Padhona

Mathematics · Statistics · NCERT Exercises

Exercise 13.3

Complete, independently verified solutions for NCERT Exercise 13.3.

Mathematics · Chapter 13 · NCERT Exercises

Download PDF
Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

Jump navigator

Choose a question

Question

Question 1

median mean mode · electricity consumption

The monthly electricity consumption of 68 consumers is distributed below. Find the median, mean and mode, then compare them.

Consumption (units)

Consumers

65–85

4

85–105

5

105–125

13

125–145

20

145–165

14

165–185

8

185–205

4

Solution

Board-exam working:

Mean calculation:

Class

fᵢ

xᵢ

fᵢxᵢ

Cumulative f

65–85

4

75

300

4

85–105

5

95

475

9

105–125

13

115

1495

22

125–145

20

135

2700

42

145–165

14

155

2170

56

165–185

8

175

1400

64

185–205

4

195

780

68

∑fi=68,∑fixi=9320\sum f_i=68,\qquad \sum f_ix_i=9320
xˉ=932068=137.0588…\bar{x}=\frac{9320}{68}=137.0588\ldots

Median calculation:

n2=34\frac n2=34

The first cumulative frequency above 34 is 42, so the median class is 125–145.

l=125,cf=22,f=20,h=20l=125,\quad cf=22,\quad f=20,\quad h=20
Median=125+34−2220×20=137\text{Median}=125+\frac{34-22}{20}\times20=137

Mode calculation; the modal class is also 125–145.

l=125,h=20,f1=20,f0=13,f2=14l=125,\quad h=20,\quad f_1=20,\quad f_0=13,\quad f_2=14
Mode=125+20−132(20)−13−14×20\text{Mode}=125+\frac{20-13}{2(20)-13-14}\times20
=125+14013=135.7692…=125+\frac{140}{13}=135.7692\ldots

Final answer

Median=137;xˉ≈137.06;Mode≈135.77 units\boxed{\text{Median}=137;\quad \bar{x}\approx137.06;\quad \text{Mode}\approx135.77\text{ units}}

Question

Question 2

median · missing frequencies

The median of the distribution is 28.5. Find x and y.

Class

Frequency

0–10

5

10–20

x

20–30

20

30–40

15

40–50

y

50–60

5

Total

60

Solution

Board-exam working:

From the total frequency:

5+x+20+15+y+5=605+x+20+15+y+5=60
x+y=15(1)x+y=15\qquad\text{(1)}

Since 28.5 lies in 20–30, that is the median class.

l=20,f=20,cf=5+x,h=10,n2=30l=20,\quad f=20,\quad cf=5+x,\quad h=10,\quad \frac n2=30
28.5=20+30−(5+x)20×1028.5=20+\frac{30-(5+x)}{20}\times10
8.5=25−x28.5=\frac{25-x}{2}
17=25−x⇒x=817=25-x\Rightarrow x=8

Use equation (1).

8+y=15⇒y=78+y=15\Rightarrow y=7

Final answer

x=8,y=7\boxed{x=8,\qquad y=7}

Question

Question 3

median age · less-than cumulative frequency

An insurance agent recorded the ages of 100 policy holders. Policies are issued from age 18 to below 60. Calculate the median age.

Age

Policy holders

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

Board-exam working:

Convert cumulative frequencies to ordinary frequencies by subtraction.

Age class

Frequency

Cumulative frequency

18–20

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

n=100,n2=50n=100,\qquad \frac n2=50

The first cumulative frequency above 50 is 78, so the median class is 35–40.

l=35,cf=45,f=33,h=5l=35,\quad cf=45,\quad f=33,\quad h=5
Median=35+50−4533×5\text{Median}=35+\frac{50-45}{33}\times5
=35+2533=35.7575…=35+\frac{25}{33}=35.7575\ldots

Final answer

Median age≈35.76 years\boxed{\text{Median age}\approx35.76\text{ years}}

Question

Question 4

median leaves · continuous classes

The lengths of 40 leaves were measured to the nearest millimetre. Find the median length.

