Statistics WORKSHEET

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:Let us solve the problem using direct method

Number of plants	Number of houses (f_i)	Class mark (x_i)	f_i x_i
0 - 2 2 - 4 4 - 6 6 - 8 8 - 10 10 - 12 12 - 14	1 2 1 5 6 2 3	1 3 5 7 9 11 13	1 6 5 35 54 22 39
	å f_i = 20		å f_ix_i = 162

Mean (

= 8.1

Thus the mean number of plants per house = 8.1

The method used here is direct method because the numerical values of x_i and f_i are small.

2. 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:(Direct method)
The following problem is solved using direct method

Daily wages (in Rs.)	Number of workers (f_i)	Class mark (x_i)	f_i x_i
100 - 120 120 - 140 140 - 160 160 - 180 180 - 200	12 14 8 6 10	110 130 150 170 190	1320 1820 1200 1020 1900
Total	å f_i = 50		å f_ix_i = 7260

The mean of the given data

= 145.2
The mean daily wages of the workers = Rs. 145.20.

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:The mean pocket allowance = Rs. 18
Let us solve the problem using assumed mean method

Daily pocket allowance (in Rs.)	Number of children	x_i	d_i = x_i- A = x_i - 18	f_id_i
11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25	7 6 9 13 f 5 4	12 14 16 18 20 22 24	-6 -4 -2 0 2 4 6	-42 -24 -18 0 2f 20 24
	å f_i = 44 + f			å f_id_i = - 40 + 2f

Here å f_id_i = -40 + 2f
å f_i = 44 + f
Mean

= A +

= 18
= 18 +

= 18
Þ

= 18 - 18

= 0
   -40 + 2f = 0
        Þ 2f = 40
     f =

= 20.

4. Thirty women are examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, clothing a suitable method.

Numbers of heart beats per minute	65-68	68-71	71-74	74-77	77-80	80-83	83-36
Number of women	2	4	3	8	7	4	2

Solution:The given problem is solved using assumed mean method

Number of heart beats per minute	Number of women (f_i)	Class mark (x_i)	d_i = x_i - A = x_i - 75.5	f_id_i
65-68 68-71 71-74 74-77 77-80 80-83 83-86	2 4 3 8 7 4 2	66.5 69.5 72.5 75.5 78.5 81.5 84.5	-9 -6 -3 0 3 6 9	-18 -24 -9 0 21 24 18
	å f_i = 30			å f_id_i = 12

Mean(x) = 75.5 +

= 75.5 + 0.4
= 75.9
(x) = A +

where A = 75.5

The mean heart beats per minute = 75.9.

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

Solution:

Number of mangoes	Number of mangoes	Number of boxes (f_i)	Class marks (x_i)	di = x_i - A = x_i - 57	f_id_i
50 - 52 53 - 55 56 - 58 59 - 61 62 - 64	49.5 - 52.5 52.5 - 55.5 55.5 - 58.5 58.5 - 61.5 61.5 - 64.5	15 110 135 115 25	51 54 57 60 63	-6 -3 0 3 6	-90 -330 0 345 150
		å f_i = 400			å f_id_i = 75

Mean (

) = A +

= 57 +

= 57 + 0.1875
= 57.1875
The mean number of mangoes in a packing box = 57.19

The method adopted for solving this problem is assumed mean method.

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:

Daily expenditure (in Rs)	Number of households (f_i)	Class Mark (x_i)	d_i =	f_iu_i
100 - 150	4	125	-2	-8
150 - 200	5	175	-1	-5
200 - 250	12	225	0	0
250 - 300	2	275	1	2
300 - 350	2	325	2	4
	å f_i = 25			å f_iu_i = -7

Mean

= A +

= 225 +

= 225 - 14
= 211

\ The mean of daily expenditure on food = Rs. 211.

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 ppm)

Frequency

0.00 - 0.04

0.04 - 0.08

0.08 - 0.12

0.12 - 0.16

0.16 - 0.20

0.20 - 0.24
4

9

9

2

4

2

Find the mean concentration of SO₂ in the air.

