Open In App
Related Articles

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Improve Article
Improve
Save Article
Save
Like Article
Like

NCERT Solutions Class 10 Maths Chapter 14 Statistics – This article is a useful resource containing free NCERT Solutions for Class 10 Maths Chapter 14 Statistics. These NCERT solutions have been developed by the subject matter experts at GFG to assist students in easily solving questions related to Statistics from the NCERT textbook.

These NCERT Solutions for Class 10 Maths Chapter 14 Statistics cover all four exercises of the NCERT Class 10 Maths Chapter 14 according to the latest CBSE syllabus 2023-24 and guidelines, which are as follows:

Class 10 Maths NCERT Solutions Chapter 14 Statistics Exercises

NCERT Class 10 Maths Chapter 14 Statistics will help the students learn important statistical concepts like mean, mode, standard deviation, and the graphical depiction of cumulative frequency distribution.

The solutions to all the problems in this Chapter 14 Statistics exercises from the NCERT textbook have been properly covered in the NCERT Solutions for Class 10 Maths.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.1

Question 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: 

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

Step 3: Now we will apply the general formula to calculate the mean

 \large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

Now, Let’s see the detailed solution: 

No.of Plants

(Class Interval)

No. of Houses

(Frequency) (fi)

Class Mark

(xi)

fi * xi

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

Now, after creating this table we will be able to find the mean very easily – 

\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

= 16

= 8.1

Hence, we come to the conclusion that the number of plants per house is 8.1. Since the numeral value of frequency(fi) and the class mark(xi) is small so we use DIRECT METHOD to find the mean number of plants per house.

Question 2. 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.

Solution:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20.

ui = (xi – A)/h 

=> ui = (xi – 150)/20

Step 3: Now we will apply the Assumed Mean Formula to calculate the mean

\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

Now, Let’s see the detailed solution: 

Daily wages

(Class interval)

Number of workers

frequency (fi)

Mid-point (xi)

ui = (xi – 150)/20

fiui

100-120

12

110

-2

-24

120-140

14

130

-1

-14

140-160

8

150

0

0

160-180

6

170

1

6

180-200

10

190

2

20

Total

Sum ∑fi = 50

 

 

Sum ∑fiui = -12

So, the formula to find out the mean is:

Mean = \large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

           = 150 + (20 × -12/50) 

           = 150 – 4.8

           = 145.20

Thus, mean daily wage of the workers = Rs. 145.20.

Question 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 ₹)

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:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency. As a certain frequency is missing and we have an odd number of class intervals hence, we will assume the middle-Class Mark as our Assumed Mean(A).

Step 3: Now we will apply the general formula to calculate the mean

\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

Now, Let’s see the detailed solution: 

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 = A

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 = \large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

          = (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.

Question 4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarized 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

Solution:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 75.5 and class size is h = 3.

di = (xi – A) 

=> di = (xi – 75.5)

Step 3: Now we will apply the Assumed Mean Formula to calculate the mean

\large \overline{x}=A+\frac{\sum f_id_i}{\sum f_i}

Now, Let’s see the detailed solution: 

Class Interval

Number of women (fi)

Mid-point (xi)

di = (xi – 75.5)

fidi

65-68

2

66.5

-9

-18

68-71

4

69.5

-6

-24

71-74

3

72.5

-3

-9

74-77

8

75.5 = A

0

0

77-80

7

78.5

3

21

80-83

4

81.5

6

24

83-86

2

84.5

9

18

 

Sum ∑fi = 30

 

 

Sum ∑fiui = 12

Mean = \large \overline{x}=A+\frac{\sum f_id_i}{\sum f_i}

= 75.5 + (12/30)

= 75.5 + 2/5

= 75.5 + 0.4

= 75.9

Therefore, the mean heartbeats per minute for these women is 75.9

Question 5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers 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:

Step 1: In the above table we find that the class intervals are not continuous and hence to make them a continuous set of data we add  0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals is 1. Then find the Mid Point by using the formula 

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, let us assume the mean value, A = 57 and class size is h = 3.

