NCERT Solution for Class 10 Mathematics Chapter 14 - Statistics RR ACADEMY MEERUT MATHS BY RAKESH CHATURVEDI SIR

NCERT Solution for Class 10 Mathematics Chapter 14 - Statistics 

Excercise 14.1

Question 1

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 1

Let us find class marks (xi) for each interval by using the relation. 
 
Now we may compute xi and fixi as following

Number of plants	Number of houses (fi)	xi	fixi
0 - 2	1	1	1— 1 = 1
2 - 4	2	3	2 — 3 = 6
4 - 6	1	5	1 — 5 = 5
6 - 8	5	7	5 — 7 = 35
8 - 10	6	9	6 — 9 = 54
10 - 12	2	11	2 —11 = 22
12 - 14	3	13	3 — 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 2

Consider the following distribution of daily wages of 50 worker 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

Solution 2

Let us find class mark for each interval by using the relation. 

 
Class size (h) of this data = 20
Now taking 150 as assured mean (a) we may calculate di, ui and fiui as following.

Daily wages (in Rs)	Number of workers (fi)	xi	di = xi - 150		fiui
100 -120	12	110	- 40	-2	- 24
120 - 140	14	130	- 20	-1	- 14
140 - 160	8	150	0	0	0
160 -180	6	170	20	1	6
180 - 200	10	190	40	2	20
Total	50				-12

From the table we may observe that 
 
So mean daily wages of the workers of the factory is Rs.145.20

Question 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. 

Daily pocket allowance (in Rs)	11 - 13	13 - 15	15 -17	17 - 19	19 - 21	21 - 23	23 - 25
Number of workers	7	6	9	13	f	5	4

Solution 3

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

 
Given that mean pocket allowance  = Rs.18
Now taking 18 as assured mean (a) we may calculate di and fidi as following.

Daily pocket allowance (in Rs.)	Number of children fi	Class mark xi	di = xi - 18	fidi
11 - 13	7	12	- 6	- 42
13 - 15	6	14	- 4	- 24
15 - 17	9	16	- 2	- 18
17 - 19	13	18	0	0
19 - 21	f	20	2	2 f
21 - 23	5	22	4	20
23 - 25	4	24	6	24
Total				2f - 40

From the table we may obtain 
 
Hence the missing frequency f is 20.

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 di, ui, fiui 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 heart beats per minute for these women are 75.9 beats per minute.

Question 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?

Solution 5

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 di, ui, fiui as following - 

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 may observe that

 
Clearly, mean number of mangoes kept in a packing box is 57.19.
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 6

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 6

We may calculate cla

mark (xi) for each interval by using the relation 

 
Class size = 50

Now taking 225 as assumed mean (a) we may calculate di, ui, fiui as following

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	25				-7

Now we may observe that - 

Question 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 pmm)	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 7

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 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 8

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

  

Now taking 16 as assumed mean (a) we may calculate di and fidi as following 

Number of days	Number of students  fi	xi	di = xi - 16	fidi
0 - 6	11	3	-13	-143
6 -10	10	8	-8	-80
10 - 14	7	12	-4	-28
14 - 20	4	16	0	0
20 - 28	4	24	8	32
28 - 38	3	33	17	51
38 - 40	1	39	23	23
Total	40			-145

Now we may observe that

 
So, mean number of days is 12.38 days, for which a student was absent.

Question 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 - 95
Number of cities	3	10	11	8	3

Solution 9

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 di, ui, and fiui as following 

Literacy rate  (in %)	Number of cities  fi	xi	di = xi - 70		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%.

NCERT Solution for Class 10 Mathematics Chapter 14 - Statistics Page/Excercise 14.2

Question 1

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 1

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

 
Now taking 30 as assumed mean (a) we may calculate di and fidi as following. 

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 2

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 2

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 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 3

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 

 

Now we may find class mark as 

 
Class size (h) of give data = 500
Now taking 2750 as assumed mean (a) we may calculate di, ui and fiui as following

Expenditure  (in Rs)	Number of families fi	xi	di = xi - 2750		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 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. 

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 4

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		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 5

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

Number of students  per teacher	Number of  states/U.T
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 5

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 6

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 6

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.

