Mode = 30.6, Mean = 29.2.
Q5
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.
Runs scored 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
| Runs scored | 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 |
Solution:
Max frequency = 18. Class = 4000-5000.
\(l = 4000, h = 1000, f_1 = 18, f_0 = 4, f_2 = 9\).
\[ \text{Mode} = 4000 + \left( \frac{18 - 4}{36 - 4 - 9} \right) \times 1000 = 4000 + \frac{14000}{23} = 4000 + 608.7 = 4608.7 \]
Answer: 4608.7 runs.
Q6
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. 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
| 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 |
Solution:
Max frequency = 20. Class = 40-50.
\(l = 40, h = 10, f_1 = 20, f_0 = 12, f_2 = 11\).
\[ \text{Mode} = 40 + \left( \frac{20 - 12}{40 - 12 - 11} \right) \times 10 = 40 + \frac{80}{17} = 40 + 4.7 = 44.7 \]
Answer: 44.7 cars.
Exercise 13.3
Q1
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption (units) 65-85 85-105 105-125 125-145 145-165 165-185 185-205 Number of consumers 4 5 13 20 14 8 4
| Monthly consumption (units) | 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:
1. Median:
Total consumers \(N = 68 \Rightarrow \frac{N}{2} = 34\).
Cumulative Frequencies (cf):
- 65-85: 4
- 85-105: 4 + 5 = 9
- 105-125: 9 + 13 = 22
- 125-145: 22 + 20 = 42 (This is the Median Class because 42 > 34)
\(l = 125, h = 20, f = 20, cf = 22\).
\[ \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \]
\[ = 125 + \left( \frac{34 - 22}{20} \right) \times 20 = 125 + 12 = 137 \]
2. Mean:
Using Step Deviation Method (\(a = 135, h = 20\)):
Mean \(\bar{x} = 137.05\).
3. Mode:
Maximum frequency is 20. Modal class is 125-145.
\(l = 125, h = 20, f_1 = 20, f_0 = 13, f_2 = 14\).
\[ \text{Mode} = 125 + \left( \frac{20 - 13}{40 - 13 - 14} \right) \times 20 = 125 + \frac{140}{13} = 125 + 10.76 = 135.76 \]
Answer: Median = 137 units, Mean = 137.05 units, Mode = 135.76 units.
Q2
If the median of the distribution given below is 28.5, find the values of \(x\) and \(y\).
Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 Total Frequency 5 x 20 15 y 5 60
| Class Interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | Total |
|---|---|---|---|---|---|---|---|
| Frequency | 5 | x | 20 | 15 | y | 5 | 60 |
Solution:
Given total frequency \(N = 60\).
Sum of frequencies: \(5 + x + 20 + 15 + y + 5 = 60 \Rightarrow 45 + x + y = 60 \Rightarrow x + y = 15 \dots (1)\)
Given Median = 28.5. Since 28.5 lies between 20 and 30, the Median Class is 20-30.
\(l = 20, h = 10, f = 20, N/2 = 30\).
Cumulative frequency of preceding class (\(cf\)) = \(5 + x\).
\[ \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \]
\[ 28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10 \]
\[ 8.5 = \frac{25 - x}{2} \]
\[ 17 = 25 - x \Rightarrow x = 8 \]
Substitute \(x\) in equation (1): \(8 + y = 15 \Rightarrow y = 7\).
Answer: \(x = 8, y = 7\).
Q3
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.
Age (in years) Number of policy holders Below 20 2 Below 25 6 Below 30 24 Below 35 45 Below 40 78 Below 45 89 Below 50 92 Below 55 98 Below 60 100
| Age (in years) | Number of policy holders |
|---|---|
| Below 20 | 2 |
| Below 25 | 6 |
| Below 30 | 24 |
| Below 35 | 45 |
| Below 40 | 78 |
| Below 45 | 89 |
| Below 50 | 92 |
| Below 55 | 98 |
| Below 60 | 100 |
Solution:
We convert the cumulative frequency distribution to a simple frequency distribution.
| Class Interval | Frequency (\(f\)) | Cumulative Frequency (\(cf\)) |
|---|---|---|
| 15-20 | 2 | 2 |
| 20-25 | 6 - 2 = 4 | 6 |
| 25-30 | 24 - 6 = 18 | 24 |
| 30-35 | 45 - 24 = 21 | 45 |
| 35-40 | 78 - 45 = 33 | 78 |
| 40-45 | 89 - 78 = 11 | 89 |
| 45-50 | 92 - 89 = 3 | 92 |
| 50-55 | 98 - 92 = 6 | 98 |
| 55-60 | 100 - 98 = 2 | 100 |
\(N = 100, N/2 = 50\). The cumulative frequency just greater than 50 is 78.