Step 3: Since the frequency values are big, hence we are using the STEP-DEVIATION METHOD.

\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

Now, Lets see the detailed solution: 

Class Interval

Number of boxes (fi)

Mid-point (xi)

di = xi – A

ui=(xi – A)/h

fiui

49.5-52.5

15

51

-6

-2

-30

52.5-55.5

110

54

-3

-1

-110

55.5-58.5

135

57 =A

0

0

0

58.5-61.5

115

60

3

1

115

61.5-64.5

25

63

6

2

50

 

Sum ∑fi = 400

 

 

 

Sum ∑fiui = 25

Mean = \large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

= 57 + 3 * (25/400)

= 57 + 0.1875

= 57.19

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

Question 6. The table below shows the daily expenditure on the food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Daily Expenditure (in ₹)

100-150

150-200

200-250

250-300

300-350

Number of Households

4

5

12

2

2

Solution:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 225 and class size is h = 50.

di = (xi – A) 

=> di = (xi – 225)

ui = (xi – A)/h

=> ui = (xi – 225)/50

Step 3: Now we will apply the Step Deviation Formula to calculate the mean

\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

Now, Let’s see the detailed solution: 

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 = \large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

= 225 + 50 (-7/25)

= 225 – 14

= 211

Therefore, the mean daily expenditure on food is ₹211

Question 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:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

Step 3: Now we will apply the general formula to calculate the mean

\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

Now, Let’s see the detailed solution: 

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

 

Sum ∑fi = 30

 

Sum ∑fixi = 2.96

The formula to find out the mean is

Mean = \large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

= 2.96/30

= 0.099 ppm

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

Question 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:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

Step 3: Now we will apply the general formula to calculate the mean

\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

Now, Let’s see the detailed solution: 

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 = \large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}

= 499/40

= 12.48 days

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

Question 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:

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 70 and class size is h = 10.

di = (xi – A) 

=> di = (xi – 70)

ui = (xi – A)/h

=> ui = (xi – 70)/10

Step 3: Now we will apply the Step Deviation Formula to calculate the mean

\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

Now, Let’s see the detailed solution: 

Class Interval

Frequency (fi)

Class Mark(xi)

di = xi – a

ui = di/h

fiui

45-55

3

50

-20

-2

-6

55-65

10

60

-10

-1

-10

65-75

11

70 = A

0

0

0

75-85

8

80

10

1

8

85-95

3

90

20

2

6

 

Sum ∑fi  = 35

 

 

 

Sum ∑fiui = -2

So, 

Mean = \large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}

= 70 + (-2/35) × 10

= 69.42

Therefore, the mean literacy rate = 69.42%.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.2

Question 1. 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.

Solution:

The greatest frequency in the given table is 23, so the modal class = 35 – 45,

l = 35,

Class width = 10, and the frequencies are

fm = 23, f1 = 21 and f2 = 14

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 35+\left[\frac{(23-21)}{(46-21-14)}\right]×10

= 35 + (20/11) = 35 + 1.8

= 36.8

Hence, the mode of the given data is 36.8 year

Now, we find the mean. So for that first we need to find the midpoint.

xi = (upper limit + lower limit)/2

Class Interval Frequency (fi) Mid-point (xi) fixi
5-15 6 10 60
15-25 11 20 220
25-35 21 30 630
35-45 23 40 920
45-55 14 50 700
55-65 5 60 300
  Sum fi = 80   Sum fixi = 2830

Mean = \bar{x}       = ∑fixi /∑fi

= 2830/80

= 35.37 years

Question 2. The following data gives information on the observed lifetimes (in hours) of 225 electrical components:

Lifetime (in hours) 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components.

