Chapter 9 – Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 9A
If the mean of 5 observation x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.
If
the mean of 25 observations is 27 and each observation is decreased by 7,
what will be the new mean?
Compute
the mean of the following data:
Class | 1-3 | 3-5 | 5-7 | 7-9 |
Frequency | 12 | 22 | 27 | 19 |
_{Find the mean, using direct method:}
Class | Frequency |
0 – 10 10 – 20 20- 30 30 – 40 40 – 50 50 – 60 | 7 5 6 12 8 2 |
We have
Class | Frequency_{} | Mid Value _{} | _{} |
0-10 10-20 20-30 30-40 40-50 50-60 | 7 5 6 12 8 2 | 5 15 25 35 45 55 | 35 75 150 420 360 110 |
_{} | _{} |
_{}Mean _{}
_{Find the mean, using direct method:}
Class | Frequency |
25 – 35 35 – 45 45 – 55 55 – 65 65 – 75 | 6 10 8 12 4 |
We have
Class | Frequency _{} | Mid – value _{} | _{} |
25 – 35 35 – 45 45 – 55 55 – 65 65 – 75 | 6 10 8 12 4 | 30 40 50 60 70 | 180 400 400 720 280 |
_{} | _{} |
_{} Mean, _{}
_{Find the mean, using direct method:}
Class | Frequency |
0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 | 6 9 15 12 8 |
We have
Class | Frequency _{} | Mid Value _{} | _{} |
0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 | 6 9 15 12 8 | 50 150 250 350 450 | 300 1350 3750 4200 3600 |
_{} = 50 | _{} |
_{} Mean, _{}
Using
an appropriate method, find the mean of the following frequency distribution:
Class interval | 84-90 | 90-96 | 96-102 | 102-108 | 108-114 | 114-120 |
Frequency | 8 | 10 | 16 | 23 | 12 | 11 |
Which
method did you use, and why?
If
the mean of the following frequency distribution is 24, find the value of p.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 3 | 4 | p | 3 | 2 |
The
following distribution shows the daily pocket allowance of children of a
locality. If the mean pocket allowance is Rs.18, find the missing frequency f.
Daily pocket allowance (in ) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |
Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
If
the mean of the following frequency distribution is 54, find the value of p.
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
Frequency | 7 | p | 10 | 9 | 13 |
The
mean of the following data is 42. Find the missing frequencies x and y if the
sum of frequencies is 100.
Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Frequency | 7 | 10 | x | 13 | y | 10 | 14 | 9 |
The
daily expenditure of 100 families are given below.
Calculate f_{1} and f_{2} if the mean daily expenditure is Rs.188.
Expenditure (in Rs.) | 140-160 | 160-180 | 180-200 | 200-220 | 220-240 |
Number of families | 5 | 25 | f_{1} | f_{2} | 5 |
The mean of the following frequency distribution is 57.6 and the sum of the observations is 50.
Class | Frequency |
0 – 20 | 7 |
20 – 40 | _{} |
40 – 60 | 12 |
60 – 80 | _{} |
80 – 100 | 8 |
100 – 120 | 5 |
Find _{}and _{}.
We have
_{}
Class | Frequency _{} | Mid Value | _{} |
0 – 20 | 7 | 10 | 70 |
20 – 40 | _{} | 30 | 30_{} |
40 – 60 | 12 | 50 | 600 |
60 – 80 | _{} =18 – | 70 | 1260 – 70_{} |
80 – 100 | 8 | 90 | 720 |
100 – 120 | 5 | 110 | 550 |
_{} = 50 | _{} |
_{}
During
a medical check-up, the number of heartbeats per
minute of 30 patients were recorded and summarized as follows:
Number of heart-beats per | 65-68 | 68-71 | 71-74 | 74-77 | 77-80 | 80-83 | 83-86 |
Number of patients | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Find
the mean heartbeats per minute for these patients, choosing a suitable
method.
