RD SHARMA SOLUTION CHAPTER- 7 Statistics| CLASS 10TH MATHEMATICS-EDUGROWN

Chapter 7 Statistics Exercise Ex. 7.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Chapter 7 Statistics Exercise Ex. 7.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 7 Statistics Exercise Ex. 7.3

Question 1

Solution 1

Question 2

A survey was conducted by a group of students as a part of their environment awareness programme, 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 2

Let us find class marks (xi) for each interval by using the relation.

Now we may compute x_i and f_ix_ias following

Number of plants	Number of houses (fi)	xi	fixi
0 – 2	1	1	1 x 1 = 1
2 – 4	2	3	2 x 3 = 6
4 – 6	1	5	1 x 5 = 5
6 – 8	5	7	5 x 7 = 35
8 – 10	6	9	6 x 9 = 54
10 – 12	2	11	2 x 11 = 22
12 – 14	3	13	3 x 13 = 39
Total	20		162

From the table we may observe that
 

So, mean number of plants per house is 8.1.
We have used here direct method as values of class marks (xi) and fi are small.

Question 3

Solution 3

Question 4

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats 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 4

We may find class mark of each interval (xi) by using the relation.

Class size h of this data = 3
Now taking 75.5 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following.

Number of heart beats per minute	Number of women fi	xi	di = xi -75.5		fiui
65 – 68	2	66.5	– 9	– 3	– 6
68 – 71	4	69.5	– 6	– 2	– 8
71 – 74	3	72.5	– 3	– 1	– 3
74 – 77	8	75.5	0	0	0
77 – 80	7	78.5	3	1	7
80 – 83	4	81.5	6	2	8
83 – 86	2	84.5	9	3	6
Total	30				4

Now we may observe from table that

So mean hear beats per minute for these women are 75.9 beats per minute.Question 6

Find the mean of the following frequency distribution:

Solution 6

Question 7

Find the mean of the following frequency distribution:

Solution 7

Question 8

Find the mean of the following frequency distribution:

Solution 8

Question 9

Find the mean of the following frequency distribution:

Solution 9

Question 10

Find the mean of the following frequency distribution:

Solution 10

Question 11

Find the mean of the following frequency distribution:

Solution 11

Question 12

Find the mean of the following frequency distribution:

Solution 12

Question 13

Find the mean of the following frequency distribution:

Solution 13

Question 14

Find the mean of the following frequency distribution:

Solution 14

Question 15

For the following distribution, calculate mean using all suitable methods :

Size of item	1-4	4-9	9-16	16-27
Frequency	6	12	26	20

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find out the missing frequency.

Solution 19

Question 20

Solution 20

Question 21

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?Solution 21

Number of mangoes	Number of boxes fi
50 – 52	15
53 – 55	110
56 – 58	135
59 – 61	115
62 – 64	25

We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals. So we have to add  to upper class limit and subtract  from lower class limit of each interval.
And class mark (xi) may be obtained by using the relation

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as follows:

Class interval	fi	xi	di = xi – 57		fiui
49.5 – 52.5	15	51	-6	-2	-30
52.5 – 55.5	110	54	-3	-1	-110
55.5 – 58.5	135	57	0	0	0
58.5 – 61.5	115	60	3	1	115
61.5 – 64.5	25	63	6	2	50
Total	400				25

Now, we have:
 
Clearly mean number of mangoes kept in a packing box is 57.19.

Note: We have chosen step deviation method here as values of fi, di are big and also there is a common multiple between all di.