Solution:

According to the given question

The modal class is 60 – 80

l = 60, and the frequencies are

fm = 61, f1 = 52, f2 = 38 and h = 20

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 60+\left[\frac{(61-52)}{(122-52-38)}\right]×20

60+\frac{(9 \times 20)}{32}

= 60 + 45/8 = 60 + 5.625

Hence, the modal lifetime of the components is 65.625 hours.

Question 3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure Number of families
1000-1500 24
1500-2000 40
2000-2500 33
2500-3000 28
3000-3500 30
3500-4000 22
4000-4500 16
4500-5000 7

Solution:

According to the question

Modal class = 1500-2000,

l = 1500,and the frequencies are

fm = 40 f1 = 24, f2 = 33 and

h = 500

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 1500+\left[\frac{(40-24)}{(80-24-33)}\right]×500

1500+\frac{16×500}{23}

= 1500 + 8000/23 = 1500 + 347.83

So, the modal monthly expenditure of the families is 1847.83 Rupees 

Now, we find the mean. So for that first we need to find the midpoint.

xi = (upper limit + lower limit)/2

Let us considered a mean, A be 2750

Class Interval fi xi di = xi – a ui = di/h fiui
1000-1500 24 1250 -1500 -3 -72
1500-2000 40 1750 -1000 -2 -80
2000-2500 33 2250 -500 -1 -33
2500-3000 28 2750 0 0 0
3000-3500 30 3250 500 1 30
3500-4000 22 3750 1000 2 44
4000-4500 16 4250 1500 3 48
4500-5000 7 4750 2000 4 28
  fi = 200       fiui = -35

Mean = \overline{x} = a +\frac{∑f_iu_i}{∑f_i}×h

On substituting the values in the given formula

2750+\frac{-35}{200}×500

= 2750 – 87.50

= 2662.50

Hence, the mean monthly expenditure of the families is  2662.50 Rupees

Question 4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures

No of Students per teacher Number of states / U.T
15-20 3
20-25 8
25-30 9
30-35 10
35-40 3
40-45 0
45-50 0
50-55 2

Solution:

According to the question

Modal class = 30 – 35,

l = 30,

Class width (h) = 5, and the frequencies are

fm = 10, f1 = 9 and f2 = 3

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 30+\frac{(10-9)}{(20-9-3)}×5

= 30 + 5/8 = 30 + 0.625

= 30.625

Hence, the mode of the given data is 30.625

Now, we find the mean. So for that first we need to find the midpoint.

xi = (upper limit + lower limit)/2

Class Interval Frequency (fi) Mid-point (xi) fixi
15-20 3 17.5 52.5
20-25 8 22.5 180.0
25-30 9 27.5 247.5
30-35 10 32.5 325.0
35-40 3 37.5 112.5
40-45 0 42.5 0
45-50 0 47.5 0
50-55 2 52.5 105.5
  Sum fi = 35   Sum fixi = 1022.5

Mean = \bar{x} = \frac{∑f_ix_i }{∑f_i}

= 1022.5/35 

= 29.2

Hence, the mean is 29.2

Question 5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Run Scored Number of Batsman
3000-4000 4
4000-5000 18
5000-6000 9
6000-7000 7
7000-8000 6
8000-9000 3
9000-10000 1
10000-11000 1

Find the mode of the data.

Solution:

According to the question

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000, and the frequencies are

fm = 18, f1 = 4 and f2 = 9

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 4000+\frac{(18-4)}{(36-4-9)}×1000

Mode = 4000 + 14000/23 = 4000 + 608.695

= 4608.695

Hence, the mode of the given data is 4608.7 runs

Question 6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

Number of cars Frequency
0-10 7
10-20 14
20-30 13
30-40 12
40-50 20
50-60 11
60-70 15
70-80 8

Solution:

According to the question

Modal class = 40 – 50, l = 40,

Class width (h) = 10, and the frequencies are

fm = 20, f1 = 12 and f2 = 11

Now, we find the mode using the given formula

Mode l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h

On substituting the values in the formula, we get

Mode = 40+\frac{(20-12)}{(40-12-11)}×10

Mode = 40 + 80/17 = 40 + 4.7 = 44.7

Hence, the mode of the given data is 44.7 cars

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.3

Question 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers in a locality. Find the median, mean, and mode of the data and compare them.