Find the mean, using assumed mean method:
Marks | No, of students |
0 – 10 10 – 20 20 -30 30 – 40 40 – 50 50 – 60 | 12 18 27 20 17 6 |
We have, Let A = 25 be the assumed mean
Marks | Frequency _{} | Mid value _{} | Deviation _{} | _{} |
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 | 12 18 27 20 17 6 | 5 15 25 = A 35 45 55 | -20 -10 0 10 20 30 | -240 -180 0 200 340 180 |
_{} = 100 | _{} = 300 |
_{}
Hence mean = 28
Find the mean, using assumed mean method:
Class | Frequency |
100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 | 10 20 30 15 5 |
Let the assumed mean be 150, h = 20
Marks | Frequency _{} | Mid value _{} | Deviation d_{i} = _{} – 150 | _{} di |
100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 | 10 20 30 15 5 | 110 130 150=A 170 190 | -40 -20 0 20 40 | -400 -400 0 300 200 |
_{} = 80 | _{} di=-300 |
_{}
Hence, Mean = 146.25
Find the mean, using assumed mean method:
Class | Frequency |
0 – 20 | 20 |
20 – 40 | 35 |
40 – 60 | 52 |
60 – 80 | 44 |
80 – 100 | 38 |
100 – 120 | 31 |
Let A = 50 be the assumed mean, we have
Marks | Frequency _{} | Mid value _{} | Deviation _{} | _{} |
0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120 | 20 35 52 44 38 31 | 10 30 50 = A 70 90 110 | -40 -20 0 20 40 60 | -800 -700 0 880 1520 1860 |
_{}= 220 | _{} |
_{}
The
following table gives the literacy rate (in percentage) in 40 cities. Find
the mean literacy rate, choosing a suitable method.
Literacy rate (%) | 45-55 | 55-65 | 65-75 | 75-85 | 85-95 |
Number of cities | 4 | 11 | 12 | 9 | 4 |
Find
the mean of the following frequency distribution using step-deviation method.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 7 | 10 | 15 | 8 | 10 |
Find
the mean of the following data, using step-deviation method:
Class | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
Frequency | 6 | 10 | 16 | 15 | 24 | 8 | 7 |
The
weights of tea in 70 packets are shown in the following table:
Weight (in grams) | 200-201 | 201-202 | 202-203 | 203-204 | 204-205 | 205-206 |
Number of packets | 13 | 27 | 18 | 10 | 1 | 1 |
Find
the mean weight of packets using step-deviation method.
Find
the mean of the following frequency distribution using a suitable method:
Class | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 25 | 40 | 42 | 33 | 10 |
In
a annual examination marks (out of 90) obtained by students of class X in
mathematics are given below:
Marks obtained | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 | 75-90 |
Number of students | 2 | 4 | 5 | 20 | 9 | 10 |
Find
the mean marks.
Find
the arithmetic mean of the following frequency distribution using
step-deviation method:
Age (in year) | 18-24 | 24-30 | 30-36 | 36-42 | 42-48 | 48-54 |
Number of workers | 6 | 8 | 12 | 8 | 4 | 2 |
Find the arithmetic mean of each of the following frequency distribution using step-deviation method:
Class | Frequency |
500 – 520 520 – 540 540 – 560 560 – 580 580 – 600 600 – 620 | 14 9 5 4 3 5 |
Let h = 20 and assume mean = 550, we prepare the table given below:
Age | Frequency _{} | Mid value _{} | _{} | _{} |
500 – 520 520 – 540 540 – 560 560 – 580 580 – 600 600 – 620 | 14 9 5 4 3 5 | 510 530 550 = A 570 590 610 | -2 -1 0 1 2 3 | -27 -9 0 4 6 15 |
_{} = 40 | _{} |
Thus, A = 550, h = 20, and _{} = 40, _{}
_{}
Hence the mean of the frequency distribution is 544
Find the mean age from the following frequency distribution:
Age(in years) | No. of persons |
25 – 29 30 – 34 35 – 39 40 – 44 45 – 49 50 – 54 55 – 59 | 4 14 22 16 6 5 3 |
Hint: change the given series to the exclusive series
The given series is an inclusive series, making it an exclusive series, we have
Class | Frequency _{} | Mid value _{} | _{} | _{} |
24.5 – 29.5 29.5 – 34.5 34.5 – 39.5 39.5 – 44.5 44.5 – 49.5 49.5 – 54.5 54.5 – 59.5 | 4 14 22 16 6 5 3 | 27 32 37 42 = A 47 52 57 | -3 -2 -1 0 1 2 3 | -12 -28 -22 0 6 10 9 |
_{} = 70 | _{} |
Thus, A = 42, h = 5, _{} = 70 and _{}
_{}
Hence, Mean = 39.36 years
The following table shows the age distribution of patients of malaria in a village during a particular month:
Age(in years) | No. of cases |
5 – 14 15 – 24 24 – 34 35 – 44 45 – 54 55 – 64 | 6 11 21 23 14 5 |
Find the average age of the patients.
