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Mathematics · Statistics

NCERT Examples

All 8 worked examples from Statistics, with independently verified solutions and direct navigation.

Mathematics · Chapter 13 · NCERT Examples

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NCERT Example

Example 1

mean · ungrouped frequency distribution · direct method

The marks obtained by 30 Class X students in a 100-mark Mathematics paper are given below. Find the mean marks.

Marks obtained by 30 students

Marks obtained

Number of students

10

1

20

1

36

3

40

4

50

3

56

2

60

4

70

4

72

1

80

1

88

2

92

3

95

1

Solution

Board-exam working:

Multiply every observation by its corresponding frequency.

Direct-method calculation

Marks (xᵢ)

Frequency (fᵢ)

fᵢxᵢ

10

1

10

20

1

20

36

3

108

40

4

160

50

3

150

56

2

112

60

4

240

70

4

280

72

1

72

80

1

80

88

2

176

92

3

276

95

1

95

∑fi=30,∑fixi=1779\sum f_i=30,\qquad \sum f_ix_i=1779
xˉ=∑fixi∑fi=177930=59.3\bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{1779}{30}=59.3

Final answer

xˉ=59.3 marks\boxed{\bar{x}=59.3\text{ marks}}

NCERT Example

Example 2

mean · grouped data · direct assumed mean step deviation

The percentage distribution of female teachers in primary schools of rural areas of various States and Union Territories is given below. Find the mean percentage by the direct, assumed-mean and step-deviation methods.

Female teachers (%)

States/U.T. (fᵢ)

15–25

6

25–35

11

35–45

7

45–55

4

55–65

4

65–75

2

75–85

1

Solution

Board-exam working:

The class marks are the midpoints of the intervals. Take the assumed mean as 50 and class size as 10.

Class marks and deviations (a = 50, h = 10)

Class

fᵢ

xᵢ

dᵢ

uᵢ

15–25

6

20

−30

−3

25–35

11

30

−20

−2

35–45

7

40

−10

−1

45–55

4

50

0

0

55–65

4

60

10

1

65–75

2

70

20

2

75–85

1

80

30

3

Products used by the three methods

Class

fᵢxᵢ

fᵢdᵢ

fᵢuᵢ

15–25

120

−180

−18

25–35

330

−220

−22

35–45

280

−70

−7

45–55

200

0

0

55–65

240

40

4

65–75

140

40

4

75–85

80

30

3

∑fi=35,∑fixi=1390,∑fidi=−360,∑fiui=−36\sum f_i=35,\quad \sum f_ix_i=1390,\quad \sum f_id_i=-360,\quad \sum f_iu_i=-36

Direct method:

xˉ=139035=39.714…\bar{x}=\frac{1390}{35}=39.714\ldots

Assumed-mean method:

xˉ=50+−36035=39.714…\bar{x}=50+\frac{-360}{35}=39.714\ldots

Step-deviation method:

xˉ=50+10(−3635)=39.714…\bar{x}=50+10\left(\frac{-36}{35}\right)=39.714\ldots

Final answer

xˉ≈39.71%\boxed{\bar{x}\approx39.71\%}

NCERT Example

Example 3

mean · unequal class intervals · step deviation

The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean and state what it signifies.

Wickets

Bowlers (fᵢ)

20–60

7

60–100

5

100–150

16

150–250

12

250–350

2

350–450

3

Solution

Board-exam working:

The class sizes vary. Use class marks with assumed mean 200 and convenient divisor 20.

Class

fᵢ

xᵢ

dᵢ = xᵢ − 200

uᵢ = dᵢ/20

fᵢuᵢ

20–60

7

40

−160

−8

−56

60–100

5

80

−120

−6

−30

100–150

16

125

−75

−3.75

−60

150–250

12

200

0

0

0

250–350

2

300

100

5

10

350–450

3

400

200

10

30

∑fi=45,∑fiui=−106\sum f_i=45,\qquad \sum f_iu_i=-106
xˉ=200+20(−10645)\bar{x}=200+20\left(\frac{-106}{45}\right)
=200−47.111…=152.888…=200-47.111\ldots=152.888\ldots

Thus these 45 bowlers took about 152.89 wickets on average.

Final answer

xˉ≈152.89 wickets\boxed{\bar{x}\approx152.89\text{ wickets}}

NCERT Example

Example 4

mode · ungrouped data

The wickets taken by a bowler in 10 cricket matches were 2, 6, 4, 5, 0, 2, 1, 3, 2 and 3. Find the mode.

