# NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1

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**Access NCERT Solutions Maths Class 10 Chapter 14 – Statistics**

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**NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.1**

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**Q.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 |

Which method did you use for finding the mean, and why?

**Ans**

$\begin{array}{l}\text{Here, we observe that class marks and frequencies are small}\\ \text{quantities.}\\ \text{So, we use direct method to compute the mean and proceed}\\ \text{as below.}\end{array}$

Number of plants | Number of houses (f)_{i} |
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 | 20 | 162 |

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}=\frac{162}{20}=8.1\\ \text{Therefore, mean number of plants per house is 8.1.}\end{array}$

**Q.2 **Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in ₹) | 500-120 | 520-140 | 540-160 | 560-180 | 580-200 |

Number of workers | 12 | 14 | 8 | 6 | 10 |

Find the mean daily wages of the workers of the factory by using an appropriate method.

**Ans**

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=20\\ \text{Here, we take assumed mean(a)}=550\text{and proceed as below.}\end{array}$

Daily wages (in ₹) |
Number of workers (f)_{i} |
x_{i} |
d = _{i}x – 550_{i} |
${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-550}{\mathrm{h}}$ | f_{i}u_{i} |

500 – 520 | 12 | 510 | -40 | -2 | -24 |

520 – 540 | 14 | 530 | -20 | -1 | -14 |

540 – 560 | 8 | 550 | 0 | 0 | 0 |

560 – 580 | 6 | 570 | 20 | 1 | 6 |

580 – 600 | 10 | 590 | 40 | 2 | 20 |

Total | 50 | -12 |

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=550+\left(\frac{-12}{50}\right)\times 20=545.2\end{array}$

$\begin{array}{l}\text{Therefore, mean daily wages of the workers is}\u20b9545.20\end{array}$

**Q.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 ₹) | 11-13 | 13-15 | 15-17 | 17-19 | 18-21 | 21-23 | 23-25 |

Number of children | 7 | 6 | 9 | 13 | f |
5 | 4 |

**Ans**

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=2\\ \text{Given mean(a)}=18\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}\text{.}\end{array}$

Daily pocket allowance (in ₹) | Number of children (f)_{i} |
x_{i} |
d= _{i} x – 18_{i} |
f_{i}d_{i} |

11 – 13 | 7 | 12 | -6 | -42 |

13 – 15 | 6 | 14 | -4 | -24 |

15 – 17 | 9 | 16 | -2 | -18 |

17 – 19 | 13 | 18 | 0 | 0 |

19 – 21 | f |
20 | 2 | 2f |

21 – 23 | 5 | 22 | 4 | 20 |

23 – 25 | 4 | 24 | 6 | 24 |

Total | $\sum {\mathrm{f}}_{\mathrm{i}}=44+\mathrm{f}$ | 2f – 40 |

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{or}18=18+\frac{2\mathrm{f}-40}{44+\mathrm{f}}\\ \text{or \hspace{0.17em}}\mathrm{f}=20\\ \text{Therefore, missing frequency}\mathrm{f}\text{is}20.\end{array}$

**Q.4 **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 |

**Ans**

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=3\\ \text{Here, we take assumed mean(a)}=75.5\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

Number of heartbeats per minute | Number of women (f)_{i} |
x_{i} |
d = _{i}x – 75.5_{i} |
${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-75.5}{\mathrm{h}}$ | f_{i}u_{i} |

65-68 | 2 | 66.5 | -9 | -3 | -6 |

68-71 | 4 | 69.5 | -6 | -2 | -8 |

71-74 | 3 | 72.5 | -3 | -1 | -3 |

74-77 | 8 | 75.5 | 0 | 0 | 0 |

77-80 | 7 | 78.5 | 3 | 1 | 7 |

80-83 | 4 | 81.5 | 6 | 2 | 8 |

83-86 | 2 | 84.5 | 9 | 3 | 6 |

Total | 30 | 4 |

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=75.5+\left(\frac{4}{30}\right)\times 3=75.9\\ \text{Therefore, mean heartbeats per minute is}75.9.\end{array}$

**Q.5 **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?

**Ans**

$\begin{array}{l}\text{The given class intervals are not continuous. So, we add 0.5 to upper class limit and subtract 0.5 to lower class limit}\\ \text{of each interval. We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=3\\ \text{Here, we take assumed mean(a)}=57\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

Class interval | f_{i} |
x_{i} |
d_{i} = x_{i} – 57 |
${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-57}{\mathrm{h}}$ | {f}_{i}{u}_{i} |

49.5 – 52.5 | 15 | 51 | -6 | -2 | -30 |

52.5 – 55.5 | 110 | 54 | -3 | -1 | -110 |

55.5 – 58.5 | 135 | 57 | 0 | 0 | 0 |

58.5 – 61.5 | 115 | 60 | 3 | 1 | 115 |

61.5 – 64.5 | 25 | 63 | 6 | 2 | 50 |

Total | 400 | 25 |

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=57+\left(\frac{25}{400}\right)\times 3=57.1875\approx 57.19\\ \text{Therefore, mean number of mangoes is}57.19.\\ \text{We have chosen step deviation method as values of}{\mathrm{f}}_{\mathrm{i}}\text{,}{\mathrm{d}}_{\mathrm{i}}\text{are big}\\ \text{and there is a common multiple between all}{\mathrm{d}}_{\mathrm{i}}.\end{array}$