The given series is an inclusive series making it an exclusive series,we get
class | Frequency _{} | Mid value _{} | _{} | _{} |
4.5 – 14.5 14.5 – 24.5 24.5 – 34.5 34.5 – 44.5 44.5 – 54.5 54.5 – 64.5 | 6 11 21 23 14 5 | 9.5 19.5 29.5=A 39.5 49.5 59.5 | -2 -1 0 1 2 3 | -12 -11 0 23 28 15 |
_{} = 80 | _{} |
Thus, A = 29.5, h = 10, _{} = 80 and _{}
_{}
Hence, Mean = 34.87 years
Weight
of 60 eggs were recorded as given below:
Weight (in grams) | 75-79 | 80-84 | 85-89 | 90-94 | 95-99 | 100-104 | 105-109 |
Number of eggs | 4 | 9 | 13 | 17 | 12 | 3 | 2 |
Calculate
their mean weight to the nearest gram.
The
following table shows the marks scored by 80 students in an examination:
Marks | Less than 5 | Less than 10 | Less than 15 | Less than 20 | Less than 25 | Less than 30 | Less than 35 | Less than 40 |
Number of students | 3 | 10 | 25 | 49 | 65 | 73 | 78 | 80 |
Calculate
the mean marks correct to 2 decimal places.
Chapter 9 – Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 9B
In
a hospital, the ages of diabetic patients were recorded as follows. Find the
median age.
Age (in years) | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 |
Number of patients | 5 | 20 | 40 | 50 | 25 |
Compute the median from the following data:
Marks | No. of students |
0 – 7 7 – 14 14 – 21 21 – 28 28 – 35 35 – 42 42 – 49 | 3 4 7 11 0 16 9 |
We prepare the frequency table, given below
Marks | No. of students _{} | C.F. |
0 – 7 7 – 14 14 – 21 21 – 28 28 – 35 35 – 42 42 – 49 | 3 4 7 11 0 16 9 | 3 7 14 25 25 41 50 |
N = _{} = 50 |
Now, _{}
The cumulative frequency is 25 and corresponding class is 21 – 28.
Thus, the median class is 21 – 28
l = 21, h = 7, f = 11, c = C.F. preceding class 21 – 28 is 14 and = 25
_{}
Hence the median is 28.
The following table shows the daily wages of workers in a factory:
Daily wages | No. of workers |
0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 | 40 32 48 22 8 |
Find the median daily wage income of the workers.
We prepare the frequency table given below:
Daily wages | Frequency _{} | C.F. |
0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 | 40 32 48 22 8 | 40 72 120 142 150 |
N = _{} = 150 |
Now, N = 150, therefore _{}
The cumulative frequency just greater than 75 is 120 and corresponding class is 200 – 300.
Thus, the median class is 200 – 300
l = 200, h = 100, f = 48
c = C.F. preceding median class = 72 and _{}
_{}
Hence the median of daily wages is Rs. 206.25.
Calculate the median from the following frequency distribution:
Class | Frequency |
5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35- 40 40 – 45 | 5 6 15 10 5 4 2 2 |
We prepare the frequency table, given below:
Class | Frequency _{} | C.F |
5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35- 40 40 – 45 | 5 6 15 10 5 4 2 2 | 5 11 26 36 41 45 47 49 |
_{} = 49 |
Now, N = 49_{ }
The cumulative frequency just greater than 24.5 is 26 and corresponding class is 15 – 20.
Thus, the median class is 15 – 20
_{} l = 15, h = 5, f = 15
c = CF preceding median class = 11 and _{}
_{}
Median of frequency distribution is 19.5
Given below is the number of units of electricity consumed in a week in a certain locality:
Consumption (in units) | No. of consumers |
65 – 85 85 – 105 105 – 125 125 – 145 145 – 165 165 – 185 185 – 205 | 4 5 13 20 14 7 4 |
Calculate the median.
We prepare the cumulative frequency table as given below:
Consumption | Frequency _{} | C.F |
65 – 85 85 – 105 105 – 125 125 – 145 145 – 165 165 – 185 185 – 205 | 4 5 13 20 14 7 4 | 4 9 22 42 56 63 67 |
N = _{} = 67 |
Now, N = 67 _{}
The cumulative frequency just greater than 33.5 is 42 and the corresponding class 125 – 145.