Question 22

The table below shows the daily expenditure on food of 25 households 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 a suitable method.Solution 22

We may calculate class mark (xi) for each interval by using the relation


Class size = 50

Now taking 225 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as follows:

Daily expenditure (in Rs)	fi	xi	di = xi – 225		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
Total					-7

Now we may observe that –

Question 23

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 23

Concentration of SO2 (in ppm)	Frequency	Class mark xi	di = xi – 0.14		fiui
0.00 – 0.04	4	0.02	-0.12	-3	-12
0.04 – 0.08	9	0.06	-0.08	-2	-18
0.08 – 0.12	9	0.10	-0.04	-1	-9
0.12 – 0.16	2	0.14	0	0	0
0.16 – 0.20	4	0.18	0.04	1	4
0.20 – 0.24	2	0.22	0.08	2	4
Total	30				-31

Question 25

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 – 95
Number of cities	3	10	11	8	3

Solution 25

We may find class marks by using the relation



Class size (h) for this data = 10

Now taking 70 as assumed mean (a) we may calculate d_i, u_i, and f_iu_i as follows:  

Literacy rate (in %)	Number of cities fi	xi	di= xi – 70	ui =di/10	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
Total	35				-2

Now we may observe that

So, mean literacy rate is 69.43%.Question 26

The following is the cumulative frequency distribution (of less than type) of 1000 persons each of age 20 years and above. Determine the mean age.

Age below (in years)	30	40	50	60	70	80
Number of persons	100	220	350	750	950	1000

Solution 26

Here, we have cumulative frequency distribution less than type. First we convert it into an ordinary frequency distribution.

Question 27

If the means of the following frequency distribution is 18, find the missing frequency.

Class interval:	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Frequency:	3	6	9	13	f	5	4

Solution 27

Question 28

Find the missing frequencies in the following distribution, if the sum of the frequencies is 120 and the mean is 50.

Class:	0-20	20-40	40-60	60-80	80-100
Frequency:	17	f1	32	f2	19

Solution 28

Question 29

The daily income of a sample of 50 employees are tabulated as follows:

Income (in Rs.):	1-200	201-400	401-600	601-800
No. of employees:	14	15	14	7

Find the mean daily income of employees.Solution 29

Question 5

Find the mean of the following frequency distribution:

Class interval	3 – 5	5 – 7	7 – 9	9 – 11	11 – 13
Frequency	5	10	10	7	8

Solution 5

Let the assumed mean be A = 8

Here, h = 2

Class interval	Mid valuexi	di = xi – 8		fi	fiui
3 – 55 – 77 – 99 – 1111 – 13	4681012	-4-2024	-2-1012	5101078	-10-100716
				N = 40

Question 24 (i)

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 – 12	12 – 18	18 – 24	24 – 30	30 – 36	36 – 42
Number of students	10	11	7	4	4	3	1

Solution 24 (i)

We may find class mark of each interval by using the relation

Number of days	Number of students fi	xi	di = xi – 21	fidi
0 – 66 – 1212 – 1818 – 2424 – 3030 – 3636 – 42	101174431	391521273339	-18-12-6061218	-180-132-420243618
	40

Assumed mean A = 21

Hence, the number of days a student was absent is 14.1.Question 30

The marks obtained by 110 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	28	23	18	8	3

Find the mean marks of the students.Solution 30

We may find class mark of each interval by using the relation

Marks	Number of students fi	xi	di = xi – 47.5	fidi
30 – 3535 – 4040 – 4545 – 5050 – 5555 – 6060 – 65	141628231883	32.537.542.547.552.557.562.5	-15-10-5051015	-210-160-1400908045
	110

Assumed mean A = 47.5

Hence, mean marks of the students is 44.81.

Chapter 7 Statistics Exercise Ex. 7.4

Question 1

Solution 1

Question 2

Solution 2

The median height of the students is Rs 167.13.Question 3

Solution 3

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Find the following tables gives the distribution of the life time of 400 neon lamps:

Life time (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 life time of a lamp.Solution 8

We can find cumulative frequencies with their respective class intervals as below –

Life time	Number of lamps (fi)	Cumulative frequency
1500 – 2000	14	14
2000 – 2500	56	14 + 56 = 70
2500 – 3000	60	70 + 60 = 130
3000 – 3500	86	130 + 86 = 216
3500 – 4000	74	216 + 74 = 290
4000 – 4500	62	290 + 62 = 352
4500 – 5000	48	352 + 48 = 400
Total (n)	400

