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Class 10 NCERT Solutions – Chapter 14 Statistics – Exercise 14.1

  • Last Updated : 15 Dec, 2020

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

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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%.




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