Table of Contents
Exercise 14.1 Solutions
Class X Mathematics
State Council of Educational Research and Training, Telangana, Hyderabad
Problem 1
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 |
Solution:
To find the mean, we’ll use the midpoint of each class interval and multiply by frequency.
| Class Interval | Midpoint (xi) | Frequency (fi) | fixi |
|---|---|---|---|
| 0-2 | 1 | 1 | 1 |
| 2-4 | 3 | 2 | 6 |
| 4-6 | 5 | 1 | 5 |
| 6-8 | 7 | 5 | 35 |
| 8-10 | 9 | 6 | 54 |
| 10-12 | 11 | 2 | 22 |
| 12-14 | 13 | 3 | 39 |
| Total | 20 | 162 | |
Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{162}{20} = 8.1\)
Mean number of plants per house = 8.1
Problem 2
Consider the following distribution of daily wages of 50 workers of a factory.
| Daily wages in Rupees | 200 – 250 | 250 – 300 | 300 – 350 | 350 – 400 | 400 – 450 |
|---|---|---|---|---|---|
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
We’ll use the step-deviation method with assumed mean a = 325 and class width h = 50
| Class Interval | Midpoint (xi) | Frequency (fi) | di = xi – a | ui = di/h | fiui |
|---|---|---|---|---|---|
| 200-250 | 225 | 12 | -100 | -2 | -24 |
| 250-300 | 275 | 14 | -50 | -1 | -14 |
| 300-350 | 325 (a) | 8 | 0 | 0 | 0 |
| 350-400 | 375 | 6 | 50 | 1 | 6 |
| 400-450 | 425 | 10 | 100 | 2 | 20 |
| Total | -12 | ||||
Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 325 + \left(\frac{-12}{50}\right) \times 50 = 325 – 12 = 313\)
Mean daily wages = ₹313
Problem 3
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 Rupees) | 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:
Let’s calculate the sum of frequencies and products of midpoints and frequencies.
| Class Interval | Midpoint (xi) | Frequency (fi) | fixi |
|---|---|---|---|
| 11-13 | 12 | 7 | 84 |
| 13-15 | 14 | 6 | 84 |
| 15-17 | 16 | 9 | 144 |
| 17-19 | 18 | 13 | 234 |
| 19-21 | 20 | f | 20f |
| 21-23 | 22 | 5 | 110 |
| 23-25 | 24 | 4 | 96 |
| Total | 44 + f | 752 + 20f | |
Given mean = 18
\(\frac{752 + 20f}{44 + f} = 18\)
752 + 20f = 792 + 18f
2f = 40
f = 20
The missing frequency \( f = 20 \)
Problem 4
Thirty women were examined in a hospital by a doctor and their heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method.
| Number of heart beats/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 |
Solution:
We’ll use the direct method to find the mean.
| Class Interval | Midpoint (xi) | Frequency (fi) | fixi |
|---|---|---|---|
| 65-68 | 66.5 | 2 | 133 |
| 68-71 | 69.5 | 4 | 278 |
| 71-74 | 72.5 | 3 | 217.5 |
| 74-77 | 75.5 | 8 | 604 |
| 77-80 | 78.5 | 7 | 549.5 |
| 80-83 | 81.5 | 4 | 326 |
| 83-86 | 84.5 | 2 | 169 |
| Total | 30 | 2277 | |
Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2277}{30} = 75.9\)
Mean heart beats per minute = 75.9
Problem 5
In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges.
| Number of oranges | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 |
|---|---|---|---|---|---|
| Number of baskets | 15 | 110 | 135 | 115 | 25 |
Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose?
Solution:
We’ll use the step-deviation method with assumed mean a = 22 and class width h = 5
| Class Interval | Midpoint (xi) | Frequency (fi) | di = xi – a | ui = di/h | fiui |
|---|---|---|---|---|---|
| 10-14 | 12 | 15 | -10 | -2 | -30 |
| 15-19 | 17 | 110 | -5 | -1 | -110 |
| 20-24 | 22 (a) | 135 | 0 | 0 | 0 |
| 25-29 | 27 | 115 | 5 | 1 | 115 |
| 30-34 | 32 | 25 | 10 | 2 | 50 |
| Total | 25 | ||||
Mean = \(a + \left(\frac{\sum f_iu_i}{\sum f_i}\right) \times h = 22 + \left(\frac{25}{400}\right) \times 5 = 22 + \frac{125}{400} = 22 + 0.3125 = 22.3125\)
Mean number of oranges per basket = 22.31 (approx)
We chose the step-deviation method because the class intervals are equal and the frequencies are large.
Problem 6
The table below shows the daily expenditure on food of 25 households in a locality.
| Daily expenditure (in Rupees) | 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.
Solution:
We’ll use the direct method to find the mean.
| Class Interval | Midpoint (xi) | Frequency (fi) | fixi |
|---|---|---|---|
| 100-150 | 125 | 4 | 500 |
| 150-200 | 175 | 5 | 875 |
| 200-250 | 225 | 12 | 2700 |
| 250-300 | 275 | 2 | 550 |
| 300-350 | 325 | 2 | 650 |
| Total | 25 | 5275 | |
Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{5275}{25} = 211\)
Mean daily expenditure on food = ₹211
Problem 7
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.
Solution:
We’ll use the direct method to find the mean.
| Class Interval | Midpoint (xi) | Frequency (fi) | fixi |
|---|---|---|---|
| 0.00-0.04 | 0.02 | 4 | 0.08 |
| 0.04-0.08 | 0.06 | 9 | 0.54 |
| 0.08-0.12 | 0.10 | 9 | 0.90 |
| 0.12-0.16 | 0.14 | 2 | 0.28 |
| 0.16-0.20 | 0.18 | 4 | 0.72 |
| 0.20-0.24 | 0.22 | 2 | 0.44 |
| Total | 30 | 2.96 | |
Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2.96}{30} \approx 0.0987\)
Mean concentration of SO2 = 0.099 ppm (approx)
Problem 8
A class teacher has the following attendance record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term.
| Number of days | 35-38 | 38-41 | 41-44 | 44-47 | 47-50 | 50-53 | 53-56 |
|---|---|---|---|---|---|---|---|
| Number of students | 1 | 3 | 4 | 4 | 7 | 10 | 11 |
Solution:
We’ll use the direct method to find the mean.
| Class Interval | Midpoint (xi) | Frequency (fi) | fixi |
|---|---|---|---|
| 35-38 | 36.5 | 1 | 36.5 |
| 38-41 | 39.5 | 3 | 118.5 |
| 41-44 | 42.5 | 4 | 170 |
| 44-47 | 45.5 | 4 | 182 |
| 47-50 | 48.5 | 7 | 339.5 |
| 50-53 | 51.5 | 10 | 515 |
| 53-56 | 54.5 | 11 | 599.5 |
| Total | 40 | 1961 | |
Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{1961}{40} = 49.025\)
Mean number of days present = 49.03 days (approx)
Problem 9
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 |
Solution:
We’ll use the direct method to find the mean.
| Class Interval | Midpoint (xi) | Frequency (fi) | fixi |
|---|---|---|---|
| 45-55 | 50 | 3 | 150 |
| 55-65 | 60 | 10 | 600 |
| 65-75 | 70 | 11 | 770 |
| 75-85 | 80 | 8 | 640 |
| 85-95 | 90 | 3 | 270 |
| Total | 35 | 2430 | |
Mean = \(\frac{\sum f_ix_i}{\sum f_i} = \frac{2430}{35} \approx 69.43\)
Mean literacy rate = 69.43%