Now we may observe that cumulative frequency just greater than  is 216 belonging to class interval 3000 – 3500.
Median class = 3000 – 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500


So, median life time of lamps is 3406.98 hours.Question 9

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 9

We may find cumulative frequencies with their respective class intervals as below

Weight (in kg)	40 – 45	45 – 50	50 – 55	55 – 60	60 – 65	65 – 70	70 – 75
Number of students (f)	2	3	8	6	6	3	2
c.f.	2	5	13	19	25	28	30

Cumulative frequency just greater than   is 19, belonging to class interval 55 – 60.
Median class = 55 – 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5



So, median weight is 56.67 kg.Question 10

Solution 10

Question 11

You are given that the median value is 46 and the total number of items is 230.(i) Using the median formula fill up missing frequencies.(ii) Calculate the AM of the completed distribution.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 16

Solution 16

Question 17

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Age (in years)	Number of policy holders
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 17

Here class width is not same. There is no need to adjust the frequencies according to class intervals. Now given frequency table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years, we can define class intervals with their respective cumulative frequency as below

Age (in years)	Number of policy holders (fi)	Cumulative frequency (cf)
18 – 20	2	2
20 – 25	6 – 2 = 4	6
25 – 30	24 – 6 = 18	24
30 – 35	45 – 24 = 21	45
35 – 40	78 – 45 = 33	78
40 – 45	89 – 78 = 11	89
45 – 50	92 – 89 = 3	92
50 – 55	98 – 92 = 6	98
55 – 60	100 – 98 = 2	100
Total (n)

Now from table we may observe that n = 100.

Cumulative frequency (cf) just greater than  is 78 belonging to interval 35 – 40
So, median class = 35 – 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45


So, median age is 35.76 years.Question 18

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm)	Number or leaves fi
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.Solution 18

The given data is not having continuous class intervals. We can observe that difference between two class intervals is 1. So, we have to add and subtract

  to upper class limits and lower class limits.
Now continuous class intervals with respective cumulative frequencies can be represented as below:

Length (in mm)	Number or leaves fi	Cumulative frequency
117.5 – 126.5	3	3
126.5 – 135.5	5	3 + 5 = 8
135.5 – 144.5	9	8 + 9 = 17
144.5 – 153.5	12	17 + 12 = 29
153.5 – 162.5	5	29 + 5 = 34
162.5 – 171.5	4	34 + 4 = 38
171.5 – 180.5	2	38 + 2 = 40

From the table we may observe that cumulative frequency just greater then
  is 29, belonging to class interval 144.5 – 153.5.
Median class = 144.5 – 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17



So, median length of leaves is 146.75 mm.Question 19

Solution 19

Question 20

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

Solution 20

Question 21

The median of the following data is 50. Find the values of p and q, if the sum of all the frequencies is 90.

Marks:	20-30	30-40	40-50	50-60	60-70	70-80	80-90
Frequency:	p	15	25	20	q	8	10

Solution 21

Question 4

Calculate the median salary of the following data giving salaries of 280 persons:

Salary (in thousands):	5 – 10	10 – 15	15 – 20	20 – 25	25 – 30	30 – 35	35 – 40	40 – 45	45 – 50
Frequency	49	133	63	15	6	7	4	2	1

Solution 4

The cumulative frequency table is

Salary	Frequencyfi	Cumulative frequency(c.f.)
5 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535 – 4040 – 4545 – 50	49133631567421	49182245260266273277279280

The cumulative frequency greater than and nearest to 140 is 182.

So, median class is 10 – 15

Lower limit (l) = 10, class size (h) = 5, c.f. = 49

Frequency of median class = 133

Hence, mean median salary is Rs. 13421.