Monthly consumption(in units) No. of customers
65-85 4
85-105 5
105-125 13
125-145 20
145-165 14
165-185 8
185-205 4

Solution:

 Total number of consumer n = 68 

n/2 =34

So, the median class is 125-145 with cumulative frequency = 42

Here, l = 125, n = 68, Cf = 22, f = 20, h = 20

Now we find the median:

Median = l +\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h        

=125+\frac{(34−22)}{20} × 20

= 125 + 12 = 137

Hence, the median is 137

Now we find the mode:

Modal class = 125 – 145,

Frequencies are

f1 = 20, f0 = 13, f2 = 14 & h = 20

Mode l+ \left[\frac{(f1-f0)}{(2f1-f0-f2)}\right]×h

On substituting the values in the given formula, we get

Mode = 125 + \frac{(20-13)}{(40-13-14)}×20

= 125 + 140/13 

= 125 + 10.77

= 135.77

Hence, the mode is 135.77

Now we find the mean:

Class Interval fi xi di = xi – a ui = di/h fiui
65-85 4 75 -60 -3 -12
85-105 5 95 -40 -2 -10
105-125 13 115 -20 -1 -13
125-145 20 135 0 0 0
145-165 14 155 20 1 14
165-185 8 175 40 2 16
185-205 4 195 60 3 12
  Sum fi = 68       Sum fiui = 7

\bar{x} =a+h \frac{∑f_iu_i}{∑f_i}

= 135 + 20(7/68)

= 137.05

Hence, the mean is 137.05

Now, on comparing the median, mean, and mode, we found that mean, median and mode are more/less equal in this distribution.

Question 2. If the median of a distribution given below is 28.5 then, find the value of x & y.

Class Interval Frequency
0-10 5
10-20 x
20-30 20
30-40 15
40-50 y
50-60 5
Total 60

Solution:

According to the question

The total number of observations are n = 60

Median of the given data = 28.5

n/2 = 30  

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class, l = 20,

Cf = 5 + x,

f = 20 & h = 10

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

28.5 = 20+\frac{(30−5−x)}{20} × 10

8.5 = (25 – x)/2

17 = 25 – x

Therefore, x = 8

From the cumulative frequency, we can identify the value of x + y as follows:

60 = 5 + 20 + 15 + 5 + x + y

On substituting the values of x, we will find the value of y

60 = 5 + 20 + 15 + 5 + 8 + y

y = 60 – 53

y = 7

So the value of a is 8 and y is 7

Question 3. The Life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons whose age is 18 years onwards but less than 60 years.

Age (in years) Number of policyholder
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

Solution: 

According to the given question the table is 

Class interval Frequency Cumulative frequency
15-20 2 2
20-25 4 6
25-30 18 24
30-35 21 45
35-40 33 78
40-45 11 89
45-50 3 92
50-55 6 98
55-60 2 100

Given data: n = 100 and n/2 = 50

Median class = 35 – 45

Then, l = 35, cf = 45, f = 33 & h = 5

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 35+\frac{(50-45)}{33} × 5

= 35 + 5(5/33) 

= 35.75

Hence, the median age is 35.75 years.

Question 4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as in the following table:

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

Find the median length of leaves.             

Solution:

The data in the given table are not continuous to reduce 0.5 in the lower limit and add 0.5 in the upper limit.

