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Statistics For Class 10th

Class 10
Statistics For Class 10th

Statistics for class 10th students -Welcome to the world of statistics! As a student in 10th grade, you may be wondering why statistics is even relevant to your studies. The truth is, statistics is an essential branch of mathematics that helps us make sense of the world around us. By analyzing data and drawing conclusions, we can better understand patterns and trends in everything from social issues to scientific experiments. In this article, we will explore the fundamentals of statistics for class 10th students, including concepts such as mean, median, and mode, as well as how to represent data using graphs and charts. So, let's dive in and explore the exciting world of statistics!

Answers to Maths NCERT Class 10 Chapter 14 – Statistics

Exercise 14.1 Page: 270 - Statistics for class 10th students

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

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:

To find the mean value, we will use the direct method because the numerical value of fi and xi are 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

 

Sum f= 20

 

Sum fixi = 162

The formula to find the mean is:

Mean = x̄ = ∑fxi /∑fi

= 162/20

= 8.1

Therefore, the mean number of plants per house is 8.1. - Statistics for class 10th students

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs.)

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:

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 = 550.

Class interval (h) = 20

So, u= (xi – a)/h

u= (xi – 550)/20

Substitute and find the values as follows:

Daily wages

(Class interval)

Number of workers

frequency (fi)

Mid-point (xi)

u= (xi – 550)/20

fiui

500-520

12

510

-2

-24

520-540

14

530

-1

-14

540-560

8

550 = a

0

0

560-580

6

570

1

6

580-600

10

590

2

20

Total

Sum f= 50

   

Sum fiui = -12

So, the formula to find out the mean is:

Mean = x̄ = a + h(∑fiui /∑f) = 550 + [20 × (-12/50)] = 550 – 4.8 = 545.20

Thus, mean daily wage of the workers = Rs. 545.20

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. -Statistics for class 10th students

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

Solution:

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

Given, mean x̄ = 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

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 = x̄ = ∑fixi /∑f= (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.

4. Thirty women were examined in a hospital by a doctor, and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

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

Solution:

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

x= (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the uand fiui as 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 = a

0

0

77-80

7

78.5

1

7

80-83

4

81.5

2

8

83-86

2

84.5

3

6

 

Sum fi= 30

   

Sum fiu= 4

Mean = x̄ = a + h(∑fiui /∑f)

= 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

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number 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? -Statistics for class 10th students

Solution:

The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 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)

ui = (xi – 57)/h

fiui

49.5-52.5

15

51

-2

-30

52.5-55.5

110

54

-1

-110

55.5-58.5

135

57 = a

0

0

58.5-61.5

115

60

1

115

61.5-64.5

25

63

2

50

 

Sum fi = 400

   

Sum fiui = 25

The formula to find out the Mean is:

Mean = x̄ = a + h(∑fiui /∑f)

= 57 + 3(25/400)

= 57 + 0.1875

= 57.19

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

6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure(in c)

100-150

150-200

200-250

250-300

300-350

Number of households

4

5

12

2

2

Find the mean daily expenditure on food by a suitable method.

Solution:

Find the midpoint of the given interval using the formula.

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

Let us assume the mean (a) = 225

Class size (h) = 50

Class Interval

Number of households (fi)

Mid-point (xi)

di = xi – A

u= 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 = x̄ = a + h(∑fiui /∑f)

= 225 + 50(-7/25)

= 225 – 14

= 211

Therefore, the mean daily expenditure on food is 211.

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:

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

Concentration of SO(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

Total

Sum fi = 30

 

Sum (fixi) = 2.96

The formula to find out the mean is

Mean = x̄ = ∑fixi /∑fi

= 2.96/30

= 0.099 ppm

Therefore, the mean concentration of SO2 in the air is 0.099 ppm. - Statistics for class 10th students

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:

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

 

Sum fi = 40

 

Sum fixi = 499

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.

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

85-98

Number of cities

3

10

11

8

3

Solution:

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.

Class interval (h) = 10

So, u= (x– a)/h

u= (x– 70)/10

Substitute and find the values as follows:

Class Interval

Frequency (fi)

(xi)

u= (x– 70)/10

fiui

45-55

3

50

-2

-6

55-65

10

60

-1

-10

65-75

11

70 = a

0

0

75-85

8

80

1

8

85-95

3

90

2

6

 

Sum fi = 35

   

Sum fiui = -2

So, Mean = x̄ = a + (∑fiui /∑fi) × h

= 70 + (-2/35) × 10

= 69.43

Therefore, the mean literacy part = 69.43%

Exercise 14.2 Page: 275

1. The following table shows the ages of the patients admitted to a hospital during a year:

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. - Statistics for class 10th students

Solution:

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,

Lower limit of modal class = l = 35,

class width (h) = 10,

fm = 23,

f1 = 21 and f2 = 14

The formula to find the mode is

Mode = l + [(f– f1)/ (2f– f– f2)] × h

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

= 35 + (20/11)

= 35 + 1.8

= 36.8 years

So the mode of the given data = 36.8 years

Calculation of Mean:

First find the midpoint using the formula, x= (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

 

Sum fi = 80

 

Sum fixi = 2830

The mean formula is

Mean = x̄ = ∑fixi /∑fi

= 2830/80

= 35.375 years

Therefore, the mean of the given data = 35.375 years

2. The following data gives the information on the observed lifetimes (in hours) of 225

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. - Statistics for class 10th students

Solution:

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

Lower limit of modal class = l = 60,

The frequencies are:

fm = 61, f1 = 52, f2 = 38 and h = 20

The formula to find the mode is

Mode = l+ [(f– f1)/(2f– f– f2)] × h

Substitute the values in the formula, we get

Mode = 60 + [(61 – 52)/ (122 – 52 – 38)] × 20

Mode = 60 + [(9 × 20)/32]

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

Therefore, modal lifetime of the components = 65.625 hours.