Chapter 7 Statistics Exercise Ex. 7.5

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

The following table shows the ages of the patients admitted in 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.Solution 7

We may compute class marks (xi) as per the relation

 
Now taking 30 as assumed mean (a) we may calculate d_i and f_id_i as follows.

Age (in years)	Number of patients fi	class mark xi	di= xi – 30	fidi
5 – 15	6	10	-20	-120
15 – 25	11	20	-10	-110
25 – 35	21	30	0	0
35 – 45	23	40	10	230
45 – 55	14	50	20	280
55 – 65	5	60	30	150
Total	80			430

From the table we may observe that


Clearly, mean of this data is 35.38. It represents that on an average the age of a patient admitted to hospital was 35.38 years.
As we may observe that maximum class frequency is 23 belonging to class interval 35 – 45.
So, modal class = 35 – 45
Lower limit (l) of modal class = 35
Frequency (f₁) of modal class = 23
Class size (h) = 10
Frequency (f₀) of class preceding the modal class = 21
Frequency (f₂) of class succeeding the modal class = 14


Clearly mode is 36.8.It represents that maximum number of patients admitted in hospital were of 36.8 years.Question 8

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Lifetimes (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.Solution 8

From the data given as above we may observe that maximum class frequency is 61 belonging to class interval 60 – 80.
So, modal class = 60 – 80
Lower class limit (l) of modal class = 60
Frequency (f₁) of modal class = 61
Frequency (f₀) of class preceding the modal class = 52
Frequency (f₂) of class succeeding the modal class = 38
Class size (h) = 20


 
So, modal lifetime of electrical components is 65.625 hours.Question 9

Solution 9

Question 10

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.

Number 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 10

We may observe from the given data that maximum class frequency is 10 belonging to class interval 30 – 35.
So, modal class = 30 – 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f₁) of modal class = 10
Frequency (f₀) of class preceding modal class = 9
Frequency (f₂) of class succeeding modal class = 3


It represents that most of states/U.T have a teacher – student ratio as 30.6

Now we may find class marks by using the relation


Now taking 32.5 as assumed mean (a) we may calculate di, ui and fiui as following.

Number of students per teacher	Number of states/U.T (fi)	xi	di = xi – 32.5	ui	fiui
15 – 20	3	17.5	-15	-3	-9
20 – 25	8	22.5	-10	-2	-16
25 – 30	9	27.5	-5	-1	-9
30 – 35	10	32.5	0	0	0
35 – 40	3	37.5	5	1	3
40 – 45	0	42.5	10	2	0
45 – 50	0	47.5	15	3	0
50 – 55	2	52.5	20	4	8
Total	35				-23

So mean of data is 29.2
It represents that on an average teacher – student ratio was 29.2.Question 11

Solution 11

Question 12

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

Number of cars	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 6	60 – 70	70 – 80
Frequency	7	14	13	12	20	11	15	8

Solution 12

From the given data we may observe that maximum class frequency is 20 belonging to 40 – 50 class intervals.
So, modal class = 40 – 50
Lower limit (l) of modal class = 40
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 12
Frequency (f₂) of class succeeding modal class = 11
Class size = 10

So mode of this data is 44.7 cars.Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

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 16

We may observe from the given data that maximum class frequency is 40 belonging to 1500 – 2000 intervals.
So, modal class = 1500 – 2000
Lower limit (l) of modal class = 1500
Frequency (f₁) of modal class = 40
Frequency (f₀) of class preceding modal class = 24
Frequency (f₂) of class succeeding modal class = 33
Class size (h) = 500

 

So modal monthly expenditure was Rs. 1847.83
Now we may find class mark as


Class size (h) of give data = 500
Now taking 2750 as assumed mean (a) we may calculate d_i, u_i and f_iu_i as follows:

Expenditure (in Rs)	Number of familiesfi	xi	di = xi – 2750	ui	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
Total	200				-35

