13. Statistics

SEBA Class 10 Maths Chapter 13 Statistics

Chapter 13. Statistics

Class 10 Maths Chapter 13. Statistics Multiple Choice Questions and Solutions :

Question : The median of the distribution 41 , 39 , 48 , 52 , 46 , 62 , 54 is : [SEBA 2013]

(a) 46 (b) 48 (c) 52 (d) 54

Solution: (b) 48

[ Here ,

We arrange the data in ascending order , we have

39 , 41 , 46 , 48 , 52 , 54 , 62

The median

]

Question : The mode of the distribution : 4 , 5 , 6 , 3 , 4 , 5 , 2 , 8 , 4 , 2 is ; [SEBA 2014]

(a) 5 (b) 4.5 (c) 4 (d) 8

Solution : (c) 4

[ Here 4 occurs most frequency i.e., three times . So, the mode is 4 .]

Question : The wickets taken by a bowler in 10 cricket matches are as follows : [SEBA 2015]

2 , 6 , 4 , 5 , 0 , 3 , 1 , 3 , 2 , 3

Mode the data is :

(a) 1 (b) 2 (c) 3 (d) 5

Solution : (c) 3

[ Here 3 occurs most frequency i.e., three times . So, the mode is 3 .]

Question : Mean or average of 1 , 2 , 3 , 4 , ……… , is : [SEBA 2017]

(a) (b) (c) (d)

Solution : (a)

[ We have , Mean

]

Question : The median and mode of a data are 33 and 45 respectively . Then its mean is : [SEBA 2016 , 2023]

(a) 30 (b) 33 (c) 27 (d) 35

Solution : (c) 27

[ We know that ,

]

Question : The wickets taken by a bowler in 8 cricket matches are as follows : [SEBA 2018]

3 , 2 , 0 , 1 , 6 , 2 , 1 , 2

The mode of the data is :

(a) 3 (b) 6 (c) 2 (d) 1

Solution : (c) 2

[ Here 2 occurs most frequency i.e., three times . So, the mode is 2 .]

Question : If the mean of 15, 12 , , 10, 16 , 19 is 12 ,then the value of .

(a) 1 (b) 7 (c) 2 (d) 0

Solution: (d) 0 .

[ We have ,

; The value of is 0 . ]

Question : The mean of first 10 odd numbers is :

(a) 10 (b) 100 (c) 20 (d) 30

Solution : (a) 10

[ We have ,

Mean

]

Question : The median class of the following data is :

Classes	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
Frequency	4	4	8	10	12	8	4

(a) 30 – 40 (b) 40 – 50 (c) 20 – 30 (d) 50 – 60

Solution: (b) 40 – 50

[ Here ,

This observation lies in the class 30 – 40 .

The median is 12 , which lies in the class 40 – 50 . ]

Question : The median class of the following data is :

Class interval	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequency	14	11	13	15	17

(a) 30 – 40 (b) 10 – 20 (c) 50 – 60 (d) 20 – 30

Solution: (a) 30 – 40

[ we construct the table :

Class interval

Frequency

C.f

10 – 20

20 – 30

30 – 40

40 – 50

50 - 60

Here,

This observation lies in the class 30 – 40 . ]

Question : The model class of the following data is :

Class interval	15 – 25	25 – 35	35 – 45	45 – 55	55 – 65	65 – 75
Frequency	6	11	7	4	4	2

(a) 6 (b) 7 (c) 4 (d) 11

Solution: (d) 11

[ The mode is the most frequency occurring observation . ]

Question : The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its :

(a) mean (b) median (c) mode (d) all the three above

Solution: (b) median .

Question : The median of the observations 11 , 12 , 14 , 18 , , , 30 , 32 , 35 , 41 arranged in ascending order is 24 , then the value of :

(a) 31 (b) 11 (c) 18 (d) 21

Solution: (d) 21

[ Here, .

A/Q, Median

]

Question : The following table gives the literacy rate ( in percentage ) of 35 cities .

Literacy rate(in )	45 – 55	55 – 65	65 – 75	75 – 85	85 – 95
Number of cities	3	10	11	8	3

The upper limit of the median class in the given data is :

(a) 55 (b) 75 (c) 65 (d) 85

Solution: (c) 65

[ Here , ; = 17.5 .

We observe that the cumulative frequency just greater than 17.5 is 24 and the corresponding class is 60 – 70 . The upper limit is 65 . ]

Class 10 Statistics Fill in the blanks

Q1. The cumulative frequency table is useful in determining the .

Answer : Median .

Class 10 Statistics Answers following the questions

Question : If upper class limit and lower class limit are 55 and 40 , then find the class mark .

If the class interval of the data is 40 – 55 , then find the class mark .

Solution : We know that , Class mark

Question : If the class mark and lower class limit are 70 and 60 respectively , then find upper class limit .