3. The following data gives the distribution of total monthly household expenditure of 200

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

mean monthly expenditure:

Expenditure (in Rs.)

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

Solution:

Given data:

Modal class = 1500-2000,

l = 1500,

Frequencies:

fm = 40 f1 = 24, f2 = 33 and

h = 500

Mode formula:

Mode = l + [(f– f1)/ (2f– f– 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, x=(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 = a

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

 

fi = 200

     

fiui = -35

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 = Rs. 2662.50

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures - Statistics for class 10th students

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

Solution:

Given data:

Modal class = 30 – 35,

l = 30,

Class width (h) = 5,

fm = 10, f1 = 9 and f2 = 3

Mode Formula:

Mode = l + [(f– f1)/ (2f– f– f2)] × h

Substitute the values in the given formula

Mode = 30 + [(10 – 9)/ (20 – 9 – 3)] × 5

= 30 + (5/8)

= 30 + 0.625

= 30.625

Therefore, the mode of the given data = 30.625

Calculation of mean:

Find the midpoint using the formula, x=(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

105.0

 

Sum fi = 35

 

Sum fixi = 1022.5

Mean = x̄ = ∑fixi /∑fi

= 1022.5/35

= 29.2 (approx)

Therefore, mean = 29.2 - Statistics for class 10th students

5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.

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.

Solution:

Given data:

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000,

fm = 18, f1 = 4 and f2 = 9

Mode Formula:

Mode = l + [(f– f1)/ (2f– f– f2)] × h

Substitute the values

Mode = 4000 + [(18 – 4)/ (36 – 4 – 9)] × 1000

= 4000 + (14000/23)

= 4000 + 608.695

= 4608.695

= 4608.7 (approximately)

Thus, the mode of the given data is 4608.7 runs.

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

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

Solution:

Given Data:

Modal class = 40 – 50, l = 40,

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

Mode = l + [(f– f1)/(2f– f– f2)] × h

Substitute the values

Mode = 40 + [(20 – 12)/ (40 – 12 – 11)] × 10

= 40 + (80/17)

= 40 + 4.7

= 44.7

Thus, the mode of the given data is 44.7 cars.

Exercise 14.3 Page: 287

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:



= 125 + [(34 − 22)/20] × 20

= 125 + 12

= 137

Therefore, median = 137

To calculate the mode:

Modal class = 125-145,

fm or f= 20, f= 13, f= 14 & h = 20

Mode formula:

Mode = l+ [(f– f0)/(2f– f– 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 = a

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 f= 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, 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

CI

0-10

10-20

20-30

30-40

40-50

50-60

Frequency

5

x

20

15

y

5

Cumulative frequency

5

5+x

25+x

40+x

40+x+y

45+x+y

Where, N/2 = 30

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class, = 20,

cf = 5 + x,

f = 20 & h = 10



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 = 45 + x + y

Now, substitute the value of x, to find y

60 = 45 + 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. - Statistics for class 10th students

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 = 35-40

Then, l = 35, cf = 45, f = 33 & h = 5

Median = 35 + [(50 – 45)/33] × 5

= 35 + (25/33)

= 35.76

Therefore, the median age = 35.76 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: - Statistics for class 10th students

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 the leaves.

(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, . . ., 171.5 – 180.5.)

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



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 lifetime 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



Median = 3000 + [(200 – 130)/86] × 500

= 3000 + (35000/86)

= 3000 + 406.98

= 3406.98

Therefore, the median lifetime of the lamps = 3406.98 hours

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the 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 median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames. - Statistics for class 10th students

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 + [(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

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

58

16-19

4

17.5

70

 

Sum fi = 100

 

Sum fixi = 832

Mean = x̄ = ∑fxi /∑fi

Mean = 832/100 = 8.32

Therefore, mean = 8.32

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

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



Median = 55 + [(15 – 13)/6] × 5

= 55 + (10/6)

= 55 + 1.666

= 56.67

Therefore, the median weight of the students = 56.67

Exercise 14.4 Page: 293

1. The following distribution gives the daily income of 50 workers in a factory.

Daily income (in Rs.)

100-120

120-140

140-160

160-180

180-200

Number of workers

12

14

8

6

10

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive. - Statistics for class 10th students

Solution

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

Daily Income

Cumulative Frequency

(or)

Number of workers

Less than 120

12

Less than 140

26

Less than 160

34

Less than 180

40

Less than 200

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

2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

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 verify the result by using the formula.

Solution:

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 and join them to get a smooth curve. The curve obtained is known as less than type ogive.

Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the median by making a table.

 

Class interval

Number of students(Frequency)

Cumulative Frequency

Less than 38

0 – 38

0

0

Less than 40

38 – 40

3 – 0 = 3

3

Less than 42

40 – 42

5 – 3 = 2

5

Less than 44

42 – 44

9 – 5 = 4

9

Less than 46

44 – 46

14 – 9 = 5

14

Less than 48

46 – 48

28 – 14 = 14

28

Less than 50

48 – 50

32 – 28 = 4

32

Less than 52

50 – 52

35 – 22 = 3

35

Here, N = 35 and N/2 = 35/2 = 17.5

Median class = 46 – 48

Here, = 46, h = 2, cf= 14, f = 14

The mode formula is given as:

= 46 + [(17.5 – 14)/ 14] × 2

= 46 + 0.5

= 46 + 0.5 = 46.5

Thus, median is verified.

3. The following table gives production yield per hectare of wheat of 100 farms of a village.

Production Yield (in kg/ha)

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.

Solution:

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 the graph paper. The graph obtained is known as more than type ogive curve.

 

 
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