NCERT Solutions for class 10 Maths Chapter 14 - Statistics Exercise 14.1

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 13.

NCERT Solutions for Class 10 Maths Chapter 14 – Statistics Exercise 14.1 have been provided here for students to prepare well for the Board exam. The solutions of 10th Class Maths have been prepared by our subject experts and are in accordance with the NCERT syllabus and guidelines prescribed by CBSE.

These solutions of 10th class Maths are helpful for students to score well in the board examination, as it works for them as a reference tool to do the revision. The materials of NCERT Class 10 Solutions are in the context of the upcoming session of 2023-24.

NCERT Solutions Class 10 Maths Chapter 14 Statistics Exercise 14.1

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Access Answers of Maths NCERT Class 10 Chapter 14 – Statistics Exercise 14.1

1. A survey was conducted by a group of students as a part of their environment awareness program, 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?

Solution:

To find the mean value, we will use the direct method because the numerical value of fi and xi are small.

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

No. of plants

(Class interval)

No. of houses

Frequency (fi)

Mid-point (xi) fixi
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
Sum fi = 20 Sum fixi = 162

The formula to find the mean is:

Mean = x̄ = ∑fi xi /∑fi

= 162/20

= 8.1

Therefore, the mean number of plants per house is 8.1.

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

Daily wages (in Rs.) 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.

Solution:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large, so let us assume the mean value, a = 550.

Class interval (h) = 20

So, ui = (xi – a)/h

ui = (xi – 550)/20

Substitute and find the values as follows:

Daily wages

(Class interval)

Number of workers

frequency (fi)

Mid-point (xi) ui = (xi – 550)/20 fiui
500-520 12 510 -2 -24
520-540 14 530 -1 -14
540-560 8 550 = a 0 0
560-580 6 570 1 6
580-600 10 590 2 20
Total Sum fi = 50 Sum fiui = -12

So, the formula to find out the mean is:

Mean = x̄ = a + h(∑fiui /∑fi ) = 550 + [20 × (-12/50)] = 550 – 4.8 = 545.20

Thus, mean daily wage of the workers = Rs. 545.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Daily Pocket Allowance(in c) 11-13 13-15 15-17 17-19 19-21 21-23 23-35
Number of children 7 6 9 13 f 5 4

Solution:

To find out the missing frequency, use the mean formula.

Given, mean x̄ = 18

Class interval Number of children (fi) Mid-point (xi) fixi
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 fi = 44+f Sum fixi = 752+20f

The mean formula is

Mean = x̄ = ∑fixi /∑fi = (752 + 20f)/ (44 + f)

Now substitute the values and equate to find the missing frequency (f)

⇒ 18 = (752 + 20f)/ (44 + f)

⇒ 18(44 + f) = (752 + 20f)

⇒ 792 + 18f = 752 + 20f

⇒ 792 + 18f = 752 + 20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f

⇒ f = 20

So, the missing frequency, f = 20.

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 heart beats 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

Solution:

From the given data, let us assume the mean as a = 75.5

xi = (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the ui and fiui as follows:

Class Interval Number of women (fi) Mid-point (xi) ui = (xi – 75.5)/h fiui
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 = a 0 0
77-80 7 78.5 1 7
80-83 4 81.5 2 8
83-86 2 84.5 3 6
Sum fi= 30 Sum fiui = 4

Mean = x̄ = a + h(∑fiui /∑fi )

= 75.5 + 3 × (4/30)

= 75.5 + (4/10)

= 75.5 + 0.4

= 75.9

Therefore, the mean heart beats per minute for these women is 75.9

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?

Solution:

The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 1.

Here, assumed mean (a) = 57

Class size (h) = 3

Here, the step deviation is used because the frequency values are big.

Class Interval Number of boxes (fi) Mid-point (xi) ui = (xi – 57)/h fiui
49.5-52.5 15 51 -2 -30
52.5-55.5 110 54 -1 -110
55.5-58.5 135 57 = a 0 0
58.5-61.5 115 60 1 115
61.5-64.5 25 63 2 50
Sum fi = 400 Sum fiui = 25

The formula to find out the Mean is:

Mean = x̄ = a + h(∑fiui /∑fi )

= 57 + 3(25/400)

= 57 + 0.1875

= 57.19

Therefore, the mean number of mangoes kept in a packing box is 57.19

6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure(in c) 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:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

Let us assume the mean (a) = 225

Class size (h) = 50

Class Interval Number of households (fi) Mid-point (xi) di = xi – A ui = di/50 fiui
100-150 4 125 -100 -2 -8
150-200 5 175 -50 -1 -5
200-250 12 225 = a 0 0 0
250-300 2 275 50 1 2
300-350 2 325 100 2 4
Sum fi = 25 Sum fiui = -7

Mean = x̄ = a + h(∑fiui /∑fi )

= 225 + 50(-7/25)

= 225 – 14

= 211

Therefore, the mean daily expenditure on food is 211.

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) Frequency
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

Find the mean concentration of SO2 in the air.

Solution:

To find out the mean, first find the midpoint of the given frequencies as follows:

Concentration of SO2 (in ppm) Frequency (fi) Mid-point (xi) fixi
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 Sum fi = 30 Sum (fixi) = 2.96

The formula to find out the mean is

Mean = x̄ = ∑fixi /∑fi

= 2.96/30

= 0.099 ppm

Therefore, the mean concentration of SO2 in the air is 0.099 ppm.

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

Solution:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

Class interval Frequency (fi) Mid-point (xi) fixi
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
Sum fi = 40 Sum fixi = 499

The mean formula is,

Mean = x̄ = ∑fixi /∑fi

= 499/40

= 12.48 days

Therefore, the mean number of days a student was absent = 12.48.

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 85-98
Number of cities 3 10 11 8 3

Solution:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large, so let us assume the mean value, a = 70.

Class interval (h) = 10

So, ui = (xi – a)/h

ui = (xi – 70)/10

Substitute and find the values as follows:

Class Interval Frequency (fi) (xi) ui = (xi – 70)/10 fiui
45-55 3 50 -2 -6
55-65 10 60 -1 -10
65-75 11 70 = a 0 0
75-85 8 80 1 8
85-95 3 90 2 6
Sum fi = 35 Sum fiui = -2

So, Mean = x̄ = a + (∑fiui /∑fi) × h

= 70 + (-2/35) × 10

= 69.43

Therefore, the mean literacy part = 69.43%


In this exercise, students will solve questions based on finding the mean number based upon the statistics given, with the help of the following:

  • Direct Method
  • Assumed Mean Method
  • Step-deviation Method

A group of data collection will be provided to the students to find their mean and the missing frequency. You will study a more advanced version in upcoming exercises. Use the material provided by us to practise the exercise problems and solutions for Class 10 Maths, Chapter 14. Also, we are collectively providing here some more materials such as notes, books, question papers, etc., for students to help them practise well.

The questions in Ex.14.1 are solved as per the pattern explained in the example questions before the exercise. NCERT solutions are one of the best learning materials, where problems are solved in a detailed way, following each and every step and method.

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