Thus, the median class is 125 – 145
_{} l = 125, h = 20, _{} and c = CF preceding the median class = 22, = 33.5
_{}
Hence median of electricity consumed is 136.5
Calculate the median from the following data:
Height(in cm) | No. of boys |
135 – 140 140 – 145 145 – 150 150 – 155 155 – 160 160 – 165 165 – 170 170 – 175 | 6 10 18 22 20 15 6 3 |
Frequency table is given below:
Height | Frequency _{} | C.F |
135 – 140 140 – 145 145 – 150 150 – 155 155 – 160 160 – 165 165 – 170 170 – 175 | 6 10 18 22 20 15 6 3 | 6 16 34 56 76 91 97 100 |
N = _{} =100 |
N = 100, _{}
The cumulative frequency just greater than 50 is 56 and the corresponding class is 150 – 155
Thus, the median class is 150 – 155
l = 150, h = 5, f = 22, c = C.F.preceding median class = 34
_{}
Hence, Median = 153.64
Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
Class | Frequency _{} |
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 | 5 25 x 18 7 |
The frequency table is given below. Let the missing frequency be x.
Class | Frequency _{} | C.F |
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 | 5 25 x 18 7 | 5 30 30 + x 48 + x 55 + x |
Median = 24 _{} Median class is 20 – 30
_{}
l = 20, h = 10, f = x, c = C.F. preceding median class = 30
_{}
Hence, the missing frequency is 25.
The
median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.
Class | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 |
Frequencies | 12 | a | 12 | 15 | b | 6 | 6 | 4 |
In
the following data the median of the runs scored by 60 top batsmen of the
world in one-day international cricket matches is 5000. Find the missing
frequencies x and y.
Runs scored | 2500-3500 | 3500-4500 | 4500-5500 | 5500-6500 | 6500-7500 | 7500-8500 |
Number of batsmen | 5 | x | y | 12 | 6 | 2 |
If the median of the following frequency distribution is 32.5, find the value of _{}.
Let _{}be the frequencies of class intervals 0 – 10 and 40 – 50
_{}
Median is 32.5 which lies in 30 – 40, so the median class is 30 – 40
l = 30, h = 10, f = 12, N = 40 and _{}
_{}
Calculate the median for the following data:
Age(in years) | Frequency |
19 – 25 26 – 32 33 – 39 40 – 46 47 – 53 54 – 60 | 35 96 68 102 35 4 |
The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get
Class | Frequency _{} | C.F |
18.5 – 25.5 25.5 – 32.5 32.5 – 39.5 39.5 – 46.5 46.5 – 53.5 53.5 – 60.5 | 35 96 68 102 35 4 | 35 131 199 301 336 340 |
f_{i} = N = 340 |
_{}
The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5 – 39.5.
_{}Median class is 32.5 – 39.5
_{}l = 32.5, h = 7, f = 68, c = C.F. of preceding median class = 131
_{}
Hence median is 36.5 years
Find the median wages for the following frequencies distribution:
Wages per day (in Rs) | Frequency |
61 – 70 71 – 80 81 – 90 91 – 100 101 – 110 111 – 120 | 5 15 20 30 20 8 |
Given series is in inclusive form converting it into exclusive form and preparing the cumulative frequency table, we get
Wages per day (in Rs) | Frequency _{} | C.F |
60.5 – 70.5 70.5 – 80.5 80.5 – 90.5 90.5 – 100.5 100.5 – 110.5 110.5 – 120.5 | 5 15 20 30 20 8 | 5 20 40 70 90 98 |
f_{i} = N =98 |
_{}
The cumulative frequency just greater than 49 is 70 and corresponding class is 90.5 – 100.5.
_{}median class is 90.5 – 100.5
_{}l = 90.5, h = 10, f = 30, c = CF preceding median class = 40
_{}
Hence, Median = Rs 93.50
Find the median from the following table:
Class | Frequency |
1 – 5 6 – 10 11 – 15 16 – 20 21 – 25 26 – 30 31 – 35 35 – 40 40 – 45 | 7 10 16 32 24 16 11 5 2 |
The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get
Marks | Frequency _{} | C.F |
0.5 – 5.5 5.5 – 10.5 10.5 – 15.5 15.5 – 20.5 20.5 – 25.5 25.5 – 30.5 30.5 – 35.5 35.5 – 40.5 40.5 – 45.5 | 7 10 16 32 24 16 11 5 2 | 7 17 33 65 89 105 116 121 123 |
f_{i} = N =123 |
_{}
The cumulative frequency just greater than 61.5 is 65.
