# Chapter 14 – Statistics Questions and Answers: NCERT Solutions for Class 10 Maths

Class 10 Maths NCERT book solutions for Chapter 14 - Statistics Questions and Answers.

Education Blogs Chapter 14 – Statistics Questions and Answers: NCERT Solutions for Class 10...

Class 10 Maths NCERT book solutions for Chapter 14 - Statistics Questions and Answers.

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?

In order to find the mean value, we will use direct method because the numerical value of fiand xiare small.

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

No. of plants

(Class interval)

No. of houses

Frequency (fi)

Mid-point (xi)

fixi

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

The formula to find the mean is:

Mean = x̄ = ∑fixi/∑fi

= 162/20

= 8.1

Daily wages (in Rs.)

100-120

120-140

140-160

160-180

180-200

Number of workers

12

14

8

6

10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150 and class interval is h = 20.

So, ui= (xi– A)/h = ui = (xi– 150)/20

Substitute and find the values as follows:

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

So, the formula to find out the mean is:

Mean = x̄ = A + h∑fiui/∑fi=150 + (20 × -12/50) = 150 – 4.8 = 145.20

Thus, mean daily wage of the workers = Rs. 145.20

Daily Pocket Allowance(in c)

11-13

13-15

15-17

17-19

19-21

21-23

23-35

Number of children

7

6

9

13

f

5

4

To find out the missing frequency, use the mean formula.

Here, the value of mid-point (xi) meanx̄= 18

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

The mean formula is

Mean = x̄ = ∑fixi/∑fi= (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.

Number of heart beats 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

From the given data, let us assume the mean as A = 75.5

xi= (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the uiand fiuias follows:

Class Interval

Number of women (fi)

Mid-point (xi)

ui= (xi– 75.5)/h

fiui

65-68

2

66.5

-3

-6

68-71

4

69.5

-2

-8

71-74

3

72.5

-1

-3

74-77

8

75.5

0

0

77-80

7

78.5

1

7

80-83

4

81.5

3

8

83-86

2

84.5

3

6

Mean = x̄ = A + h∑fiui/∑fi

= 75.5 + 3×(4/30)

= 75.5 + 4/10

= 75.5 + 0.4

= 75.9

Therefore, the mean heart beats per minute for these women is 75.9

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?

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals are 1

Here, assumed mean (A) = 57

Class size (h) = 3

Here, the step deviation is used because the frequency values are big.

Class Interval

Number of boxes (fi)

Mid-point (xi)

di= xi– A

fidi

49.5-52.5

15

51

-6

90

52.5-55.5

110

54

-3

-330

55.5-58.5

135

57 = A

0

0

58.5-61.5

115

60

3

345

61.5-64.5

25

63

6

150

The formula to find out the Mean is:

Mean = x̄ = A +h ∑fidi/∑fi

= 57 + 3(75/400)

= 57 + 0.1875

= 57.19

Therefore, the mean number of mangoes kept in a packing box is 57.19

Daily expenditure(in c)

100-150

150-200

200-250

250-300

300-350

Number of households

4

5

12

2

2

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

Let is assume the mean (A) = 225

Class size (h) = 50

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

0

0

0

250-300

2

275

50

1

2

300-350

2

325

100

2

4

Mean = x̄ = A +h∑fiui/∑fi

= 225+50(-7/25)

= 225-14

= 211

Therefore, the mean daily expenditure on food is 211

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 SO2in the air.

To find out the mean, first find the midpoint of the given frequencies as follows:

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

0.20

0.40

Total

Sum fi= 30

The formula to find out the mean is

Mean = x̄ = ∑fixi/∑fi

= 2.96/30

= 0.099 ppm

Therefore, the mean concentration of SO2in air is 0.099 ppm.

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

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

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

The mean formula is,

Mean = x̄ = ∑fixi/∑fi

= 499/40

= 12.48 days

Therefore, the mean number of days a student was absent = 12.48.

literacy rate.

Literacy rate (in %)

45-55

55-65

65-75

75-85

85-98

Number of cities

3

10

11

8

3

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 70 and class interval is h = 10.