Solution:Let us solve the problem by Step-deviation method

Concentration of SO₂(in ppm)	Frequency (f_i)	x_i	d_i = x_i - 0.1	u_i =	f_iu_i
0.0 - 0.04 0.04 - 0.08 0.08 - 0.12 0.12 - 0.16 0.16 - 0.20 0.20 - 0.24	4 9 9 2 4 2	0.02 0.06 0.1 0.14 0.18 0.22	-0.08 -0.04 0 0.04 0.08 0.12	-2 -1 0 1 2 3	-8 -9 0 2 8 6
Total	å f = 30				å f_iu_i = -1

Mean x = A +

=0.1 +

= 0.1 - (0.00133)
= 0.09867

\ The mean concentration of SO₂ in the air is 0.09867 ppm.

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:

Number of day	Number of students (f_i)	x_i	d_i = x_i – A = x_i – 17	U_i =	f_iu_i
0 – 6	11	3	-14	-7	-77
6 – 10	10	8	-9	-4.5	-45
10 – 14	7	12	-5	-2.5	-17.5
14 – 20	4	17	0	0	0
20 – 28	4	24	7	3.5	14
28 – 38	3	33	16	8	24
38 – 40	1	39	22	11	11
	å f_i =40				å f_iu_i = -90.5

Mean (x) = A +

´ h = 17+

= 17 – 4.525
= 12.475

The mean number of days a student was absent = 12.48 days.

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:

Literacy rate (in %)	Number of cities f_i	Class mark (x_i)	ui = =	f_iu_i
45 – 55	3	50	-2	-6
55 – 65	10	60	-1	-10
65 – 75	11	70	0	0
75 – 85	8	80	1	8
85 – 95	3	90	2	6
	å f_i = 35			å f_iu_i = -2

Mean (x) = A +

´ h
= 70 +

= 70 – 0.571 = 69.428
= 69.43%

The mean literacy rate (in percentage) = 69.43%.

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

Age (in years)	5 – 55	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:

Age (in years)	Number of patients (f_i)	x_i	u_i = u_i =	f_iu_i
5 – 15	6	10	-3	-18
15 – 25	11	20	-2	-22
25 – 35	21	30	-1	-21
35 – 45	23	40	0	0
45 – 55	14	50	1	44
55 – 65	5	60	2	10
	å f_i = 80			å f_iu_i = -37

Mean

= A +

´ h
= 40 +

= 40 – 4.625
= 35.375
Mode = l +

´ h

Here the maximum class frequency is 23, and the class corresponding frequency is 35 – 45. So the modal class is 35 – 75.

Now, lower limit(l) of modal class = 35
Class size(h) = 10
Frequency (f₁) of the modal class = 23
Frequency (f₀) of class preceding the model class = 21
Frequency (f₂) of class succeeding the model class = 14
Mode = l +

´ h
= 35 +

= 35 +

= 35 + 1.82
= 36.82.

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

Lifetime (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:

Life time
(in hours)

Frequency (f_i)

0 – 20

10

20 – 40

35

40 – 60

52

60 – 80

61

80 – 100

38

100 – 120

29

Here maximum frequency is 61, the class corresponding to this 60 – 80. So, the model class is 60 – 80.

Now lower limit of the model class (l) = 60
frequency(f₁) of the modal class = 61
frequency(f₀) of the preceding the model class = 52
frequency(f₂) of class succeeding the model class = 38
class size (h) = 20
Mode = + ´ h
        = 60 + ´ 20
        = 60 + ´ 20
        = 60 +
        = 60 + 5.625
        = 65.625

\ Modal lifetime of the components = 65.63.

12. 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:

Expenditure (in Rs.)	Number of families	Mid value (x_i)	u_i = =	f_iu_i
1000 - 1500	24	1250	-4	-96
1500 - 2000	40	1750	-3	-120
2000 - 2500	33	2250	-2	-66
2500 - 3000	28	2750	-1	-28
3000 - 3500	30	3250	0	0
3500 - 4000	22	3750	1	22
4000 - 4500	16	4250	2	32
4500 - 5000	7	4750	3	21
	å f_i = 200		å u_i = -4	f_iu_i = 235

Mode l +

´ h

The highest frequency is 40. The modal class is 1500 - 2000.