_{}The corresponding median class is 15.5 – 20.5.
Then the median class is 15.5 – 20.5
_{}l = 15.5, h = 5, f = 32, c = C.F. preceding median class = 33
_{}
Hence, Median = 19.95
Find the median from the following data:
Marks | No. of students |
Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below 80 | 12 32 57 80 92 116 164 200 |
Marks | Frequency _{} | C.F |
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 | 12 20 25 23 12 24 48 36 | 12 32 57 80 92 116 164 200 |
N = _{} |
_{}
The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 – 60.
Thus the median class is 50 – 60
_{}l = 50, h = 10, f = 24, c = C.F. preceding median class = 92, = 100
_{}
Hence, Median = 53.33
Chapter 9 – Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 9C
Find
the mode of the following frequency distribution:
Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Frequency | 12 | 35 | 45 | 25 | 13 |
Compute
the mode of the following data:
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
Frequency | 25 | 16 | 28 | 20 | 5 |
Heights
of students of Class X are given in the following frequency distribution:
Height (in cm) | 150-155 | 155-160 | 160-165 | 165-170 | 170-175 |
Number of students | 15 | 8 | 20 | 12 | 5 |
Find
the modal height.
Also,
find the mean height. Compare and interpret the two measures of central
tendency.
Find the mode of the following distribution:
Class interval | Frequency |
10 – 14 14 – 18 18 – 22 22 – 26 26 – 30 30 – 34 34 – 38 38 – 42 | 8 6 11 20 25 22 10 4 |
As the class 26 – 30 has maximum frequency so it is modal class
_{}
Hence, mode = 28.5
Given below is the distribution of total household expenditure of 200 manual workers in a city:
Expenditure | No. of manual workers |
1000 – 1500 1500 – 2000 2000 – 2500 2500 – 3000 3000 – 3500 3500 – 4000 4000 – 4500 4500 – 5000 | 24 40 31 28 32 23 17 5 |
Find the average expenditure done by maximum number of manual workers.
As the class 1500 – 2000 has maximum frequency, so it os modal class
_{}
Hence the average expenditure done by maximum number of workers = Rs. 1820
Calculate the mode from the following data:
Monthly salary (in Rs) | No. of employees |
0 – 5000 5000- 10000 10000 – 15000 15000 – 20000 20000 – 25000 25000 – 30000 | 90 150 100 80 70 10 |
As the class 5000 – 10000 has maximum frequency, so it is modal class
_{}
Hence, mode = Rs. 7727.27
Compute the mode from the following data:
Age (in years) | No. of patients |
0 – 5 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 | 6 11 18 24 17 13 5 |
As the class 15 – 20 has maximum frequency so it is modal class.
_{}
_{}
Hence mode = 17.3 years
Compute the mode from the following series:
Size | Frequency |
45 – 55 55 – 65 65 – 75 75 – 85 85 – 95 95 – 105 105 – 115 | 7 12 17 30 32 6 10 |
As the class 85 – 95 has the maximum frequency it is modal class
_{}
Hence, mode = 85.71
Compute the mode of the following data:
Class Interval | Frequency |
1 – 5 6 – 10 11 – 15 16 – 20 21 – 25 26 – 30 31 – 35 36 – 40 41 – 45 46 – 50 | 3 8 13 18 28 20 13 8 6 4 |
The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get
Class | Frequency |
0.5 – 5.5 5.5- 10.5 10.5 – 15. 5 15.5 – 20.5 20.5 – 25. 5 25.5 – 30.5 30.5 – 35.5 35.5 – 40.5 40.5 – 45.5 45.5 – 50.5 | 3 8 13 18 28 20 13 8 6 3 |
As the class 20.5 – 25.5 has maximum frequency, so it is modal class
_{}
_{}
_{}
Hence, mode = 23.28
The
age wise participation of students in the Annual Function of a school is
shown in the following distribution.
Age (in years) | 5-7 | 7-9 | 9-11 | 11-13 | 13-15 | 15-17 | 17-19 |
Number of students | x | 15 | 18 | 30 | 50 | 48 | x |
Find
the missing frequencies when the sum of frequencies is 181. Also, find the
mode of the data.
