NCERT Solutions for class 10 Maths Chapter 14- Statistics Exercise 14.3

NCERT Solutions for class 10 Maths Chapter 14- Statistics Exercise 14.3 have been provided here, prepared by our subject experts. These solutions are designed in accordance with NCERT syllabus and guidelines prescribed by CBSE.

NCERT solutions of 10th class Maths, are in the context of the upcoming session of 2019-2020. These are helpful for students, to do the revision at the time of exam and also to score well. Go through the solutions presented here for each question of exercise 14.3.

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Access other exercise solutions of class 10 Maths Chapter 14- Statistics

Exercise 14.1 Solutions 6 Question ( 6 long)

Exercise 14.2 Solutions 7 Question ( 7 long)

Exercise 14.4 Solutions 3 Question ( 3 long)

Access Answers of Maths NCERT class 10 Chapter 14 – Statistics Exercise 14.3

1. The following frequency distribution gives the monthly consumption of an 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:

Find the cumulative frequency of the given data as follows:

Class Interval

Frequency

Cumulative frequency

65-85

4

4

85-105

5

9

105-125

13

22

125-145

20

42

145-165

14

56

165-185

8

64

185-205

4

68

N=68

From the table, it is observed that, n = 68 and hence n/2=34

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

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

Median is calculated as follows:

Ncert solutions class 10 chapter 14-1

=125+((34−22)/20) × 20

=125+12 = 137

Therefore, median = 137

To calculate the mode:

Modal class = 125-145, 

f1=20, f0=13, f2=14 & h = 20

Mode formula:

Mode = l+ [(f1-f0)/(2f1-f0-f2)]×h

Mode = 125 + ((20-13)/(40-13-14))×20

=125+(140/13)

=125+10.77

=135.77

Therefore, mode = 135.77

Calculate 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

x̄ =a+h ∑fiui/∑fi =135+20(7/68)

Mean=137.05

In this case, mean, median and mode are more/less equal in this distribution.

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: 

Given data, n = 60

Median of the given data = 28.5

Where, 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

Ncert solutions class 10 chapter 14-2

Substitute the values

28.5=20+((30−5−x)/20) × 10

8.5 = (25 – x)/2

17 = 25-x

Therefore, x =8

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

Since,

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

Now, substitute the value of x, to find y

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

y = 60-53

y = 7

Therefore, the value of x = 8 and y = 7.

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

Age (in years)

Number of policy holder

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:

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

Ncert solutions class 10 chapter 14-3

Median = 35+((50-45)/33) × 5

= 35 + (5/33)5 

= 35.75

Therefore, the median age = 35.75 years.

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:

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

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

So, the data obtained are:

n = 40 and n/2 = 20

Median class = 144.5-153.5

then, l = 144.5,

cf = 17, f = 12 & h = 9

Ncert solutions class 10 chapter 14-4

Median = 144.5+((20-17)/12)×9

= 144.5+(9/4)

= 146.75 mm

Therefore, the median length of the leaves = 146.75 mm.

5. The following table gives the distribution of a life time 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:

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

Data:

n = 400 &n/2 = 200

Median class = 3000 – 3500

Therefore, l = 3000, Cf = 130,

f = 86 & h = 500

Ncert solutions class 10 chapter 14-5

Median = 3000 + ((200-130)/86) × 500

= 3000 + (35000/86)

= 3000 + 406.97

= 3406.97

Therefore, the median life time of the lamps = 3406.97 hours

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: 

To calculate median:

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

Given:

n = 100 &n/2 = 50

Median class = 7-10

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

Ncert solutions class 10 chapter 14-6

Median = 7+((50-36)/40) × 3

Median = 7+42/40

Median=8.05

Calculate the Mode:

Modal class = 7-10,

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

Ncert solutions class 10 chapter 14-7

Mode = 7+((40-30)/(2×40-30-16)) × 3

= 7+(30/34)

= 7.88

Therefore mode = 7.88

Calculate 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 = x̄ = ∑fi xi /∑f

Mean = 825/100 = 8.25

Therefore, mean = 8.25

 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:

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

Given: n = 30 and n/2= 15

Median class = 55-60

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

Ncert solutions class 10 chapter 14-8

Median = 55+((15-13)/6)×5

Median=55 + (10/6) = 55+1.666

Median =56.67

Therefore, the median weight of the students = 56.67


In this exercise, students will solve questions based on finding the median of a grouped data, which is a measure of central tendency which gives the value of the middle-most observation in the data.

A group of collection of data will be provided to the students to find the median for cases having odd and even number of observations. To solve complete problems for class 10 Maths chapter 14 exercise-wise, students can visit us any time. Also get few more study materials such as notes, books, question papers, etc. to practice well.

The problems in Ex.14.3 is solved as per the guidance and formulas, explained in the example questions before the exercise. NCERT solutions serve as the best materials for students who are looking for solutions for all the subjects class-wise and chapter-wise.

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