14 : Statistics

Statistics

Chapter 14. STATISTICS

Class 10 Maths Chapter 14 Statistics : Multiple Choice Questions , Answer following the Questions , Fill in the blanks , 2 Marks Questions , 3 Marks Question ,4 Marks Question and Solutions :

Class 10 Statistics Multiple Choice Questions and Answers

SECTION = A

Question : In a frequency distribution , the class mark of a class is 10 and its width is 5 .The lower limit of class is :

(a) 5 (b) 7.5 (c) 10 (d) 12.5

Solution: (b) 7.5

[ let and be the lower term and upper term respectively .

So, Class mark

and Class width

From , we get

The class interval is ]

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 mean of first 10 odd numbers is :

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

Solution : (a) 10

[ We have,

Mean ]

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 . ]

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 . ]

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 . OR 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 : If the class mark and the class size of the data are 40 and 20 respectively , then find the class interval of the data .

Solution : let lower class limit and upper class limit of the data are and respectively.

A/Q, [ Class mark ]

and [ Class size U.C.L – L.C.L ]

From we get ,

Therefore, the class interval of the data is 30 – 50 .

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 .

Q6. 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 obs. obs. obs. obs.

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

Solution : 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 2 Marks Questions and Answers

SECTION = B

Question : Find the mean of the following data :

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

25 – 40

40 – 55

55 – 70

70 – 85

85 – 100

17.5

32.5

47.5

62.5

77.5

92.5

97.5

332.5

375.5

465.5

555.0

Total

Mean

Class 10 Statistics 3 Marks Questions and Answers

SECTION = C

Question : The mean of the following frequency distribution is 62.8. Find the missing frequency . [CBSE 2007]

Class	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100	100 – 120
Frequency	5	8		12	7	8

Solution : We construct frequency distribution table :

Class interval
0 – 20	5	10	50
20 – 40	8	30	240
40 – 60		50
60 – 80	12	70	840
80 – 100	7	90	630
100 – 120	8	110	880
Total

Mean

The value of is 10 .

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 .

Class 10 Statistics 4 Marks Questions and Answers

SECTION = D

Question : Find mean , median and mode of the following data : [CBSE 2008]

Classes	Frequency
0 – 10	6
10 – 20	8
20 – 30	10
30 – 40	15
40 – 50	5
50 – 60	4
60 – 70	2

Solution : We construct the frequency distribution table :

Class interval
0 – 10	5	6	30
10 – 20	15	8	120
20 – 30	25	10	250
30 – 40	35	15	525
40 – 50	45	5	225
50 – 60	55	4	220
60 – 70	65	2	130
Total		50	1500

Using direct method :

Mean

For mode : Here , ,,

Using mode formula :

Mode

We have, 3 Median = Mode + 2 Mean

3 Median

Median

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 : Find the missing frequencies in following frequency distribution table , if median is 32 .

Class interval	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	Total
Number of students	10		25	30		10	100

Solution: We construct the frequency distribution table :

Class interval

No . of students

C.F.

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

50 – 60

Total

The median is 32 , which lies in the class 30 – 40 .

Here , , , ,

Using the formula :

Median

Putting the value of in , we get

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 .

Class 10 Statistics 5 Marks Questions and Answers

SECTION = E

Question : If the mean of the following frequency distribution is 145 , find the missing frequencies and .

Class interval	0 – 50	50 – 100	100 – 150	150 – 200	200 – 250	250 – 300	Total
Frequency	8	12		25		5	80

Solution: We construct the table :

Class interval

Frequency ()

Class mark ()

0 – 50

50 – 100

100 – 150

150 – 200

200 – 250

250 – 300

125

175

225

275

125

900

125

4375

225

1375

Total

Here,

Mean

From and , we get

From , we have

Therefore, the value of and .

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

120 – 140

140 – 160

160 – 180

180 – 200

110

130

150

170

190

1320

1820

1200

1020

1900

Total

Using the formula of mean :

Mean ()

Using the formula of mode :

Here , ; ; ; ;

Mode ×

Using the formula of median :

Here , , will be in 120 – 140 .

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

; ; ;

Median