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NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Last Updated : 16 Nov, 2023
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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)5-1515-2525-3535-4545-5555-65
Number of patients6112123145

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 IntervalFrequency (fi)Mid-point (xi)fixi
5-1561060
15-251120220
25-352130630
35-452340920
45-551450700
55-65560300
 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)0-2020-4040-6060-8080-100100-120
Frequency103552613829

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:

ExpenditureNumber of families
1000-150024
1500-200040
2000-250033
2500-300028
3000-350030
3500-400022
4000-450016
4500-50007

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 Intervalfixidi = xi – aui = di/hfiui
1000-1500241250-1500-3-72
1500-2000401750-1000-2-80
2000-2500332250-500-1-33
2500-3000282750000
3000-3500303250500130
3500-40002237501000244
4000-45001642501500348
4500-5000747502000428
 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 teacherNumber of states / U.T
15-203
20-258
25-309
30-3510
35-403
40-450
45-500
50-552

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 IntervalFrequency (fi)Mid-point (xi)fixi
15-20317.552.5
20-25822.5180.0
25-30927.5247.5
30-351032.5325.0
35-40337.5112.5
40-45042.50
45-50047.50
50-55252.5105.5
 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 ScoredNumber of Batsman
3000-40004
4000-500018
5000-60009
6000-70007
7000-80006
8000-90003
9000-100001
10000-110001

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 carsFrequency
0-107
10-2014
20-3013
30-4012
40-5020
50-6011
60-7015
70-808

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
65-854
85-1055
105-12513
125-14520
145-16514
165-1858
185-2054

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 Intervalfixidi = xi – aui = di/hfiui
65-85475-60-3-12
85-105595-40-2-10
105-12513115-20-1-13
125-14520135000
145-1651415520114
165-185817540216
185-205419560312
 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 IntervalFrequency
0-105
10-20x
20-3020
30-4015
40-50y
50-605
Total60

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 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Solution: 

According to the given question the table is 

Class intervalFrequencyCumulative frequency
15-2022
20-2546
25-301824
30-352145
35-403378
40-451189
45-50392
50-55698
55-602100

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
118-1263
127-1355
136-1449
145-15312
154-1625
163-1714
172-1802

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 IntervalFrequencyCumulative frequency
117.5-126.533
126.5-135.558
135.5-144.5917
144.5-153.51229
153.5-162.5534
162.5-171.5438
171.5-180.5240

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
1500-200014
2000-250056
2500-300060
3000-350086
3500-400074
4000-450062
4500-500048

Find the median lifetime of a lamp.

Solution:

According to the question

Class IntervalFrequencyCumulative
1500-20001414
2000-25005670
2500-300060130
3000-350086216
3500-400074290
4000-450062352
4500-500048400

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 letters1-44-77-1010-1313-1616-19
Number of surnames630401644

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 IntervalFrequencyCumulative Frequency
1-466
4-73036
7-104076
10-131692
13-16496
16-194100

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 Intervalfixifixi
1-462.515
4-7305.5165
7-10408.5340
10-131611.5184
13-16414.551
16-19417.570
 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)40-4545-5050-5555-6060-6565-7070-75
Number of students2386632

Solution:

According to the question

Class IntervalFrequencyCumulative frequency
40-4522
45-5035
50-55813
55-60619
60-65625
65-70328
70-75230

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 Rupees100-120120-140140-160160-180180-200
Number of workers12148610

Solution:

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

Daily incomeFrequencyCumulative Frequency
Less than 1201212
Less than 1401426
Less than 160834
Less than 180640
Less than 2001050

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 kgNumber of students
Less than 380
Less than 403
Less than 425
Less than 449
Less than 4614
Less than 4828
Less than 5032
Less than 5235

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 intervalNumber of students(Frequency)Cumulative Frequency
Less than 3800
Less than 403 – 0 = 33
Less than 425 – 3 = 28
Less than 449 – 5 = 49
Less than 4614 – 9 = 514
Less than 4828 – 14 = 1428
Less than 5032 – 28 = 432
Less than 5235 – 22 = 335

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 Yield50-5555-6060-6565-7070-7575-80
Number of farms2812243816

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 50100
More than or equal to 55100 – 2 = 98
More than or equal to 6098 – 8 = 90
More than or equal to 6590 – 12 = 78
More than or equal to 7078 – 24 = 54
More than or equal to 7554 – 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.



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