Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Long Answer Questions: Statistics - 1

Class 10 Maths Chapter 13 Question Answers - Statistics

III. LONG ANSWER TYPE QUESTIONS

Q1. The median of the following frequency distribution is 35. Find the value of x.

 

Class Interval

Frequency

Cumulative

frequency

0-20

7

7

20-40

8

15

40-60

12

27

60-80

10

37

80-100

8

45

100-120

5

50

Total

50

 

 

Also find the modal class.

Sol. Let us prepare the cumulative frequency table:

Class intervals

f

cf

0-10

2

2 + 0 = 2

10-20

3

2 + 3 = 5

20-30

x

5 + x = (5 + x)

30-40

6

(5 + x) + 6 = 11 + x

40-50

5

(11 + x) + 5 = 16 + x

50-60

3

(16 + x) + 3 = 19 + x

60-70

2

(19 + x) + 2 = 21 + x

Total

(21 + x)

 

 

Here,  Class 10 Maths Chapter 13 Question Answers - Statistics

Obviously,  Class 10 Maths Chapter 13 Question Answers - Statistics lies in the class interval 30−40.


l = 30, Cf = x + 5, f = 6 and h = 10 

Class 10 Maths Chapter 13 Question Answers - Statistics

Class 10 Maths Chapter 13 Question Answers - Statistics
Class 10 Maths Chapter 13 Question Answers - Statistics

⇒ 5 × 12 = 110 − 10x
⇒ 10x = 110 − 60
⇒ 10x =50 ⇒ x = 5 

(i) ∴ The required value of x is 5.

(ii) ∵ The maximum frequency is 6
∴ The modal class is 30−40.


Q2. The mean of the following data is 53, find the missing frequencies.

Age (in years)

0-20

20-40

40-60

60-80

80-100

Total

Number of people

15

f

21

f2

17

100


Sol. 

 

Age (in years)

Number of People (fi)

Mid value (xi)

fi × xi

0-20

15

10

15 x 10 = 150

20-40

fa

30

f1 x 30 = 30 f1

40-60

21

50

21 x 50 = 1050

60-80

f2

70

h x 70 = 70 f2

80-100

17

90

17 x 90 = 1530

Total

100

 

2730 + 30 f1 + 70 f2

 

Since, 15 + f1 + 21 + f2 + 17 = 100
∴ 53 + (f1 + f2) = 100

⇒ f1 + f2 = 100 − 53 = 47              ...(1)

Class 10 Maths Chapter 13 Question Answers - Statistics

⇒ 2730 + 30 f1 + 70 f2 = 5300
⇒ 30 f1 + 70 f2 = 5300 − 2730 = 2570
⇒ 3 f1 + 7 f2 = 257
⇒ 3 (47 − f2) + 7 f2 = 257     |∵ f1 = 47 − f2
⇒ 141 − 3 f2 + 7 f2 = 257
⇒ 4 f2 = 116 or f2 = 29
∴ f1 = 41 − 29 = 18
Thus, f1 = 18 and f2 = 29.


Q3. The mean of the following distribution is 18.

 

Class interval

11-13

13-15

15-17

17-19

19-21

21-23

23-25

Frequencies

3

6

9

13

f

5

4


Find f

Sol. Let the assumed mean (a) = 18

Class 10 Maths Chapter 13 Question Answers - Statistics

∴ We have the following table:

Class intervalxifiClass 10 Maths Chapter 13 Question Answers - Statisticsfi ui

11-13

12

3

-3

3 x (-3) = -9

13-15

14

6

-2

6 x (-2) = -12

15-17

16

9

-1

9 x (-1) = -9

17-19

18

13

0

13 x 0 = 0

19-21

20

f

1

f x 1 = f

21-23

22

5

2

5 x 2 = 10

23-25

24

4

3

3 x 3 = 9

 

 

∑ fi = 40 + f

 

∑fi ui = f− 8

 

Class 10 Maths Chapter 13 Question Answers - Statistics
Class 10 Maths Chapter 13 Question Answers - Statistics
Class 10 Maths Chapter 13 Question Answers - Statistics

⇒  Class 10 Maths Chapter 13 Question Answers - Statistics

Thus, the required frequency = 8

Q4. The percentage of marks obtained by 100 students in an examination are given below:

Marks

30 - 35

35 - 40

40 - 45

45 - 50

50 -55

55 - 60

60-65

Frequency

14

16

18

23

18

8

3

 

Find the median of the above data.

