Mathematics solution NCERT
Class 10 - Chapter 13: Statistics
Exercise 13.1
Q1. A survey was conducted by a group of students as a part of their environment awareness programme. 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 (f) | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
| Class Interval | f | Class Mark (x) | fx |
|---|---|---|---|
| 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 | 20 | - | 162 |
Mean = Σfx / Σf
= 162 / 20
= 8.1
Mean number of plants per house = 8.1
Method Used: Direct Method.
Reason: The class marks and frequencies are small numbers, so calculations are easy and direct.
Q2. Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages using an appropriate method.
| Daily Wages (₹) | 500-520 | 520-540 | 540-560 | 560-580 | 580-600 |
|---|---|---|---|---|---|
| Number of Workers (f) | 12 | 14 | 8 | 6 | 10 |
Using Assumed Mean Method
Assumed Mean (A) = 550
Class Width (h) = 20
| Class Interval | f | x | d' = (x-A)/h | fd' |
|---|---|---|---|---|
| 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 | 50 | - | - | -12 |
Mean = A + (Σfd' / Σf) × h
= 550 + (-12/50) × 20
= 550 - 4.8
= 545.2
Mean daily wage = ₹545.2
Q3. The mean pocket allowance is ₹18. Find the missing frequency f.
| Daily Pocket Allowance (₹) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
|---|---|---|---|---|---|---|---|
| Number of Children | 7 | 6 | 9 | 13 | f | 5 | 4 |
| Class Interval | f | x | fx |
|---|---|---|---|
| 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 |
Given Mean = 18
Mean = Σfx / Σf
18 = (752 + 20f) / (44 + f)
18(44 + f) = 752 + 20f
792 + 18f = 752 + 20f
40 = 2f
f = 20
Missing frequency = 20
Q4. Thirty women were examined in a hospital and the number of heartbeats per minute were recorded. Find the mean heartbeats per minute.
| Heartbeats per Minute | 65-68 | 68-71 | 71-74 | 74-77 | 77-80 | 80-83 | 83-86 |
|---|---|---|---|---|---|---|---|
| Number of Women (f) | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Using Assumed Mean Method
| Class Interval | f | x | d' = (x-75.5)/3 | fd' |
|---|---|---|---|---|
| 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 |
Mean = A + (Σfd'/Σf) × h
= 75.5 + (4/30) × 3
= 75.5 + 0.4
= 75.9
Mean heartbeats per minute = 75.9
Q5. 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 (f) | 15 | 110 | 135 | 115 | 25 |
Using Step-Deviation Method
Assumed Mean (A) = 57
Class Width (h) = 3
| Class Interval | f | x | d'=(x-57)/3 | fd' |
|---|---|---|---|---|
| 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 |
Mean = A + (Σfd'/Σf) × h
= 57 + (25/400) × 3
= 57 + 0.1875
= 57.19
Mean number of mangoes per box = 57.19
Method Used: Step-Deviation Method.
Reason: Frequencies are large and class intervals are equal, making calculations easier.
Q6. Find the mean daily expenditure on food.
| Daily Expenditure (₹) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
|---|---|---|---|---|---|
| Number of Households (f) | 4 | 5 | 12 | 2 | 2 |
| Class Interval | f | x | fx |
|---|---|---|---|
| 100-150 | 4 | 125 | 500 |
| 150-200 | 5 | 175 | 875 |
| 200-250 | 12 | 225 | 2700 |
| 250-300 | 2 | 275 | 550 |
| 300-350 | 2 | 325 | 650 |
| Total | 25 | - | 5275 |
Mean = Σfx / Σf
= 5275 / 25
= 211
Mean daily expenditure = ₹211
Q7. Find the mean concentration of SO₂ in the air.
| Concentration of SO₂ (ppm) | Frequency (f) |
|---|---|
| 0.00-0.04 | 4 |
| 0.04-0.08 | 9 |
| 0.08-0.12 | 9 |
| 0.12-0.16 | 2 |
| 0.16-0.20 | 4 |
| 0.20-0.24 | 2 |
| Class Interval | f | x | fx |
|---|---|---|---|
| 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 |
Mean = Σfx / Σf
= 2.96 / 30
= 0.0987
≈ 0.099 ppm
Mean concentration of SO₂ = 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 Absent | 0-6 | 6-10 | 10-14 | 14-20 | 20-28 | 28-38 | 38-40 |
|---|---|---|---|---|---|---|---|
| Number of Students (f) | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Using Direct Method
| Class Interval | f | Class Mark (x) | fx |
|---|---|---|---|
| 0-6 | 11 | 3 | 33 |
| 6-10 | 10 | 8 | 80 |
| 10-14 | 7 | 12 | 84 |
| 14-20 | 4 | 17 | 68 |
| 20-28 | 4 | 24 | 96 |
| 28-38 | 3 | 33 | 99 |
| 38-40 | 1 | 39 | 39 |
| Total | 40 | - | 499 |
Mean = Σfx / Σf
= 499 / 40
= 12.475
≈ 12.48
Mean number of days absent = 12.48 days
Q9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
| Literacy Rate (%) | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 |
|---|---|---|---|---|---|
| Number of Cities (f) | 3 | 10 | 11 | 8 | 3 |
Using Step-Deviation Method
Assumed Mean (A) = 70
Class Width (h) = 10
| Class Interval | f | x | d' = (x-A)/h | fd' |
|---|---|---|---|---|
| 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 |
Mean = A + (Σfd' / Σf) × h
= 70 + (-2/35) × 10
= 70 - 0.571
= 69.43
Mean literacy rate = 69.43%