# 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:
Let us solve the problem using direct method
 Number of plants Number of houses (fi) Class mark (xi) fi xi 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 Î£ fi = 20 Î£ fixi = 162

Mean ( = = = 8.1
Thus the mean number of plants per house = 8.1
The method used here is direct method becau
se the numerical values of xi and fi are small.

# Question-2

Consider the following distribution of daily wages of 50 workers of a factory.
 Daily wages (in  ) 100 â€“ 120 120 â€“ 140 140 â€“ 160 160 â€“ 180 180 â€“ 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  ) Number of workers (fi) Class mark (xi) fi xi 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 Î£ fi = 50 Î£ fixi = 7260

The mean of the given data ( = = = 145.2
The mean daily wages of the workers =
145.20.

# Question-3

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is 18. Find the missing frequency f.
 Daily pocket allowance (in   ) 11 â€“ 13 13 â€“ 15 15 â€“ 17 17 â€“ 19 19 â€“ 21 21 â€“ 23 23 â€“ 25 Number of workers 7 6 9 13 f 5 4

Solution:
The mean pocket allowance = 18

Let us solve the problem using assumed mean method

 Daily pocket allowance (in ) Number of children xi di = xi - A        = xi - 18 fidi 11 - 13 7 12 -6 -42 13 - 15 6 14 -4 -24 15 - 17 9 16 -2 -18 17 â€“ 19 13 18 0 0 19 - 21 f 20 2 2f 21 - 23 5 22 4 20 23 â€“ 25 4 24 6 24 Ã¥ fi = 44 + f Ã¥ fidi = - 40 + 2f

Here âˆ‘ fidi = -40 + 2f           âˆ‘ fi = 44 + f
Mean = A + = 18
= 18 + = 18
Ãž
= 18 - 18
= 0
-40 + 2f = 0         â‡’ 2f = 40
f = = 20.

# Question-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 (fi) Class mark (xi) di = xi - A          = xi - 75.5 fidi 65 - 68 2 66.5 -9 -18 68 - 71 4 69.5 -6 -24 71 - 74 3 72.5 -3 -9 74 - 77 8 75.5 0 0 77 - 80 7 78.5 3 21 80 - 83 4 81.5 6 24 83 - 86 2 84.5 9 18 Î£ fi = 30 Î£ fidi = 12

Mean ( x) =  A +  where A = 75.5
= 75.5 +
= 75.5 + 0.4

(x) = 75.9
âˆ´ The mean heart beats per minute = 75.9.

# 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 â€“ 52 53 â€“ 55 56 â€“ 58 59 â€“ 61 62 - 64 Number of boxes 15 110 135 115 25

Solution:

# Question-6

The table below shows the daily expenditure on food of 25 households in a locality.
 Daily Expenditure (in ) 100 â€“ 150 150 - 200 200 â€“ 250 250 â€“ 300 300 â€“ 350 Number of households 4 5 12 2 2
Find the mean daily expenditure on food by a suitable method.

Solution:

 Daily expenditure (in  ) Number of households (fi) Class Mark (xi) ui = fiui 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 âˆ‘ fi = 25 âˆ‘ fiui = -7

Mean = A +
= 225 +
= 225 - 14
= 211

âˆ´ The mean of daily expenditure on food = 211.

# 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:
Find the mean concentration of SO2 in the air.
 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

Solution:
Let us solve the problem by Step-deviation method
 Concentration of SO2(in ppm) Frequency (fi) xi di = xi - 0.1 ui = fiui 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 Î£ f = 30 Î£ fiui = -1

Mean (x) = A + = 0.1 +
= 0.1 - (0.00133)
= 0.09867

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

# 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 â€“ 6 6 â€“ 10 10 â€“ 14 14 â€“ 20 20 â€“ 28 28 â€“ 38 38 â€“ 40 Number of students 11 10 7 4 4 3 1

Solution:

 Number of day Number of students (fi) xi di = xi â€“ A = xi â€“ 17 Ui = di/2 fiui 0 â€“ 6 11 3 -14 -7 -77 6 â€“ 10 10 8 -9 -4.5 -45 10 â€“ 14 7 12 -5 -2.5 -17.5 14 â€“ 20 4 17 0 0 0 20 â€“ 28 4 24 7 3.5 14 28 â€“ 38 3 33 16 8 24 38 â€“ 40 1 39 22 11 11 Î£ fi = 40 Î£  fiui = -90.5

Mean (x) = A + Ã— h = 17 +

= 17 â€“ 4.525
= 12.475
The mean number of days a student was absent = 12.48 days.