Sol. 

Class 10 Maths Chapter 13 Question Answers - Statistics

Here n = 100 ⇒ n/2 = 50, which lies in the class 45 – 50, where

l1 (lower limit of the median – class) = 45
c (The cumulative frequency of the class preceding the median class) = 48
f (The frequency of the median class) = 23
h (The class size) = 5

Class 10 Maths Chapter 13 Question Answers - Statistics

Class 10 Maths Chapter 13 Question Answers - Statistics

∴ The median percentage of marks is 45.4.

Q5. The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.

 

Class interval

0 - 6

6 -12

12 -18

18 - 24

24 - 30

Frequency

4

x

5

y

1


Sol.

 

Class interval

Frequency

Cumulative

Frequency

0-6

4

4 + 0 = 4

6-12

x

4 + x = (4 + x)

12-18

5

5 + (4 + x) = 9 + x

18-24

y

y + (9 + x) = 9 + x + y

24-30

1

1 + (9 + x + y) = 10 + x + y

 

Since, n = 20
∴ 10 + x + y = 20
⇒ x + y = 20 − 10
⇒ x + y = 10.      ...(1)

Also, we have
Median = 14.4,
which lies in the class interval 12−18.
∴ The median class is 12−18, such that

l = 12, f = 5, Cf = 4 + x and h = 6

Class 10 Maths Chapter 13 Question Answers - Statistics

Class 10 Maths Chapter 13 Question Answers - Statistics
Class 10 Maths Chapter 13 Question Answers - Statistics

⇒ 24 + 6x = (9.6) × 5
⇒ 24 + 6x =48
⇒ 6 x = 48 − 24 = 24

⇒ x = 24/6 = 4

Now, from (1), we have:

x + y = 10
⇒ 4 + y = 10
⇒ y = 10 − 4 = 6
Thus, x =4 and y = 6


Q6. The distribution below gives the weights of 30 students of a class. Find the mean and the median weight of the students:

 

Weight (in kg)

Number of students

40-45

2

45-50

3

50-55

8

55-60

6

60-65

6

65-70

3

70-75

2

 

Sol. Let the assumed mean, a = 57.5
∴ h = 5

∴  Class 10 Maths Chapter 13 Question Answers - Statistics

We have the following table:

Weight of students
 (in kg)

Class

mark
(xi)

Frequency

fi

Cumulative

frequency
 Cf

Class 10 Maths Chapter 13 Question Answers - Statistics

fi ui

 

 

 

 

 

 

40-45

42.5

2

2 + 0 = 2

-3

-6

45-50

47.5

3

2 + 3 = 5

-2

-6

50-55

52.5

8

5 + 8 = 13

-1

-8

55-60

57.5

6

13 + 6 = 19

0

0

60-65

62.5

6

19 + 6 = 25

1

6

65-70

67.5

3

25 + 3 = 28

2

6

70-75

72.5

2

28 + 2 = 30

3

6

Total

 

∑ fi = 30

 

 

∑fiui = - 2

 

∴ The mean of the given data  Class 10 Maths Chapter 13 Question Answers - Statistics
Class 10 Maths Chapter 13 Question Answers - Statistics

Class 10 Maths Chapter 13 Question Answers - Statistics

For finding the median:

Class 10 Maths Chapter 13 Question Answers - Statistics

And it lies in the class 55–60.

Class 10 Maths Chapter 13 Question Answers - Statistics

Class 10 Maths Chapter 13 Question Answers - Statistics

Class 10 Maths Chapter 13 Question Answers - Statistics

Q7. The lengths of 40 leaves of a plant are measured correct upto the nearest millimetre and the data is as under:

 

Length (in mm)

Numbers of Leaves

118-126

4

126-134

5

134-142

10

142-150

12

150-158

4

158-166

5


Find the mean and median length of the leaves.