We get a new table:

Class Interval Frequency Cumulative frequency
117.5-126.5 3 3
126.5-135.5 5 8
135.5-144.5 9 17
144.5-153.5 12 29
153.5-162.5 5 34
162.5-171.5 4 38
171.5-180.5 2 40

From the given table

n = 40 and n/2 = 20

Median class = 144.5 – 153.5

l = 144.5,

cf = 17, f = 12 & h = 9

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 144.5+\frac{(20-17)}{12}×9

= 144.5 + 9/4 

= 146.75 mm

Hence, the median length of the leaves is 146.75 mm.

Question 5. The following table gives the distribution of a lifetime of 400 neon lamps.

Lifetime (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

Find the median lifetime of a lamp.

Solution:

According to the question

Class Interval Frequency Cumulative
1500-2000 14 14
2000-2500 56 70
2500-3000 60 130
3000-3500 86 216
3500-4000 74 290
4000-4500 62 352
4500-5000 48 400

n = 400 and n/2 = 200

Median class = 3000 – 3500

l = 3000, Cf = 130,

f = 86 & h = 500

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 3000 + \frac{(200-130)}{86} × 500

= 3000 + 35000/86 = 3000 + 406.97

= 3406.97

Hence, the median lifetime of the lamps is 3406.97 hours

Question 6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Number of letters 1-4 4-7 7-10 10-13 13-16 16-19
Number of surnames 6 30 40 16 4 4

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution:

According to the question

Class Interval Frequency Cumulative Frequency
1-4 6 6
4-7 30 36
7-10 40 76
10-13 16 92
13-16 4 96
16-19 4 100

n = 100 and n/2 = 50

Median class = 7 – 10

Therefore, l = 7, Cf = 36, f = 40 & h = 3

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 7+\frac{(50-36)}{40} × 3

Median = 7 + 42/40 = 8.05

Hence, the median is 8.05

Now we find the mode:

Modal class = 7 – 10,

Where, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3

Mode = l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h

On substituting the values in the given formula, we get

Mode = 7+\frac{(40-30)}{(2×40-30-16)} × 3

= 7 + 30/34 = 7.88

Hence, the mode is 7.88

Now we find the mean:

Class Interval fi xi fixi
1-4 6 2.5 15
4-7 30 5.5 165
7-10 40 8.5 340
10-13 16 11.5 184
13-16 4 14.5 51
16-19 4 17.5 70
  Sum fi = 100   Sum fixi = 825

Mean = \bar{x}= \frac{∑f_i x_i }{∑f_i}

= 825/100 = 8.25

Hence, the mean is 8.25

Question 7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.

Weight(in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75
Number of students 2 3 8 6 6 3 2

Solution:

According to the question

Class Interval Frequency Cumulative frequency
40-45 2 2
45-50 3 5
50-55 8 13
55-60 6 19
60-65 6 25
65-70 3 28
70-75 2 30

n = 30 and n/2 = 15

Median class = 55 – 60

l = 55, Cf = 13, f = 6 & h = 5

Now we find the median:

Median = l +\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 55+\frac{(15-13)}{6}×5

= 55 + 10/6 = 55 + 1.666

= 56.67

Hence, the median weight of the students is 56.67

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.4

Question 1. The following distribution gives the daily income of 50 workers if a factory. Convert the distribution above to a less-than-type cumulative frequency distribution and draw its ogive.

Daily income in Rupees 100-120 120-140 140-160 160-180 180-200
Number of workers 12 14 8 6 10

Solution:

According to the question, we convert the given distribution to a less than type cumulative frequency distribution, 

Daily income Frequency Cumulative Frequency
Less than 120 12 12
Less than 140 14 26
Less than 160 8 34
Less than 180 6 40
Less than 200 10 50

Now according to the table we plot the points that are corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40), and (200, 50) on a graph paper. Here x-axis represents the upper limit and y-axis represent the frequency. The curve obtained from the graph is known as less than type ogive curve. 

Class 10 NCERT Chapter 14 Exercise 14.4 Solution

Question 2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight in kg Number of students
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution:

According to the given table, we get the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35). Now using these points we draw an ogive, where the x-axis represents the upper limit and y-axis represents the frequency. The curve obtained is known as less than type ogive.