Now from table may observe that


So, mean monthly expenditure was Rs. 2662.50.Question 17

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

Runs scored	No 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 17

From the given data we may observe that maximum class frequency is 18 belonging to class interval 4000 – 5000.
So, modal class = 4000 – 5000
Lower limit (l) of modal class = 4000
Frequency (f₁) of modal class = 18
Frequency (f₀) of class preceding modal class = 4
Frequency (f₂) of class succeeding modal class = 9
Class size (h) = 1000

So mode of given data is 4608.7 runs.Question 18

The frequency distribution table of agriculture holdings in a village is given below:

Area of land (in hectares):	1-3	3-5	5-7	7-9	9-11	11-13
Number of families:	20	45	80	55	40	12

Find the modal agriculture holdings of the village.Solution 18

Question 19

The monthly income of 100 families are given as below:

Income in (in Rs.)	Number of families
0-5000	8
5000-10000	26
10000-15000	41
15000-20000	16
20000-25000	3
25000-30000	3
30000-35000	2
35000-40000	1

Calculate the modal income.Solution 19

Question 3 (iv)

Find the mode of the following distribution:

Class	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
Frequency	8	10	10	16	12	6	7

Solution 3 (iv)

The cumulative frequency table is

Salary	Frequencyfi
0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 70	81010161267

Modal class is 30 – 40

Lower limit (l) = 30, class size (h) = 10, f = 16, f₁ = 10 and f₂ = 12

Hence, the mode is 36.

Chapter 7 Statistics Exercise Ex. 7.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

The annual rainfall record of a city for 66 days if given in the following table:

Rainfall (in cm):	0-10	10-20	20-30	30-40	40-50	50-60
Number of days:	22	10	8	15	5	6

Calculate the median rainfall using ogives of more than type and less than type.Solution 8

Less Than Series:

Class interval	Cumulative Frequency
Less than 10	22
Less than 20	32
Less than 30	40
Less than 40	55
Less than 50	60
Less than 60	66

We plot the points (10, 22), (20, 32), (30, 40), (40, 55), (50, 60) and (60, 66) to get ‘less than type’ ogive.

More Than Series:

Class interval	Frequency
More than 0	66
More than 10	44
More than 20	34
More than 30	26
More than 40	11
More than 50	6

We plot the points (0, 66), (10, 44), (20, 24), (30, 26), (40, 11), and (50, 6) to get more than ogive.

From the graph, median = 21.25 cmQuestion 9

Change the following distribution to a ‘More than type’ distribution. Hence, draw the ‘more than type’ ogive for this distribution.

Class-interval	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90
Frequency	10	8	12	24	6	25	15

Solution 9

More than type frequency distribution is given by

Class interval	Frequencyfi	Cumulative frequency
More than 20More than 30More than 40More than 50More than 60More than 70More than 80	108122462515	100908270464015

So, more than type frequency curve is

Question 10

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

Daily income (in Rs.)	200 – 220	220 – 240	240 – 260	260 – 280	280 – 300
Number of workers	12	14	8	6	10

Convert the distribution above to a ‘less than type’ cumulative frequency distribution and draw its ogive.Solution 10

Less than type frequency distribution is given by

Class interval	Frequencyfi	Cumulative frequency
Less than 220Less than 240Less than 260Less than 280Less than 300	12148610	1226344050

So, less than type frequency curve is

Chapter 7 Statistics Exercise 7.66

Question 1

Which of the following is not a measure of central tendency?

(a) Mean

(b) Median

(c) Mode

(d) Standard deviationSolution 1

There are three main measure of central tendency the mode, the median and the mean.

Each of these measures describes a different indication of the typical or central value in the distribution.

The mode is the most commonly occuring value in a distribution.

Median is middle value of distribution.

While standard deviation is a measure of dispersion of a set of data from its mean.