Sol. For finding the mean:

Let the assumed mean a = 146
h = 8

Class 10 Maths Chapter 13 Question Answers - Statistics

Now, we have the following table:

Length
 (in mm)
Class
 mark
 (xi)

 
Frequency
 (fi)
Cumulative
 frequency
 Cf
Class 10 Maths Chapter 13 Question Answers - Statisticsfi ui

118-126

122

4

4 + 0

= 4

 

-3

-12

126-134

130

5

4 + 5

= 9

 

-2

-10

134-142

138

10

9 + 10

= 19

 

-1

-10

142-150

146

12

19 + 12

= 31

 

0

0

150-158

154

4

31 + 4

= 35

 

1

4

158-166

162

5

35 + 5

= 40

 

2

10

Total

 

∑ fi = 40

 

 

∑ fi ui =-18

 

∴ Mean  Class 10 Maths Chapter 13 Question Answers - Statistics
Class 10 Maths Chapter 13 Question Answers - Statistics
= 142.4 mm

Median: Since  Class 10 Maths Chapter 13 Question Answers - Statistics  [Median class is 142-150]

l = 142
cf = 19
f = 12 and h = 8

Class 10 Maths Chapter 13 Question Answers - Statistics
Class 10 Maths Chapter 13 Question Answers - Statistics

Q8. The table below shows the daily expenditure on food of 30 households in a locality:

 

Daily expenditure (in Rs)

Numbers of households

100-150

6

150-200

7

200-250

12

250-300

3

300-350

2


Find the mean and median daily expenditure on food.

Sol. For finding the mean

Let the assumed mean a = 225.

h = 50

Class 10 Maths Chapter 13 Question Answers - Statistics

We have the following table:

Daily
 expenditure
xific.f.Class 10 Maths Chapter 13 Question Answers - Statisticsfi ui

100-150

125

6

6 + 0 = 6

-2

(-2) x 6 = -12

150-200

175

7

6 + 7 = 13

-1

(-1) x 7 = -7

200-250

225

12

13 + 12 = 25

0

(0) x 12 = 0

250-300

275

3

25 + 3 = 28

1

(1) x 3 = 3

300-350

325

2

28 + 2 = 30

2

(2) x 2 = 4

Total

 

∑ fi = 30

 

 

∑ fiui =-12

 

∴ Mean  Class 10 Maths Chapter 13 Question Answers - Statistics

⇒  Class 10 Maths Chapter 13 Question Answers - Statistics

To find median:

Class 10 Maths Chapter 13 Question Answers - Statistics

And 15 lies in the class 200−250.
∴ Median class is 200−250.

∴ l = 200
cf = 13
f = 12 and h = 50

Thus, Class 10 Maths Chapter 13 Question Answers - Statistics

Class 10 Maths Chapter 13 Question Answers - Statistics

The document Class 10 Maths Chapter 13 Question Answers - Statistics is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Class 10 Maths Chapter 13 Question Answers - Statistics

1. What is statistics and why is it important?
Statistics is a branch of mathematics that deals with the collection, analysis, interpretation, presentation, and organization of data. It helps in making informed decisions and predictions based on data. Statistics is important because it provides a way to understand and communicate complex data, identify trends and patterns, and make reliable conclusions.
2. What are the different types of data in statistics?
In statistics, data can be classified into four types: nominal, ordinal, interval, and ratio. Nominal data represents categories that cannot be ordered, such as gender or color. Ordinal data has categories that can be ordered, like a rating scale or survey responses. Interval data includes numbers where the difference between values is meaningful, but there is no true zero point, for example, temperature in Celsius. Ratio data has a meaningful zero point and the ratios between values are meaningful, like weight or height.
3. How is mean, median, and mode used in statistics?
Mean, median, and mode are measures of central tendency used in statistics. The mean is calculated by adding up all the values and dividing by the total number of values. It represents the average value of a dataset. The median is the middle value when the data is arranged in order, or the average of the two middle values if there is an even number of data points. It represents the middle value of a dataset. The mode is the value that appears most frequently in a dataset. It represents the most common value in a dataset.
4. What is the difference between probability and statistics?
Probability and statistics are related but distinct concepts. Probability deals with the likelihood of an event occurring based on assumptions and mathematical principles. It focuses on predicting the outcome of a future event. On the other hand, statistics involves the collection, analysis, and interpretation of data to make inferences and draw conclusions about a population. It focuses on summarizing and understanding past or existing data.
5. How can statistics be misused or misrepresented?
Statistics can be misused or misrepresented in various ways. One common misuse is cherry-picking data or selectively presenting only the information that supports a particular argument or viewpoint, while ignoring contradictory data. Another way is through misleading graphs or charts that distort the scale or axis to exaggerate or downplay certain trends. Additionally, statistics can be misinterpreted by ignoring confounding factors or failing to consider the limitations of the data. It is important to critically evaluate statistical claims and consider the context in which they are presented.
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