**Q.6 **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.

**Ans**

$\begin{array}{l}\text{We find class mark for each interval by using the following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=50\\ \text{Here, we take assumed mean(a)}=225\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

Class interval | f_{i} |
x_{i} |
d_{i} = x_{i} – 225 |
${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-57}{\mathrm{h}}$ | f_{i}u_{i} |

100 – 150 | 4 | 125 | -100 | -2 | -8 |

150 – 200 | 5 | 175 | -50 | -1 | -5 |

200 – 250 | 12 | 225 | 0 | 0 | 0 |

250 – 300 | 2 | 275 | 50 | 1 | 2 |

300 – 350 | 2 | 325 | 100 | 2 | 4 |

Total | -7 |

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=225+\left(\frac{-7}{25}\right)\times 50=211\\ \text{Therefore, mean daily expenditure is}\u20b9211.\end{array}$

**Q.7 **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.

Concentration of SO_{2} (in ppm) |
Frequency |

0.00 – 0.04
0.04– 0.08 0.08 – 0.12 0.12–0.16 0.16–0.20 0.20–0.24 |
4
9 9 2 4 2 |

Find the mean concentration of SO_{2} in the air.

**Ans**

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Class size(h) of the given data}=0.04\\ \text{Here, we take assumed mean(a)}=0.14\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

Concentration of SO_{2} (in ppm) |
Frequency (f)_{i} |
x_{i} |
d = _{i}x – 0.14_{i} |
${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-0.14}{\mathrm{h}}$
b |
f_{i}u_{i} |

0.00 – 0.04 | 4 | 0.02 | -0.12 | -3 | -12 |

0.04 – 0.08 | 9 | 0.06 | -0.08 | -2 | -18 |

0.08 – 0.12 | 9 | 0.10 | -0.04 | -1 | -9 |

0.12 – 0.16 | 2 | 0.14 | 0 | 0 | 0 |

0.16 – 0.20 | 4 | 0.18 | 0.04 | 1 | 4 |

0.20 – 0.24 | 2 | 0.22 | 0.08 | 2 | 4 |

Total | 30 | -31 |

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=0.14+\left(\frac{-31}{30}\right)\times 0.04\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\approx 0.099\text{ppm}\\ {\text{Therefore, mean concentration of SO}}_{\text{2}}\text{in the air is}0.099\text{ppm}.\end{array}$

**Q.8** 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 |

**Ans**

$\begin{array}{l}\text{We find class mark for each interval by using the following}\\ \text{relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Assumed mean(a)}=17\\ \text{We proceed as below to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}\text{.}\end{array}$

Number of days | Number of students f_{i} |
x_{i} |
d = _{i}x – 17_{i} |
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 |

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=17+\frac{-181}{40}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12.475\approx 12.48\\ \text{Therefore, mean number of days for which a student was absent is}12.48.\end{array}$

**Q.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 |

**Ans**

$\begin{array}{l}\text{We find class mark for each interval by using the following relation.}\\ {\text{x}}_{\mathrm{i}}=\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\\ \text{Class size(h) of the given data}=10\\ \text{Here, we take assumed mean(a)}=70\text{and proceed as below}\\ \text{to find}{\mathrm{d}}_{\mathrm{i}},\text{}{\mathrm{u}}_{\mathrm{i}}\text{and}{\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}\text{.}\end{array}$

Literacy rate (in %) | Number of cities f_{i} |
x_{i} |
d = _{i}x – 70_{i} |
${\mathrm{u}}_{\mathrm{i}}=\frac{{\mathrm{x}}_{\mathrm{i}}-70}{\mathrm{h}}$ | f_{i}u_{i} |

45 – 55 | 3 | 50 | -20 | -2 | -6 |

55 – 65 | 10 | 60 | -10 | -1 | -10 |

65 – 75 | 11 | 70 | 0 | 0 | 0 |

75 – 85 | 8 | 80 | 10 | 1 | 8 |

85 – 95 | 3 | 90 | 20 | 2 | 6 |

Total | 35 | -2 |

$\begin{array}{l}\text{Mean}=\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{{}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}\right)\mathrm{h}=70+\left(\frac{-2}{35}\right)\times 10=69.43\\ \text{Therefore, mean literacy rate is}69.43.\end{array}$

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- Additionally, they have the option of studying in groups. They may benefit from group study if they have trouble focusing or are easily distracted by their surroundings. Members can hold one another accountable and help one another in tracking their academics and syllabus. They can also review each other’s performance in mock tests and sample papers and help each other grow. They can support one another in overcoming doubts.