Length (mm)

Leaves

118–126

3

127–135

5

136–144

9

145–153

12

154–162

5

163–171

4

172–180

2

Solution

Board-exam working:

Because measurements are to the nearest millimetre, subtract 0.5 from each lower limit and add 0.5 to each upper limit.

Continuous class

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

n=40,n2=20n=40,\qquad \frac n2=20

The median class is 144.5–153.5.

l=144.5,cf=17,f=12,h=9l=144.5,\quad cf=17,\quad f=12,\quad h=9
Median=144.5+20−1712×9\text{Median}=144.5+\frac{20-17}{12}\times9
=144.5+2.25=146.75=144.5+2.25=146.75

Final answer

Median leaf length=146.75 mm\boxed{\text{Median leaf length}=146.75\text{ mm}}

Question

Question 5

median lifetime · neon lamps

The lifetimes of 400 neon lamps are distributed below. Find the median lifetime.

Lifetime (hours)

Lamps

1500–2000

14

2000–2500

56

2500–3000

60

3000–3500

86

3500–4000

74

4000–4500

62

4500–5000

48

Solution

Board-exam working:

Class

Frequency

Cumulative frequency

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,n2=200n=400,\qquad \frac n2=200

The first cumulative frequency above 200 is 216, so the median class is 3000–3500.

l=3000,cf=130,f=86,h=500l=3000,\quad cf=130,\quad f=86,\quad h=500
Median=3000+200−13086×500\text{Median}=3000+\frac{200-130}{86}\times500
=3000+406.9767…=3406.9767…=3000+406.9767\ldots=3406.9767\ldots

Final answer

Median lifetime≈3406.98 hours\boxed{\text{Median lifetime}\approx3406.98\text{ hours}}

Question

Question 6

median mean mode · surname length

The number of letters in 100 surnames is distributed below. Determine the median, mean and modal surname length.

Letters

Surnames

1–4

6

4–7

30

7–10

40

10–13

16

13–16

4

16–19

4

Solution

Board-exam working:

Mean calculation:

Class

fᵢ

xᵢ

fᵢxᵢ

Cumulative f

1–4

6

2.5

15

6

4–7

30

5.5

165

36

7–10

40

8.5

340

76

10–13

16

11.5

184

92

13–16

4

14.5

58

96

16–19

4

17.5

70

100

xˉ=832100=8.32\bar{x}=\frac{832}{100}=8.32

Median calculation; the median class is 7–10.

l=7,cf=36,f=40,h=3,n2=50l=7,\quad cf=36,\quad f=40,\quad h=3,\quad \frac n2=50
Median=7+50−3640×3=8.05\text{Median}=7+\frac{50-36}{40}\times3=8.05

Mode calculation; the modal class is 7–10.

l=7,h=3,f1=40,f0=30,f2=16l=7,\quad h=3,\quad f_1=40,\quad f_0=30,\quad f_2=16
Mode=7+40−302(40)−30−16×3\text{Mode}=7+\frac{40-30}{2(40)-30-16}\times3
=7+3034=7.8823…=7+\frac{30}{34}=7.8823\ldots

Final answer

Median=8.05;xˉ=8.32;Mode≈7.88 letters\boxed{\text{Median}=8.05;\quad \bar{x}=8.32;\quad \text{Mode}\approx7.88\text{ letters}}

Question

Question 7

median weight · students

The weights of 30 students are distributed below. Find the median weight.

Weight (kg)

Students

40–45

2

45–50

3

50–55

8

55–60

6

60–65

6

65–70

3

70–75

2

Solution

Board-exam working:

Class

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,n2=15n=30,\qquad \frac n2=15

The first cumulative frequency above 15 is 19, so the median class is 55–60.

l=55,cf=13,f=6,h=5l=55,\quad cf=13,\quad f=6,\quad h=5
Median=55+15−136×5\text{Median}=55+\frac{15-13}{6}\times5
=55+53=56.666…=55+\frac53=56.666\ldots

Final answer

Median weight≈56.67 kg\boxed{\text{Median weight}\approx56.67\text{ kg}}