NCERT Solution for Class 10 Mathematics Chapter 14 - Statistics Page/Excercise 14.3

Question 1

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	4
85 - 105	5
105 - 125	13
125 - 145	20
145 - 165	14
165 - 185	8
185 - 205	4

Solution 1

We may find class marks by using the relation 

 
Taking 135 as assumed mean (a) we may find di, ui, fiui, according to step deviation method as following 

Monthly consumption  (in units)	Number of consumers (f i)	xi class mark	di = xi - 135		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	0	0	0
145 - 165	14	155	20	1	14
165 - 185	8	175	40	2	16
185 - 205	4	195	60	3	12
Total	68				7

From the table we may observe that 

 


Now from table it is clear that maximum class frequency is 20 belonging to class interval 125 - 145.
Modal class = 125 - 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 13
Frequency (f2) of class succeeding the modal class = 14 

 
We know that
3 median = mode + 2 mean
= 135.76 + 2 (137.058)
= 135.76 + 274.116
= 409.876
Median = 136.625
So median, mode, mean of given data is 136.625, 135.76, 137.05 respectively.

Question 2

If the median of the distribution is given below is 28.5, find the values of x and 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 2

We may find cumulative frequency for the given data as following 

Class interval	Frequency	Cumulative frequency
0 - 10	5	5
10 - 20	x	5 + x
20 - 30	20	25 + x
30 - 40	15	40 + x
40 - 5	y	40 + x + y
50 - 60	5	45 + x + y
Total (n)	60

It is clear that n = 60 

45 + x + y = 60 

x + y = 15        (1)
Median of data is given as 28.5 which lies in interval 20 - 30.
So, median class = 20 - 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10 

 

From equation (1)
    8 + y = 15
    y = 7
Hence values of x and y are 8 and 7 respectively.

Question 3

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 3

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 4

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.
(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 4

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 5

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 5

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 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	6	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.

Solution 6

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

Number of letters	Frequency (fi)	Cumulative frequency
1 - 4	6	6
4 - 7	30	30 + 6 = 36
7 - 10	40	36 + 40 = 76
10 - 13	16	76 + 16 = 92
13 - 16	4	92 + 4 = 96
16 - 19	4	96 + 4 = 100
Total (n)	100

Now we may observe that cumulative frequency just greater than  is 76 belonging to class interval 7 - 10.
Median class = 7 - 10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3 
 
Now we can find class marks of given class intervals by using relation 
 

Taking 11.5 as assumed mean (a) we can find di, ui and fiui according to step deviation method as below. 

Number of  letters	Number of  surnames	xi	xi - a		fiui
1 - 4	6	2.5	-9	-3	-18
4 - 7	30	5.5	-6	-2	-60
7 - 10	40	8.5	-3	-1	-40
10 - 13	16	11.5	0	0	0
13 - 16	4	14.5	3	1	4
16 - 19	4	17.5	6	2	8
Total	100				-106

fiui = - 106
fi = 100 
 

We know that
3 median = mode + 2 mean
3(8.05) = mode + 2(8.32)
24.15 - 16.64 = mode
7.51 = mode
So, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.51.

Question 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 7

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


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.

NCERT Solution for Class 10 Mathematics Chapter 14 - Statistics Page/Excercise 14.4

Question 1

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

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution 1

We can find frequency distribution table of less than type as following - 

Daily income  (in Rs) (upper class limits)	Cumulative frequency
Less than 120	12
Less than 140	12 + 14 = 26
Less than 160	26 + 8 = 34
Less than 180	34 + 6 = 40
Less than 200	40 + 10 = 50

Now taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis we can draw its ogive as following -
 

Question 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 verify the result by using the formula.

Solution 2

The given cumulative frequency distributions of less than type is - 

Weight  (in kg)  upper class limits	Number of students  (cumulative frequency)
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

Now taking upper class limits on x-axis and their respective cumulative frequency on y-axis we may draw its ogive as following - 

> 

Now mark the point A whose ordinate is 17.5 its x-coordinate is 46.5. So median of this data is 46.5.
We may observe that difference between two consecutive upper class limits is 2. Now we may obtain class marks with their respective frequencies as below 

Weight  (in kg)	Frequency (f)	Cumulative frequency
Less than 38	0	0
38 - 40	3 -  0 = 3	3
40 - 42	5 - 3 = 2	5
42 - 44	9 - 5 = 4	9
44 - 46	14 - 9 = 5	14
46 - 48	28 - 14 = 14	28
48 - 50	32 - 28 = 4	32
50 - 52	35 - 32 = 3	35
Total (n)	35

Now the cumulative frequency just greater than  is 28 belonging to class interval 46 -  48
Median class = 46 - 48 
Lower class limit (l) of median class = 46 
Frequency (f) of median class = 14
Cumulative frequency (cf) of class preceding median class = 14
Class size (h) = 2 
 

So median of this data is 46.5 
Hence, value of median is verified. 

Question 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 ogive.

Solution 3

We can obtain cumulative frequency distribution of more than type as following -

Production yield  (lower class limits)	Cumulative frequency
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

Now taking lower class limits on x-axis and their respective cumulative frequencies on y-axis we can obtain its ogive as following.