Solution : We know that, Class mark

Question : Find the mode of the following marks (out of 10) obtained by 20 students : 4 , 6 , 5 , 9 , 3 , 2 , 7 , 7 , 6 , 5 , 4 , 9 , 10 , 10 , 3 , 4 , 7 , 6 , 9 , 9

Solution : We arrange the given data in the following form : 2 , 3 , 3 , 4 , 4 ,4 , 5 , 5 , 6 , 6 , 6 , 7 , 7 , 7 , 9 , 9 , 9 , 9 , 10 , 10

Here, 9 occurs most frequently [ i.e., four time] . So, the mode is 9 .

Question : Find the mode of the following data : 51 , 34 , 17 , 58 , 41 , 51 , 17 , 29 , 39 , 29 , 41 , 29

Solution : We arrange the given data in the following form : 17 , 17 , 29 , 29 , 29 , 34 , 39 , 41 , 41 , 51 , 51 , 58

Here, 29 occurs most frequently [ i.e., three time] . So, the mode is 29 .

Question : Find the median of the following data : 41 , 39 , 48 , 52 , 46 , 62 , 54 , 40 , 96 , 98 ,60

Solution : First of all we arrange the data in ascending order , as follows : 39 , 40 , 41 , 46 , 48 , 52 , 54 , 60 , 62 , 96 , 98

Here is odd

Therefore, the median observation observation observation observation

Question : Find the median of the following data : 48 , 29 , 84 , 32 , 78 , 95 , 72 , 53

Solution : First of all we arrange the data in ascending order , as follows : 29 , 32 , 48 , 53 , 72 , 78 , 84 , 95

Here is even

The median

Question : If the mode of a data is 45 and mean is 27 , the find the median of the data.

Solution : We know that , 3 Median = Mode + 2 Mean

3 Median = 45 + 2 × 27 = 45 + 54 = 99

3 Median = 99

Median

Therefore, the median of the data is 33 .

Class 10 Statistics 5 Marks Questions and Answers

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

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 .

Solution: We construct the distribution table ,

Daily wages (in Rs)			No. of workers
100 – 120	110	– 40	12	– 480
120 – 140	130	– 20	14	– 280
140 – 160	150 (a)	0	8	0
160 – 180	170	20	6	120
180 – 200	190	40	10	400
Total

Using the assumed mean method ,

Mean

Therefore, the wages of the workers of the factory is Rs 145.20 .

Question : The following distribution shows the daily pocket allowance of children of a locality . The mean pocket allowance is Rs. 18 . Find the missing frequency .

Daily pocket allowance (in Rs)

Number of children

11 – 13

13 – 15

15 – 17

17 – 19

19 - 21

21 – 23

23 – 25

Solution: We construct the table ,

Daily pocket allowance (in Rs)

Number of children

11 – 13

13 – 15

15 – 17

17 – 19

19 - 21

21 – 23

23 – 25

144

234

110

Total

Mean

Therefore, the missing frequency is 20 .

Question : The marks obtained by 30 students of Class X of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students. [SEBA 2015]

Class interval	10 – 25	25 – 40	40 – 55	55 – 70	70 – 85	85 – 100
Number of students	2	3	7	6	6	6

Solution : We construct the table :

Class interval	Class marks	Number of students
10 – 25	17.5	2	35
25 – 40	32.5	3	97.5
40 – 55	47.5	7	332.5
55 – 70	62.5	6	375.5
70 – 85	77.5	6	465.5
85 – 100	92.5	6	555.0
Total

Mean

Question : The following table shows the ages of the patients admitted in a hospital during a year :

Age (in years)

Number of patients

5 – 15

15 – 25

25 – 35

35 – 45

45 – 55

55 – 65

Find the mode and the mean of the data given above . Compare and interpret the two measures of central tendency . [SEBA 2018]

Solution : We construct the table :

Age	Class mark	No. of patients
5 – 15	10	6	60
15 – 25	20	11	220
25 – 35	30	21	630
35 – 45	40	23	920
45 – 55	50	14	700
55 – 65	60	5	300
Total		80	2830

Here,

Using the formula ,

Mode

years

Using the direct method ,

Mean years

Therefore, the maximum number of patients admitted in the hospital are of the age 36.82 years , while on an average the age of a patient admitted to the hospital is 35.37 years .

Question : If median of the distribution given below is 28.5 , find the values of and . [SEBA 2016,2021,2023]

Class interval

Frequency

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

Total

Solution : We construct the distribution table :

Class interval	Frequency	C.F.
0 – 10	5	5
10 – 20
20 – 30	20
30 – 40	15
40 – 50
50 – 60	5
Total	60

Given ,

The median is 28.5 , which lies in the class 20 – 30 .

Here,

Using the formula ,

Median

Putting the value of in , we get

The value of and .

Question : The marks distribution of 30 students in a mathematics examination are given below . Find the mode of this data . [SEBA 2019]

Class interval	10 – 25	25 – 40	40 – 55	55 – 70	70 – 85	85 – 100
Number of students	2	3	7	6	6	6

Solution : We have ,

Class interval	10 – 25	25 – 40	40 – 55	55 – 70	70 – 85	85 – 100
Number of students	2	3	7	6	6	6

Here, ,

Using the formula ,

Mode

Therefore , the mode of the data is 52 .