Lower limit (l) of the modal class = 1500

Class size (h) = 500

Frequency of the modal class (f₁) = 40
Frequency of the preceding modal class (f₀) = 24
Frequency of the succeeding modal class (f₂) = 33
Mode = l +

´ h
Mode = 1500 +

´ 500
= 1500 +

= 1500 +

= 1500 + 347.83
= 1847.83
\ The model monthly expenditure of the family = Rs. 1847.83
Mean = A +

= 3250 -

= 3250 - 587.50
= 266.65

\ Mean monthly expenditure = Rs. 2662.5.

13. 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 student 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:Let us first find the mode.

In this problem, the maximum frequency is 10, and the class corresponding to this frequency is 30-35. So, the modal class is 30 - 35.

Now
Lower limit (l) of modal class = 30
Class size (h)= 5
frequency (f₁) of model class = 10
frequency (f₀) proceeding the modal class = 9
frequency (f₂) succeeding the modal class = 3
Mode = l +

Substituting the values above in the given formula,

= 30 +

= 30.625

\ The mode of the given data is 30.6

Now calculate mean by applying direct method.

Number of students / teacher	Number of states/U.T (f_i)	Classmark (x_i)	di = x_i - A = x_i - 325	f_id_i
15-20	3	17.5	-15	-45
20-25	8	22.5	-10	-80
25-30	9	27.5	-5	-45
30-35	10	32.5	0	0
35-40	3	37.5	5	15
40-45	0	42.5	10	0
45-50	0	47.5	15	0
50 - 55	2	52.5	20	40
	å f_i = 35			å f_id_i = -11.5

= A +

= 32.5 +

= 32.5 - 3.29 = 29.21

\ The mean of the given data = 29.21

Thus we interpret that the most states/ U.T. student teacher ratio of 30.6 and on an average this ratio is 29.2.

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

Runs scored	Number 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:The maximum frequency is 18, and the class corresponding to the maximum frequency is 4000 – 5000.
Now the model class is 4000 – 5000
Class size(h) = 1000
lower limit (f) of the model class = 4000
frequency(fo) of the class proceeding the modal class = 4
frequency(f₁) of the modal class = 18
frequency(f₂) of the class succeeding the modal class = 9
mode = l +
         = 4000 +
         = 4000 +
         = 4000 +
         = 4000 + 608.7
         = 4608.7

Mode = 4608.7 runs.

15. 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 - 60	60 - 70	70 – 80
Frequency	7	14	13	12	20	11	15	8

Solution:In this problem the maximum frequency is 20, the class corresponding to the maximum frequency is 40 – 50.
Now,
The model class is 40 – 50
Class size (h) = 10
lower limit (l) of the model class = 40
frequency(f₁) of the modal class = 20
frequency(f₀)of the class proceeding the modal class = 12
frequency(f₂)of the class succeeding the modal class = 11
Mode = l +
        = 40 +
        = 40 +
        = 40 +
        = 40 + 4.71
        = 44.71 cars

\ Mode = 44.71 cars.

16. 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:

Monthly consumption (in units)	Number of consumers	Cumulative frequency
65 - 85	4	4
85 - 105	5	4 + 5 = 9
105 - 125	13	9 + 13 = 22
125 - 145	20	22 + 20 = 42
145 - 165	14	42 + 14 = 56
165 - 185	8	56 + 8 = 64
185 - 205	4	64 + 4 = 68

n = 68,

= 34. Here 125 - 145 is the class whose cumulative frequency 42 is greater than 34.

Here
L - lower limit of median class = 125

= 34 where n = number of observations
c.f - cumulative frequency of class proceeding the median class = 22
f - frequency of median class = 20
h - class size = 20
Median = l +

= 125 +

= 125 + 12
= 137

\ The median of the given data is 137.

The mean can be calculated using assumed mean method.