Median Class = 35-40.
\(l = 35, cf = 45, f = 33, h = 5\).
\[ \text{Median} = 35 + \left( \frac{50 - 45}{33} \right) \times 5 \]
\[ = 35 + \frac{25}{33} = 35 + 0.76 = 35.76 \]
Answer: 35.76 years.
Q4
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table. Find the median length of the leaves.
Length (mm) 118-126 127-135 136-144 145-153 154-162 163-171 172-180 No. of leaves 3 5 9 12 5 4 2
| Length (mm) | 118-126 | 127-135 | 136-144 | 145-153 | 154-162 | 163-171 | 172-180 |
|---|---|---|---|---|---|---|---|
| No. of leaves | 3 | 5 | 9 | 12 | 5 | 4 | 2 |
Solution:
The classes are not continuous. We subtract 0.5 from the lower limit and add 0.5 to the upper limit.
Continuous classes: 117.5-126.5, 126.5-135.5, etc.
\(N = 40, N/2 = 20\).
Cumulative Frequencies: 3, 8, 17, 29, 34, 38, 40.
Median Class (cf > 20): 144.5-153.5.
\(l = 144.5, h = 9, f = 12, cf = 17\).
\[ \text{Median} = 144.5 + \left( \frac{20 - 17}{12} \right) \times 9 \]
\[ = 144.5 + \frac{3 \times 9}{12} = 144.5 + 2.25 = 146.75 \]
Answer: 146.75 mm.
Q5
The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.
Life time (hours) 1500-2000 2000-2500 2500-3000 3000-3500 3500-4000 4000-4500 4500-5000 No. of lamps 14 56 60 86 74 62 48
| Life time (hours) | 1500-2000 | 2000-2500 | 2500-3000 | 3000-3500 | 3500-4000 | 4000-4500 | 4500-5000 |
|---|---|---|---|---|---|---|---|
| No. of lamps | 14 | 56 | 60 | 86 | 74 | 62 | 48 |
Solution:
\(N = 400, N/2 = 200\).
Cumulative Frequencies:
- 1500-2000: 14
- 2000-2500: 70
- 2500-3000: 130
- 3000-3500: 216 (Median Class)
\(l = 3000, h = 500, f = 86, cf = 130\).
\[ \text{Median} = 3000 + \left( \frac{200 - 130}{86} \right) \times 500 \]
\[ = 3000 + \frac{35000}{86} \approx 3000 + 406.98 = 3406.98 \]
Answer: 3406.98 hours.
Q6
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows. Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Number of letters 1-4 4-7 7-10 10-13 13-16 16-19 Number of surnames 6 30 40 16 4 4
| 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:
1. Median:
\(N = 100, N/2 = 50\). Cumulative frequencies: 6, 36, 76... Median Class is 7-10.
\(l = 7, h = 3, f = 40, cf = 36\).
\[ \text{Median} = 7 + \left( \frac{50 - 36}{40} \right) \times 3 = 7 + \frac{42}{40} = 7 + 1.05 = 8.05 \]
2. Mean:
Using Step Deviation (\(a=8.5, h=3\)): \(\bar{x} = 8.32\).
3. Mode:
Max frequency is 40. Modal Class is 7-10.
\(l = 7, f_1 = 40, f_0 = 30, f_2 = 16\).
\[ \text{Mode} = 7 + \left( \frac{40 - 30}{80 - 30 - 16} \right) \times 3 = 7 + \frac{30}{34} = 7 + 0.88 = 7.88 \]
Answer: Median = 8.05, Mean = 8.32, Mode = 7.88.
Q7
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
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
| 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:
\(N = 30, N/2 = 15\).
Cumulative Frequencies: 2, 5, 13, 19 (Median Class 55-60), 25, 28, 30.
\(l = 55, h = 5, f = 6, cf = 13\).
\[ \text{Median} = 55 + \left( \frac{15 - 13}{6} \right) \times 5 \]
\[ = 55 + \frac{10}{6} = 55 + 1.67 = 56.67 \]
Answer: 56.67 kg.