# Question-9

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

 Literacy rate (In %) 45 â€“ 55 55 â€“ 65 65 â€“ 75 75 â€“ 85 85 â€“ 95 Number of cities 3 10 11 8 3

Solution:

Mean (x) = A + Ã— h
= 70 +
= 70 â€“ 0.571 = 69.428
= 69.43%

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

# Question-10

The following table shows the ages of the patients admitted in a hospital during a year:
 Age (in years) 5 â€“ 15 15 â€“ 25 25 â€“ 35 35 â€“ 45 45 â€“ 55 55 â€“ 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:

Mean = A + Ã— h

= 40 +
= 40 â€“ 4.625
= 35.375

Mode = l +
Ã— h

Here the maximum class frequency is 23, and the class corresponding frequency is 35 â€“ 45. So the modal class is 35 â€“ 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 +
Ã— h

= 35 +
= 35 +
= 35 +
= 35 + 1.82
= 36.82.

# Question-11

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components
 Lifetime (in hours) 0 â€“ 20 20 â€“ 40 40 â€“ 60 60 â€“ 80 80 - 100 100 â€“ 120 Frequency 10 35 52 61 38 29
Determine the modal lifetimes of the components.

Solution:

Here maximum frequency is 61, the class corresponding to this 60 â€“ 80. So, the model class is 60 â€“ 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 = l + Ã— h
= 60 + Ã— 20
= 60 + Ã— 20
= 60 +
= 60 + 5.625
= 65.625

âˆ´ Modal lifetime of the components = 65.63 hours.

# Question-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 ) Number of families 1000 â€“ 1500 24 1500 â€“ 2000 40 2000 â€“ 2500 33 2500 â€“ 3000 28 3000 â€“ 3500 30 3500 â€“ 4000 22 4000 â€“ 4500 16 4500 â€“ 5000 7

Solution:

Mode l + Ã— 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 + Ã— h
Mode = 1500 + Ã— 500
= 1500 +
= 1500 +
= 1500 + 347.83
= 1847.83 âˆ´ The model monthly expenditure of the family =
1847.83
Mean = A +
= 3250 -
= 3250 - 587.50
= 266.65

âˆ´ Mean monthly expenditure =  2662.5.

# Question-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 â€“ 20 3 20 â€“ 25 8 25 â€“ 30 9 30 â€“ 35 10 35 â€“ 40 3 40 â€“ 45 0 45 â€“ 50 0 50 â€“ 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 +
= 30 + = = 30.625

âˆ´ 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 (fi) Classmark (xi) di = xi - A = xi - 325 fidi 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 Ã¥ fi = 35 Ã¥ fidi = -11.5

= A +
= 32.5 +
= 32.5 - 3.29 = 29.21

âˆ´ 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.

# Question-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 â€“ 4000 4 4000 â€“ 5000 18 5000 â€“ 6000 9 6000 â€“ 7000 7 7000 â€“ 8000 6 8000 â€“ 9000 3 9000 â€“ 10000 1 10000 â€“ 11000 1

Find the mode of the data.

Solution:
The maximum frequency is 18, and the class corresponding to the maximum frequency is 4000 â€“ 5000.
Now the model class is 4000 â€“ 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.

# Question-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 â€“ 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 â€“ 50.
Now,
The model class is 40 â€“ 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

âˆ´ Mode = 44.71 cars.

# Question-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:

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

âˆ´ The median of the given data is 137.