Question : The median of the following data is 525 .Find the values of x and y , if the total frequency is 100 . [SEBA 2020]

Class interval

Frequency

0 – 100

100 – 200

200 – 300

300 – 400

400 – 500

500 – 600

600 – 700

700 – 800

800 – 900

900 – 1000

Solution : We construct the table :

Class interval

Frequency

Cumulative frequency

0 – 100

100 – 200

200 – 300

300 – 400

400 – 500

500 – 600

600 – 700

700 – 800

800 – 900

900 – 1000

Given, ,

The median is , which lies in the class

Here , , , ,

Using the formula : Median

Putting in equation we have,

Required the value of and are and respectively.

Question : The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs.)	100 – 120	120 – 140	140 – 160	160 – 180	180 – 200
Number of workers	12	14	8	6	10

Find the mean , mode and median of the given distribution .

Solution : We construct the table :

Daily income (in Rs.)				C.F.
100 – 120	110	12	1320	12
120 – 140	130	14	1820	26
140 – 160	150	8	1200	34
160 – 180	170	6	1020	40
180 – 200	190	10	1900	50
Total		50	7260

Here, ,

Using the formula of mean :

Mean

Using the formula of mode :

Here , ; ; ; ;

Mode ×

Using the formula of median :

Here , . So, will be in 120 – 140 .

This observation lies in the class 120 – 140 . So ,The frequency of the median class is 14 .

; ; ;

Median

Question : The table below shows the daily expenditure on food of 25 household in a locality.

Daily expenditure (in Rs.)	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 the assumed mean method .

Solution : We constructed the table :

Daily exp. (in Rs.)

No. of households

Class mark

100 – 150

150 – 200

200 – 250

250 – 300

300 – 350

125

175

225

275

325

– 100

– 50

100

– 400

– 250

100

200

Total

The assumed mean method

Required the daily expenditure on the food is Rs. 211 .

Question : The distribution below shows the number of wickets taken by bowlers in one-day cricket matches .Find the mean number of wickets by choosing a suitable method . What does the mean signify ? [Using the assumed mean method .]

Number of wickets	20 – 60	60 – 100	100 – 150	150 – 250	250 – 350	350 – 450
Number of bowlers	7	5	16	12	2	3

Solution : We constructed the table :

No. of wickets

No. of bowlers

20 – 60

60 – 100

100 – 150

150 – 250

250 – 350

350 – 450

125

200

300

400

– 160

– 120

– 75

100

200

– 1120

– 600

– 1200

200

600

Total

Using the assumed mean method :

Mean

Therefore, the number of wickets taken by these 45 bowlers in one day cricket is 152.89 .

Question : Month pocket money of 50 students of a class are given in the following distribution :

Monthly pocket money (Rs.)	0 – 50	50 – 100	100 – 150	150 – 200	200 – 250	250 – 300
Number of student	2	7	8	30	12	1

Find modal class and also give class mark of the modal class .

Solution: We construct the table :

Monthly pocket money (Rs.)

Class marks

No. of student

0 – 50

50 – 100

100 – 150

150 – 200

200 – 250

250 – 300

125

175

225

275

Here the maximum class frequency is 30 . So, the modal class is 150 – 200 .

, , , ,

Mode

Therefore, the mode is 177.5

Question : 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: We construct the table :

Weight (in kg)

Frequency

Cumulative frequency

40 – 45

45 – 50

50 – 55

55 – 60

60 – 65

65 – 70

70 – 75

Now , , . This observation lies in the class 55 – 60 .

Here , , , and

Median

Therefore, the weight of the students is 56.67 .

Question : If the mode of the following distribution is 57.5 , then find the value of :

Class interval	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90	90 – 100
Frequency	6	10	16		10	5	2

Solution: We construct the table :

Class interval	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90	90 – 100
Frequency	6	10	16		10	5	2

Here , the maximum class frequency is 16 and the modal class is 50 – 60 .

, , ,

Mode

Question : Determine the missing frequency , from the following data , when mode is 67 .

Class interval	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90
Frequency	5		15	12	7

Solution: We construct the distribution frequency table of the given data as follows :

Class interval

Frequency

40 – 50

50 – 60

60 – 70

70 – 80

80 – 90

Here , , , , and

Using mode formula :

Mode

Therefore, the value of x is 13.7 .

Question : The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality . Find the median , mean and mode of the data and compare them .

Monthly consumption (in units)

Number of consumers

65 – 85

85 – 105

105 – 125

125 – 145

145 – 165

165 – 185

185 – 205

Solution : We the frequency distribution table :

Class interval				C.F.
65 – 85	75	– 60	4	4	– 240
85 – 105	95	– 40	5	9	– 200
105 – 125	115	– 20	13	22	– 260
125 – 145	135	0	20	42	0
145 – 165	155	20	14	56	280
165 – 185	175	40	8	64	320
185 – 205	195	60	4	68	240
Total			68		140

Here,

This observation lies in the class 125 – 145 . Then,

Using the formula ,

Median

units

Using assumed mean method ,

Here,

Mean

units

Using the formula ,

Here,

Mode

units

The three measures are approximately the same in this case .