Monthly consumption (in units)	Number of consumer (f_i)	Class mark (x_i)	d_i = x_i- f_i = x_i - 135	f_id_i
65-85	4	75	-60	-240
85-105	5	95	-40	-200
105-125	13	115	-20	-260
125 - 145	20	135	0	0
145 - 165	14	155	20	280
165 - 185	8	175	40	320
185 - 205	4	195	60	240
	å f_i = 68			å f_id_i = 140

Mean(

= A +

= 135 +

= 135 + 2.06 = 137.06

\ The mean of the given data = 137.06

Let us find the mode as follows.

The highest frequency is 20 which corresponds to the class interval 125 - 145. So the modal class is 125 - 145.

Here
Lower limit (l) of the modal class = 125
Class size (h) = 20
frequency (f₁) of the modal class = 20
frequency (f₀) of the class succeeding the modal class = 13
frequency (f₂) of the class succeeding the modal class = 14
mode = l +

= 125 +

= 125 + 10.77
= 135.77

\ The mode of the data = 135.77

The three measures are approximately the same in this case.

17. If the median of the distribution 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:

Class Interval	Frequency (f_i)	Cumulative frequency
0 - 10	5	5
10 - 20	x	5 + x
20 - 30	20	25 + x
30 - 40	15	40 + x
40 - 50	y	40 + x + y
50 - 60	5	45 + x + y
Total	å f_i= 60

Here in this problem = n = 45 + x + y

It is given that total frequency is 60, thus n = 60
\ 45 + x + y = 60
Þ x + y = 60 - 45
= 15(1)

The median is 28.5 which lies in the class 20-30
So,
l (lower limit) of the median class = 20
f (frequency) of the median class = 20
cf(cumulative) frequency of the class = 5 + x

proceeding 20 - 30
h (class size) = 10
median = l +

28.5 = 20 +

28.5 - 20 +

8.5 =

Þ 17.0 = 25 - x
Þ x = 25 - 17
= 8

substitute x = 8 in (1)
x + y = 15
8 + y = 15
y = 15 - 8
= 7

\ The value of x and y are 8 and 7.

18. 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:

Age (in years)	Number of policy holders	Cumulative frequency
Below 20	2	2
Below 25	6	6
Below 30	24	24
Below 35	45	45
Below 40	78	78
Below 45	89	89
Below 50	92	92
Below 55	98	98
Below 60	100	100

Here n = 100,

= 50

The observation lies in the class 35 – 40

Lower limit (l) = 35

cf – cumulative frequency of the class proceeding the interval 35 – 40 = 45.

frequency (f) of the median class 35 – 40 = 33

class size (h) = 5
median = l +

= 35 +

= 35 + 0.758
= 35.76

\ The median age is 35.76 years.

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

Length (in mm)	Number of leaves
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:

Length (in mm)	Length (in mm) continuous class	Number of leaves (f_i)	Cumulative frequency
118 - 126	117.5 - 126.5	3	3
127 - 135	126.5 - 135.5	5	3 + 5 = 8
136 - 144	135.5 - 144.5	9	8 + 9 = 17
145 - 153	144.5 - 153.5	12	17 + 12 = 29
154 - 162	153.5 - 162.5	5	29 + 5 = 34
163 - 171	162.5 - 171.5	4	34 + 4 = 38
172 - 180	171.5 - 179.5	2	38 + 2 = 40
		å f = 40

Now n = 40,

= 20

This observation lies in the class 144.5 - 153.5

l (lower limit of the classes) = 144.5

cf (cumulative frequency of the class proceeding 144.5 - 153.5 = 17

f (frequency of the median class) = 12

h (class size) = 9
median = l +

= 144.5 +

= 144.5 + 2.25
= 146.75 mm

\ The median length of the leaves = 146.75 mm.

20. The following table 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:

Life time (in hours)	Number of lamps (f_i)	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
	å f_i = 400

n = å f_i = 400

= 200

Thus the observation lies in the class 3000 - 3500
median = l +

where l - lower limit of the median class = 3000

f = frequency of the median class = 86

cf = cumulative frequency proceeding the median class = 216

h - class size = 500
= 3000 +

= 3000 +

= 3000 + 406.98
= 3406.98

Thus the median life time of a lamp is 3406.98.