The mean can be calculated using assumed mean method.
 Monthly consumption (in units) Number of consumer (fi) Class mark (xi) di = xi- fi = xi - 135 fidi 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 âˆ‘ fi = 68 âˆ‘ fidi = 140

Mean( = A +
= 135 +
= 135 + 2.06 = 137.06

âˆ´ 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 +
= 125 + 10.77
= 135.77

âˆ´ The mode of the data = 135.77
The three measures are approximately the same in this case.

# Question-17

 Class Interval Frequency 0 â€“ 10 5 10 â€“ 20 x 20 â€“ 30 20 30 - 40 15 40 â€“ 50 y 50 â€“ 60 5 Total 60

Solution:

 Class Interval Frequency (fi) 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 Σ fi = 60

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

It is given that total frequency is 60, thus n = 60 âˆ´ 45 + x + y = 60 â‡’ 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 +

28.5 - 20 +
8.5 = â‡’ 17.0 = 25 - x â‡’ x = 25 - 17
= 8

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

âˆ´ The value of x and y are 8 and 7.

# Question-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 â€“ 40

Lower limit (l) = 35

cf â€“ cumulative frequency of the class proceeding the interval 35 â€“ 40 = 45.

frequency (f) of the median class 35 â€“ 40 = 33

class size (h) = 5
median = l +
= 35 +
= 35 +
= 35 + 0.758
= 35.76

âˆ´ The median age is 35.76 years.

# Question-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 â€“ 126 3 127 â€“ 135 5 136 â€“ 144 9 145 â€“ 153 12 154 â€“ 162 5 163 â€“ 171 4 172 â€“ 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:

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 +
= 144.5 +
= 144.5 + 2.25
= 146.75 mm

âˆ´ The median length of the leaves = 146.75 mm.

# Question-20

The following table gives the distribution of the life time of 400 neon lamps:

 Life time (in hours) Number of lamps 1500 â€“ 2000 14 2000 â€“ 2500 56 2500 â€“ 3000 60 3000 â€“ 3500 86 3500 â€“ 4000 74 4000 â€“ 4500 62 4500 â€“ 5000 48

Find the median life time of a lamp.

Solution:

 Life time (in hours) 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 Σ fi = 400

n = âˆ‘ fi = 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
median = 3000 +
= 3000 +
= 3000 + 406.98
= 3406.98

Thus the median life time of a lamp is 3406.98 hours.

# Question-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 â€“ 4 4 â€“ 7 7 â€“ 10 10 â€“ 13 13 â€“ 16 16 â€“ 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 +
= 7 + 1.05
= 8.05

âˆ´ median = 8.05
Median number of letters in the surnames = 8.05
Let us calculate the mean using step deviation method

This mean number of letters in the surname = 8.32
Mean = A +
= 11.5 +
= 11.5 - 3.18
= 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 +
= 7 +
= 7 + = 7 + 0.88
= 7.88

âˆ´ The model size of the surname = 7.88.

# Question-22

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

 Weight (in kg) 40 â€“ 45 45 â€“ 50 50 â€“ 55 55 â€“ 60 60 â€“ 65 65 â€“ 70 70 â€“ 75 Number of students 2 3 8 6 6 3 2

Solution:

 Weight (in kg) Number of students (âˆ‘ fi) Cumulative frequency 40 â€“ 45 2 2 45 â€“ 50 3 2+3 = 5 50 â€“ 55 8 5 + 8 = 13 55 â€“ 60 6 13 + 6 = 19 60 â€“ 65 6 19 + 6 = 25 65 â€“ 70 3 25 + 3 = 28 70 â€“ 75 2 28 + 2 = 30

n = 30,
Here the observation lies in the class 55 â€“ 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 =
=
= 55 + (5/3)
= 55 + 1.67
= 56.67 kg.

# Question-23

The following distribution gives the daily income of 50 workers of a factory.
 Daily Income (in ) 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 `) 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.

# Question-24

During the medical check-up of 35 students of a class, their weights were recorded as follows:
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.

 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

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.

# Question-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).