Class 10 NCERT Chapter 14 Exercise 14.4 Solution

Now, locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From this point, now we draw a perpendicular line to the x-axis and the intersection point which is perpendicular to x-axis is the median of the given data. After, locating point now we create a table to find the mode:

Class interval Number of students(Frequency) Cumulative Frequency
Less than 38 0 0
Less than 40 3 – 0 = 3 3
Less than 42 5 – 3 = 2 8
Less than 44 9 – 5 = 4 9
Less than 46 14 – 9 = 5 14
Less than 48 28 – 14 = 14 28
Less than 50 32 – 28 = 4 32
Less than 52 35 – 22 = 3 35

The class 46 – 48 has the maximum frequency, hence, this is the modal class

= 46, h = 2, f1 = 14, f0 = 5 and f2 = 4

Now we find the mode:

Mode = l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h

On substituting the values in the given formula, we get

= 46 + 0.95 = 46.95

Hence, the mode is verified.

Question 3. The following tables give the production yield per hectare of wheat of 100 farms of a village.

Production Yield 50-55 55-60 60-65 65-70 70-75 75-80
Number of farms 2 8 12 24 38 16

Change the distribution to a more than type distribution and draw its ogive.

Solution:

According to the question, we change the distribution to a more than type distribution.

Production Yield (kg/ha) Number of farms
More than or equal to 50 100
More than or equal to 55 100 – 2 = 98
More than or equal to 60 98 – 8 = 90
More than or equal to 65 90 – 12 = 78
More than or equal to 70 78 – 24 = 54
More than or equal to 75 54 – 38 = 16

Now, according to the table we draw the ogive by plotting the points. Here, the a-axis represents the upper limit and y-axis represents the cumulative frequency. And the points are(50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type ogive curve.

Class 10 NCERT Chapter 14 Exercise 14.4 Solution

Key Features of NCERT Solutions for Class 10 Maths Chapter 14 Statistics:

  • These NCERT solutions are developed by the GFG team, with a focus on students’ benefit.
  • These solutions are entirely accurate and can be used by students to prepare for their board exams. 
  • Each solution is presented in a step-by-step format with comprehensive explanations of the intermediate steps.

Also Check:

NCERT Solutions for Class 10 Maths Chapter 14 Statistics – FAQs

Q1: Why is it important to learn Statistics in NCERT Class 10 Maths Chapter 14?

Statistics help people and groups in resolving issues regardless of the industry, market, economic sector, etc. Any issue where relevant data can be gathered, examined, analysed, and presented so as to work towards an efficient resolution can benefit from the knowledge and assistance provided by Statistics. The possibilities are endless.

Q2: What topics are covered in NCERT Solutions for Class 10 Maths Chapter 14 Statistics?

NCERT Solutions for Class 10 Maths Chapter 14 Statistics covers topics such as calculation of mean, median, mode, frequency distribution and standard deviation and variance etc.

Q3: How can NCERT Solutions for Class 10 Maths Chapter 14 Statistics help me?

NCERT Solutions for Class 10 Maths Chapter 14 Statistics can help you solve the NCERT exercises in an easy mannner. If you are stuck on a problem you can find its solution here and free yourself from the frustration of being stuck on some question.

Q4: How many exercises are there in Class 10 Maths Chapter 14 Statistics?

There are 4 exercises in the NCERT Class 10 Maths Chapter 14 Statistics which covers all the important topics and sub-topics.

Q5: Where can I find NCERT Solutions for Class 10 Maths Chapter 14 Statistics?

You can find the NCERT Solutions for Class 10 Maths Chapter 14 Statistics in this article created by our team of experts at GeeksforGeeks.


Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape, GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out - check it out now!

Last Updated : 16 Nov, 2023
Like Article
Save Article
Previous
Next
Similar Reads
Complete Tutorials