21. 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	16	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:

Number of letters	Number of surnames	Cumulative frequency
1 - 4	6	6
4 - 7	30	6 + 30 = 36
7 - 10	40	36 + 40 = 76
10 - 13	16	76 + 16 = 92
13 - 16	4	9 + 4 = 96
16 - 19	4	96 + 4 = 100
	100

Here n = 100,

Here the observation lies in the class 7 - 10

l - lower limit of the class = 7

f - frequency of the median class (7 - 10) = 40

cf - (cumulative frequency of the class proceeding (7 - 10) = 36

h - class size = 3
median = l +

= 7 +

= 7 + 1.05
= 8.05

\ median = 8.05

Median number of letters in the surnames = 8.05

Let us calculate the mean using step deviation method

Number of letters	Number of surnames f_i	Class size (x_i)	u_i = =	f_iu_i
1 - 4	6	2.5	-3	-18
4 - 7	30	5.5	-2	-60
7 - 10	40	8.5	-1	-40
10 - 13	16	11.5	0	0
13 - 16	4	14.5	1	4
16 - 19	4	17.5	2	8
				å f_iu_i = -106

Mean

= A +

= 11.5 +

= 11.5 - 3.18
= 8.32

This mean number of letters in the surname = 8.32

Now we have to find the modal size of the surnames

Number of letters	Numbered surnames (f)
1 - 4	6
4 - 7	30
7 - 10	40
10 - 13	16
13 - 16	4
16 - 19	4

Here the maximum frequency is 40. The class corresponding to this frequency is 7 - 10.

l = lower limit of the model class = 7

frequency(f₁) of the modal class = 40

frequency(f₀) of the class proceeding the modal class = 30

frequency(f₂) of the class succeeding the modal class = 16

size of the class (h) = 3
mode = l +

= 7 +

= 7 + 0.88
= 7.88

\ The model size of the surname = 7.88.

22. 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:

Weight (in kg)	Number of students (å f_i)	Cumulative frequency
40 – 45	2	2
45 – 50	3	2+3 = 5
50 – 55	8	5+8 = 13
55 – 60	6	13+6 = 19
60 – 65	6	19 + 6 = 25
65 – 70	3	25+3 = 28
70 – 75	2	28+2 = 30

n = 30,

Here the observation lies in the class 55 – 60

The lower limit (l) = 55

f (frequency of the median class) = 6

cf (cumulative frequency of the class proceeding the median class = 13
Median = l +

= 55 +

= 55 + 3.46
= 58.46 kg.

23. 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 students	12	14	8	6	10

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

Solution:The less than type cumulative frequency distribution is given as follows:

Daily income (in Rs.)	Number of students	Cumulative frequency
less than 120	12	12
less than 140	14	12 + 14 = 26
less than 160	8	26 + 8 = 34
less than 180	6	34 + 6 = 40
less than 200	10	40 + 10 = 50

Draw the ogive curve by plotting the points(120, 12), (140, 26), (160, 34), (180, 40), (200, 50)

First draw the coordinate axes with upper limits of daily income along the horizontal axis and the cumulative frequency along the vertical axis.

24. 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 and verify the result by using the formula.

Solution:Draw a less than type ogive curve, by plotting the points(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35).
Here n = 35,

The x-coordinate of the point 17.5 will be the median.

25. 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 fans	2	8	12	24	38	16

Change the distribution to a more than type distribution, and draw its ogive.

Solution:

Production yield kg / ha	Number of forms	Cumulative frequency
More than or equal to 50	2	100
More than or equal to 55	8	98
More than or equal to 60	12	90
More than or equal to 65	24	78
More than or equal to 70	38	54
More than or equal to 75	16	16

Now draw the ogive by plotting the points(50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16).

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0.00 - 0.04 0.04 - 0.08 0.08 - 0.12 0.12 - 0.16 0.16 - 0.20 0.20 - 0.24	4 9 9 2 4 2

Life time (in hours)	Frequency (f_i)
0 – 20	10
20 – 40	35
40 – 60	52
60 – 80	61
80 – 100	38
100 – 120	29

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