Solution

Board-exam working:

Frequency distribution

Wickets

Number of matches

0

1

1

1

2

3

3

2

4

1

5

1

6

1

The observation with the greatest frequency is the mode.

f(2)=3, which is the maximum frequencyf(2)=3\text{, which is the maximum frequency}

Final answer

Mode=2 wickets\boxed{\text{Mode}=2\text{ wickets}}

NCERT Example

Example 5

mode · grouped data · modal class

A survey of 20 households gave the following frequency table for family size. Find the mode.

Family size

Families

1–3

7

3–5

8

5–7

2

7–9

2

9–11

1

Solution

Board-exam working:

The highest frequency is 8, so the modal class is 3–5.

l=3,h=2,f1=8,f0=7,f2=2l=3,\quad h=2,\quad f_1=8,\quad f_0=7,\quad f_2=2
Mode=l+f1−f02f1−f0−f2×h\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
=3+8−72(8)−7−2×2=3+\frac{8-7}{2(8)-7-2}\times2
=3+27=3.2857…=3+\frac{2}{7}=3.2857\ldots

Final answer

Mode≈3.286 members\boxed{\text{Mode}\approx3.286\text{ members}}

NCERT Example

Example 6

mode · mean comparison · grouped marks

For the grouped marks distribution below, find the mode and compare it with the mean.

Marks

Students

10–25

2

25–40

3

40–55

7

55–70

6

70–85

6

85–100

6

Solution

Board-exam working:

The maximum frequency is 7, so the modal class is 40–55.

l=40,h=15,f1=7,f0=3,f2=6l=40,\quad h=15,\quad f_1=7,\quad f_0=3,\quad f_2=6
Mode=40+7−32(7)−3−6×15\text{Mode}=40+\frac{7-3}{2(7)-3-6}\times15
=40+45×15=52=40+\frac45\times15=52

The mean of the same grouped data is 62. The greatest concentration is around 52 marks, while the arithmetic average is 62 marks.

Final answer

Mode=52 marks;xˉ=62 marks\boxed{\text{Mode}=52\text{ marks};\quad \bar{x}=62\text{ marks}}

NCERT Example

Example 7

median · less-than cumulative frequency · heights

The following less-than cumulative-frequency distribution gives the heights of 51 Class X girls. Find the median height.

Height (cm)

Girls

Less than 140

4

Less than 145

11

Less than 150

29

Less than 155

40

Less than 160

46

Less than 165

51

Solution

Board-exam working:

Subtract successive cumulative frequencies to obtain class frequencies.

Class

Frequency

Cumulative frequency

Below 140

4

4

140–145

7

11

145–150

18

29

150–155

11

40

155–160

6

46

160–165

5

51

n=51,n2=25.5n=51,\qquad \frac n2=25.5

The first cumulative frequency greater than 25.5 is 29, so the median class is 145–150.

l=145,cf=11,f=18,h=5l=145,\quad cf=11,\quad f=18,\quad h=5
Median=l+n2−cff×h\text{Median}=l+\frac{\frac n2-cf}{f}\times h
=145+25.5−1118×5=145+\frac{25.5-11}{18}\times5
=145+72.518=149.027…=145+\frac{72.5}{18}=149.027\ldots

Final answer

Median height≈149.03 cm\boxed{\text{Median height}\approx149.03\text{ cm}}

NCERT Example

Example 8

median · missing frequencies

The median of the following distribution is 525 and the total frequency is 100. Find x and y.

Class

Frequency

0–100

2

100–200

5

200–300

x

300–400

12

400–500

17

500–600

20

600–700

y

700–800

9

800–900

7

900–1000

4

Solution

Board-exam working:

Using the total frequency:

2+5+x+12+17+20+y+9+7+4=1002+5+x+12+17+20+y+9+7+4=100
x+y=24(1)x+y=24\qquad\text{(1)}

Since 525 lies in 500–600, that is the median class.

l=500,f=20,cf=36+x,h=100,n2=50l=500,\quad f=20,\quad cf=36+x,\quad h=100,\quad \frac n2=50
525=500+50−(36+x)20×100525=500+\frac{50-(36+x)}{20}\times100
25=5(14−x)=70−5x25=5(14-x)=70-5x
5x=45⇒x=95x=45\Rightarrow x=9

Substitute in equation (1).

9+y=24⇒y=159+y=24\Rightarrow y=15

Final answer

x=9,y=15\boxed{x=9,\qquad y=15}