So, the correct option is (d).Question 2

The algebraic sum of deviations of a frequency distribution from its mean is

(a) always positive

(b) always negative

(c) 0

(d) a non-zero numberSolution 2

Question 3

Solution 3

Question 4

For a frequency distribution, mean, median and mode are connected by the relation

(a) Mode = 3 mean – 2 median

(b) Mode = 2 median – 3 mean

(c) Mode = 3 median – 2 mean

(d) Mode = 3 median + 2 meanSolution 4

It is well known that relation between mean, median and mode 1 s

3 median = mode + 2 mean

Mode = 3 median – 2 mean

So, the correct option is (a).Question 5

Which of the following cannot be determined graphically?

(a) Mean

(b) Median

(c) Mode

(d) None of theseSolution 5

Mean is an average value of any given data which cannot be determined by a graph.

Value of median and mode can easily be calculated by graph.

Median is middle value of a distribution and mode is highest frequent value of a given distribution.

So, the correct option is (a).Question 6

The median of a given frequency distribution is found graphically with the help of

(a) Histogram

(b) Frequency curve

(c) Frequency polygon

(d) OgiveSolution 6

The median of a series may be determined through the graphical presentation of data in the forms of Ogives.

Ogive is a curve showing the cummulative frequency for a given set of data.

To get the median we present the data graphically in the form of ‘less than’ ogive  or ‘more than’ ogive

Then the point of intersection of the two graphs gives the value of the median.

So, the correct option is (d).Question 7

The mode of a frequency distribution can be detremined graphically from

(a) Histogram

(b) frequency polygon

(c) ogive

(d) frequency curveSolution 7

Histogram is used to plot the distribution of numerical data or frequency of occurrences of data.

Mode is the most commonly occurring value in the data.

So in distribution or Histogram, the value of the x-coordinate corresponding to the peak value on y – axis, is the mode.

So, the correct option is (a).Question 8

Mode is

(a) least frequent value

(b) middle most value

(c) most frequent value

(d) None of theseSolution 8

Mode is the most frequent value in the data.

Mode is the value which occurs the most number of times.

So, the correct option is (c).Question 9

Solution 9

Question 10

One of the methods of determining mode is

(a) Mode = 2 median – 3 Mean

(b) Mode = 2 Median + 3 Mean

(c) Mode = 3 Median – 2 Mean

(d) Mode = 3 Median + 2 MeanSolution 10

We know that the relation between mean, median & mode is

3 Median = Mode + 2 Mean

Hence, Mode = 3 Median – 2 Mean

So, the correct option is (c).

Chapter 7 Statistics Exercise 7.67

Question 11

If the mean of the following distribution is 2.6, then the value of y is

Variable (y) : 1   2   3   4   5

Frequency :   4   5    y   1   2

(a) 3    (b) 8     (c) 13    (d) 24Solution 11

Question 12

The relationship between mean, median and mode for a moderately skewed distribution is

(a) Mode = 2 Median – 3 Mean

(b) Mode = Median – 2 Mean

(c) Mode = 2 Median – Mean

(d) Mode = 3 Median – 2 meanSolution 12

We know that the relation between mean, median & mode is

3 Median = mode + 2 Mean

Hence, mode = 3 Median – 2 Mean

So, the correct option is (d).Question 13

Solution 13

Question 14

If the arithmetic mean of x, x +3 , x + 6, x + 9, x + 12 is 10, then x =

(a) 1

(b) 2

(c) 6

(d) 4Solution 14

Question 15

If the median of the data : 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5 then x =

(a) 27

(b) 25

(c) 28

(d) 30Solution 15

Question 16

If the median of the data: 6, 7, x – 2, x, 17, 20 written in ascending order, is 16. Then x =

(a) 15

(b) 16

(c) 17

(d) 18Solution 16

Question 17

The median of first 10 prime numbers is

(a) 11

(b) 12

(c) 13

(d) 14Solution 17

Question 18

If the mode of the data : 64, 60, 48, x, 43, 48, 43, 34 is 43 then x + 3 =

(a) 44

(b) 45

(c) 46

(d) 48Solution 18

Mode is the number in observation data is that which repeats most number of time

In the given data 48 comes twice and 43 comes twice but mode is 43.