So, ui= (xi-A)/h = ui= (xi-70)/10

Substitute and find the values as follows:

Class Interval

Frequency (fi)

(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

0

0

0

75-85

8

80

10

1

8

85-95

3

90

20

2

6

So, Mean = x̄ = A+(∑fiui/∑fi)×h

= 70+(-2/35)×10

= 69.42

Therefore, the mean literacy part = 69.42

Age (in years)

5-15

15-25

25-35

35-45

45-55

55-65

Number of patients

6

11

21

23

14

5

Find the mode and the mean of the data given above. Compare and interpret the two

measures of central tendency.

To find out the modal class, let us the consider the class interval with high frequency

Here, the greatest frequency = 23, so the modal class = 35 – 45,

l = 35,

class width (h) = 10,

fm= 23,

f1= 21 and f2= 14

The formula to find the mode is

Mode= l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

Mode = 35+(20/11) = 35+1.8

Mode = 36.8 year

So the mode of the given data = 36.8 year

Calculation of Mean:

First find the midpoint using the formula, xi= (upper limit +lower limit)/2

Class Interval

Frequency (fi)

Mid-point (xi)

fixi

5-15

6

10

60

15-25

11

20

220

25-35

21

30

630

35-45

23

40

920

45-55

14

50

700

55-65

5

60

300

The mean formula is

Mean = x̄ = ∑fixi/∑fi

= 2830/80

= 35.37 years

Therefore, the mean of the given data = 35.37 years

electrical components:

Lifetime (in hours)

0-20

20-40

40-60

60-80

80-100

100-120

Frequency

10

35

52

61

38

29

Determine the modal lifetimes of the components.

From the given data the modal class is 60–80.

l = 60,

The frequencies are:

fm= 61, f1= 52, f2= 38 and h = 20

The formula to find the mode is

Mode= l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values in the formula, we get

Mode =60+[(61-52)/(122-52-38)]×20

Mode = 60+((9 x 20)/32)

Mode = 60+(45/8) = 60+ 5.625

Therefore, modal lifetime of the components = 65.625 hours.

families of a village. Find the modal monthly expenditure of the families. Also, find the

mean monthly expenditure:

Expenditure

Number of families

1000-1500

24

1500-2000

40

2000-2500

33

2500-3000

28

3000-3500

30

3500-4000

22

4000-4500

16

4500-5000

7

Given data:

Modal class = 1500-2000,

l= 1500,

Frequencies:

fm= 40 f1= 24, f2= 33 and

h = 500

Mode formula:

Mode= l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values in the formula, we get

Mode =1500+[(40-24)/(80-24-33)]×500

Mode = 1500+((16×500)/23)

Mode = 1500+(8000/23) = 1500 + 347.83

Therefore, modal monthly expenditure of the families = Rupees 1847.83

Calculation for mean:

First find the midpoint using the formula, xi=(upper limit +lower limit)/2

Let us assume a mean, A be 2750

Class Interval

fi

xi

di = xi – a

ui = di/h

fiui

1000-1500

24

1250

-1500

-3

-72

1500-2000

40

1750

-1000

-2

-80

2000-2500

33

2250

-500

-1

-33

2500-3000

28

2750

0

0

0

3000-3500

30

3250

500

1

30

3500-4000

22

3750

1000

2

44

4000-4500

16

4250

1500

3

48

4500-5000

7

4750

2000

4

28

The formula to calculate the mean,

Mean = x̄ = a +(∑fiui/∑fi)×h

Substitute the values in the given formula

= 2750+(-35/200)×500

= 2750-87.50

= 2662.50

So, the mean monthly expenditure of the families = Rupees 2662.50

No of Students per teacher

Number of states / U.T

15-20

3

20-25

8

25-30

9

30-35

10

35-40

3

40-45

0

45-50

0

50-55

2

Given data:

Modal class = 30 – 35,

l= 30,

Class width (h) = 5,

fm= 10, f1= 9 and f2= 3

Mode Formula:

Mode= l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values in the given formula

Mode = 30+((10-9)/(20-9-3))×5

Mode = 30+(5/8) = 30+0.625

Mode = 30.625

Therefore, the mode of the given data = 30.625

Calculation of mean:

Find the midpoint using the formula, xi=(upper limit +lower limit)/2

Class Interval

Frequency (fi)

