A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data 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 |
Which method did you use for finding the mean, and why?
Mean number of plants per house = 8.1 plants. Direct method is used because the numerical values of \(x_i\) and \(f_i\) are small.
Consider the following distribution of daily wages of 50 workers of a factory.
| Daily wages (in ₹) | 500 - 520 | 520 - 540 | 540 - 560 | 560 - 580 | 580 - 600 |
|---|---|---|---|---|---|
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Mean daily wages = ₹545.20
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 pocket 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 |
Missing frequency \(f = 20\).
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.
| Number of heartbeats per minute | 65 - 68 | 68 - 71 | 71 - 74 | 74 - 77 | 77 - 80 | 80 - 83 | 83 - 86 |
|---|---|---|---|---|---|---|---|
| Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Mean heartbeats per minute = 75.9
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.
| Number of mangoes | 50 - 52 | 53 - 55 | 56 - 58 | 59 - 61 | 62 - 64 |
|---|---|---|---|---|---|
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Mean number of mangoes per box = 57.19
The table below shows the daily expenditure on food of 25 households in a locality.
| Daily expenditure (in ₹) | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 | 300 - 350 |
|---|---|---|---|---|---|
| Number of households | 4 | 5 | 12 | 2 | 2 |
Find the mean daily expenditure on food by a suitable method.
Mean daily expenditure on food = ₹211
To find out the concentration of SO2 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:
| Concentration of SO2 (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 |
Find the mean concentration of SO2 in the air.
Mean concentration of SO2 = 0.099 ppm
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 |
Mean number of days absent = 12.48 days
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 |
Mean literacy rate = 69.43 %
The following table shows the ages of the patients admitted in a hospital during a year:
| 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.
Mode of the data = 36.8 years
Mean of the data = 35.37 years
Interpretation: The maximum number of patients admitted in the hospital are of age about 36.8 years (modal age), while on an average a patient admitted is about 35.37 years old (mean age).
The following data give the information on the observed lifetimes (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 lifetime of the components.
Modal lifetime of the components = 65.625 hours
The following distribution gives the total monthly household expenditure of 200 families of a village:
| Expenditure (in ₹) | 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 |
Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.
Modal monthly expenditure = ₹ 1847.83 (approximately)
Mean monthly expenditure = ₹ 2662.5
The following distribution gives the state-wise teacher–student ratio in higher secondary schools of India:
| Number of students per teacher | 15–20 | 20–25 | 25–30 | 30–35 | 35–40 | 40–45 | 45–50 | 50–55 |
|---|---|---|---|---|---|---|---|---|
| Number of states / U.T. | 3 | 8 | 9 | 10 | 3 | 0 | 0 | 2 |
Find the mode and mean of this data. Interpret the two measures.
Mode of the data = 30.6 students per teacher
Mean of the data = 29.2 students per teacher
Interpretation: In most states / U.T., the typical teacher–student ratio is about 30.6 (modal value), while on an average the ratio is about 29.2 students per teacher (mean value).
The following distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches:
| 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 |
Find the mode of the data.
Mode of the data = 4608.7 runs (approximately)
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 following table:
| 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.
Mode of the data = 44.7 cars
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 (in units) | Number of consumers |
|---|---|
| 65 - 85 | 4 |
| 85 - 105 | 5 |
| 105 - 125 | 13 |
| 125 - 145 | 20 |
| 145 - 165 | 14 |
| 165 - 185 | 8 |
| 185 - 205 | 4 |
Median = 137 units
Mean = 137.05 units
Mode = 135.76 units
The three measures of central tendency are approximately the same in this case.
If the median of the distribution given below is 28.5, find the values of x and y.
| Class interval | Frequency |
|---|---|
| 0 - 10 | 5 |
| 10 - 20 | x |
| 20 - 30 | 20 |
| 30 - 40 | 15 |
| 40 - 50 | y |
| 50 - 60 | 5 |
| Total | 60 |
x = 8, y = 7
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 |
Median age = 35.76 years
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 (in mm) | Number of leaves |
|---|---|
| 118 - 126 | 3 |
| 127 - 135 | 5 |
| 136 - 144 | 9 |
| 145 - 153 | 12 |
| 154 - 162 | 5 |
| 163 - 171 | 4 |
| 172 - 180 | 2 |
Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes.
Median length = 146.75 mm
The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.
| Life time (in hours) | Number of lamps |
|---|---|
| 1500 - 2000 | 14 |
| 2000 - 2500 | 56 |
| 2500 - 3000 | 60 |
| 3000 - 3500 | 86 |
| 3500 - 4000 | 74 |
| 4000 - 4500 | 62 |
| 4500 - 5000 | 48 |
Median life = 3406.98 hours
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:
| Number of letters | Number of surnames |
|---|---|
| 1 - 4 | 6 |
| 4 - 7 | 30 |
| 7 - 10 | 40 |
| 10 - 13 | 16 |
| 13 - 16 | 4 |
| 16 - 19 | 4 |
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.
Median = 8.05
Mean = 8.32
Modal size = 7.88
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
| Weight (in kg) | Number of students |
|---|---|
| 40 - 45 | 2 |
| 45 - 50 | 3 |
| 50 - 55 | 8 |
| 55 - 60 | 6 |
| 60 - 65 | 6 |
| 65 - 70 | 3 |
| 70 - 75 | 2 |
Median weight = 56.67 kg