Hence if x = 43 then 43 comes thrice.

So x + 3 = 43 + 3 = 46

So, the correct option is (c).Question 19

If the mode of the data : 16, 15, 17, 16, 15, x, 19, 17, 14 is 15 then x =

(a) 15

(b) 16

(c) 17

(d) 19Solution 19

In the given data 15, 16, 17 comes twice but given 15 is mode.

Hence 15 comes more than 16, 17.

This is only possible if x = 15.

So, the correct option is (a).Question 20

The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3, p have mean m – 1 and median q. Then, p + q =

(a) 4

(b) 5

(c) 6

(d) 7Solution 20

Question 21

Solution 21

Question 22

If the mean of 6, 7, x , 8, y, 14 is 9, then

(a) x + y = 21

(b) x + y = 19

(c) x – y = 19

(d) x – y = 21Solution 22

Question 23

Solution 23

Question 24

Solution 24

Chapter 7 Statistics Exercise 7.68

Question 25

The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is

(a) 25

(b) 18

(c) 20

(d) 22Solution 25

We know, 3 Median = Mode + 2 Mean

mean = 24

mode = 12

3 median = 12 + 2 × 24

              = 12 + 48

              = 60

median = 20

So, the correct option is (c).Question 26

Solution 26

Question 27

Solution 27

Question 28

If the difference of mode and median of a data is 24, then the difference of median and mean is

(a) 12

(b) 24

(c) 8

(d) 36Solution 28

Question 29

If the arithmetic mean of 7, 8, x, 11, 14 is x then x = 

(a) 1

(b) 9.5

(c) 10

(d) 10.5Solution 29

Question 30

If mode of a series exceeds its mean by 12, then mode exceeds the median by

(a) 4

(b) 8

(c) 6

(d) 10Solution 30

Question 31

If the mean of first n natural number is 15, then  n =

(a) 15

(b) 30

(c) 14

(d) 29Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increases by

(a) 2

(b) 1.5

(c) 1

(d) 0.5Solution 35

Question 36

While computing mean of grouped data, we assume that the frequencies are

a. evenly distributed over all the classes.

b. centred at the class marks of the classes.

c. centred at the upper limit of the classes.

d. centred at the lower limit of the classes.Solution 36

While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.

Hence, correct option is (b).Question 37

In the formula , for finding the mean of grouped frequency distribution u_i =

a. 

b. h(x_i – a)

c. 

d. Solution 37

Chapter 15 – Statistics Exercise 15.69

Question 38

For the following distribution:

Class:	0-5	5-10	10-15	15-20	20-25
Frequency:	10	15	12	20	9

The sum of the lower limits of the median and modal class is

a. 15

b. 25

c. 30

d. 35Solution 38

Question 39

For the following distribution:

Below:	10	20	30	40	50	60
Number of students:	3	12	27	57	75	80

The modal class is

a. 10-20

b. 20-30

c. 30-40

d. 50-60Solution 39

Question 40

Consider the following frequency distribution:

Class:	65-85	85-105	105-125	125-145	145-165	165-185	185-205
Frequency:	4	5	13	20	14	7	4

The difference of the upper limit of the median class and the lower limit of the modal class is

a. 0

b. 19

c. 20

d. 38Solution 40

Question 41

In the formula , for finding the mean of grouped data d_i^s are deviations from a of

a. Lower limits of classes

b. Upper limits of classes

c. Mid-points of classes

d. Frequency of the class marksSolution 41

d_i‘s are the deviations from a of mid-points of classes.

Hence, correct option is (c).Question 42

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

a. Mean

b. Median

c. Mode

d. All the three aboveSolution 42

The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data given its median.

Hence, correct option is (b).Question 43

Consider the following frequency distribution:

Class:	0-5	6-11	12-17	18-23	24-29
Frequency:	13	10	15	8	11

The upper limit of the median class is

a. 17

b. 17.5

c. 18

d. 18.5Solution 43