Mid-point (xi)

fixi

15-20

3

17.5

52.5

20-25

8

22.5

180.0

25-30

9

27.5

247.5

30-35

10

32.5

325.0

35-40

3

37.5

112.5

40-45

0

42.5

0

45-50

0

47.5

0

50-55

2

52.5

105.5

Mean = x̄ = ∑fixi/∑fi

= 1022.5/35

= 29.2

Therefore, mean = 29.2

Run Scored

Number of Batsman

3000-4000

4

4000-5000

18

5000-6000

9

6000-7000

7

7000-8000

6

8000-9000

3

9000-10000

1

10000-11000

1

Find the mode of the data.

Given data:

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000,

fm= 18, f1= 4 and f2= 9

Mode Formula:

Mode= l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values

Mode = 4000+((18-4)/(36-4-9))×1000

Mode = 4000+(14000/23) = 4000+608.695

Mode = 4608.695

Mode = 4608.7 (approximately)

Thus, the mode of the given data is 4608.7 runs

Number of cars

Frequency

0-10

7

10-20

14

20-30

13

30-40

12

40-50

20

50-60

11

60-70

15

70-80

8

Given Data:

Modal class = 40 – 50, l = 40,

Class width (h) = 10, fm= 20, f1= 12 and f2= 11

Mode= l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values

Mode = 40+((20-12)/(40-12-11))×10

Mode = 40 + (80/17) = 40 + 4.7 = 44.7

Thus, the mode of the given data is 44.7 cars

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

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

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:

=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

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.

Class Interval

Frequency

0-10

5

10-20

x

20-30

20

30-40

15

40-50

y

50-60

5

Total

60

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

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.

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

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 andn/2= 50

Median class = 35-45

Then,l= 35, cf= 45, f = 33 & h = 5

= 35 + (5/33)5

= 35.75

Therefore, the median age = 35.75 years.

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.

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 andn/2= 20

Median class = 144.5-153.5

then,l= 144.5,

cf = 17, f = 12 & h = 9

= 144.5+(9/4)

= 146.75mm

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

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.

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

= 3000 + (35000/86)

= 3000 + 406.97

= 3406.97

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

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

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

= 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

Mean = x̄ = ∑fixi/∑fi

Mean = 825/100 = 8.25

Therefore, mean = 8.25

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

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 andn/2= 15

Median class = 55-60

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

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

Median =56.67

Therefore, the median weight of the students = 56.67

Daily income in Rupees

100-120

120-140

140-160

160-180

180-200

Number of workers

12

14

8

6

10

Convert the given distribution table to a less than type cumulative frequency distribution, and we get

Daily income

Frequency

Cumulative Frequency

Less than 120

12

12

Less than 140

14

26

Less than 160

8

34

Less than 180

6

40

Less than 200

10

50

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50)on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve

Weight in kg

Number of students

Less than 38

0

Less than 40

3

Less than 42

5

Less than 44

9

Less than 46

14

Less than 48

28

Less than 50

32

Less than 52

35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verifythe result by using the formula.

From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35)on a graph paper an join them to get a smooth curve. The curve obtained is known as less than type ogive.

Class interval

Number of students(Frequency)

Cumulative Frequency

Less than 38

0

0

Less than 40

3-0=3

3

Less than 42

5-3=2

5

Less than 44

9-5=4

9

Less than 46

14-9=5

14

Less than 48

28-14=14

28

Less than 50

32-28=4

32

Less than 52

35-22=3

35

The class 46 – 48 has the maximum frequency, therefore, this is modal class

Here,l= 46, h = 2,f1= 14,f0= 5 andf2= 4

The mode formula is given as:

Now, Mode =

Thus, mode is verified.

Production Yield

50-55

55-60

60-65

65-70

70-75

75-80

Number of farms

2

8

12

24

38

16

Change the distribution to a more than type distribution and draw its ogive.

Converting the given distribution to a more than type distribution, we get

Production Yield (kg/ha)

Number of farms

More than or equal to 50

100

More than or equal to 55

100-2 = 98

More than or equal to 60

98-8= 90

More than or equal to 65

90-12=78

More than or equal to 70

78-24=54

More than or equal to 75

54-38 =16

From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis 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.

Advertisements