Chapter 9 – Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 9D
Find the mean, mode and median of the following data:
Class | Frequency |
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 | 5 10 18 30 20 12 5 |
Let assumed mean be 35, h = 10, now we have
Class | Frequency _{} | Mid-value _{} | C.F | _{} | |
0-10 10-20 20-30 30-40 40-50 50-60 60-70 | 5 10 18 30 20 12 5 | 5 15 25 35 = A 45 55 65 | -3 -2 -1 0 1 2 3 | 5 15 33 63 83 95 100 | -15 -20 -18 0 20 24 15 |
N = 100 | _{} |
(i)Mean _{}
_{}
(ii)N = 100, _{}
Cumulative frequency just after 50 is 63
_{}Median class is 30 – 40
_{}l = 30, h = 10, N = 100, c = 33, f = 30
_{}
(iii)Mode = 3 × median – 2 × mean
= 3 × 35.67 – 2 × 35.6 = 107.01 – 71.2
= 35.81
Thus, Mean = 35.6, Median = 35.67 and Mode = 35.81
Find
the mean, median and mode of the following data:
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 | 120-140 |
Frequency | 6 | 8 | 10 | 12 | 6 | 5 | 3 |
Find
the mean, median and mode of the following data:
Class | 0-50 | 50-100 | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
Frequency | 2 | 3 | 5 | 6 | 5 | 3 | 1 |
Find
the mode, median and mean for the following data:
Marks obtained | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 | 75-85 |
Number of students | 7 | 31 | 33 | 17 | 11 | 1 |
A survey regarding the heights of 50 girls of a class was conducted and the following data was obtained:
Height in cm | No. of girls |
120 – 130 130 – 140 140 – 150 150 – 160 160 – 170 | 2 8 12 20 8 |
Total | 50 |
Find the mean, Median and mode of the above data.
Let the assumed mean A be 145.Class interval h = 10.
Class | Frequency _{} | Mid-Value _{} | _{} | _{} | C.F. |
120-130 130-140 140-150 150-160 160-170 | 2 8 12 20 8 | 125 135 145=A 155 165 | -2 -1 0 1 2 | -4 -8 0 20 16 | 2 10 22 42 50 |
N = 50 | _{} |
(i)Mean _{}
_{}
(ii)N = 50, _{}
Cumulative frequency just after 25 is 42
Corresponding median class is 150 – 160
Cumulative frequency before median class, c = 22
Median class frequency f = 20
_{}
(iii)Mode = 3 median – 2 mean
= 3 151.5 – 2 149.8 = 454.5 – 299.6
= 154.9
Thus, Mean = 149.8, Median = 151.5, Mode = 154.9
The following table gives the daily income of 50 workers of a factory:
Daily income(in Rs) | No. of workers |
100 – 120 120 – 140 140 – 160 160 – 180 180 – 200 | 12 14 8 6 10 |
Find the mean, mode and median of the above data
Class | Frequency_{} | Mid-value _{} | _{} | _{} | C.F. |
100-120 120-140 140-160 160-180 180-200 | 12 14 8 6 10 | 110 130 150= A 170 190 | -2 -1 0 1 2 | -24 -14 0 6 20 | 12 26 34 40 50 |
N = 50 | _{} |
Let assumed mean A = 150 and h = 20
(i)Mean _{}
_{}
(ii)_{}
Cumulative frequency just after 25 is 26
_{}Corresponding frequency median class is 120 – 140
So, l = 120, f = 14, _{}h = 20, c = 12
_{ }
(iii)Mode = 3 Median – 2 Mode
= 3 138.6 – 2 145.2
= 415.8 – 190.4
= 125.4
Hence, Mean = 145.2, Median = 138.6 and Mode = 125.4
The table below shows the daily expenditure on food of 30 households in a locality:
Daily expenditure | No. of households |
100 – 150 150 – 200 200 – 250 250 – 300 300 – 350 | 6 7 12 3 2 |
Find the mean and median daily expenditure on food
Class | Frequency_{} | Mid-value _{} | _{} | _{} | C.F. |
100-150 150-200 200-250 250-300 300-350 | 6 7 12 3 2 | 125 175 225 275 325 | -2 -1 0 1 2 | -12 -7 0 3 4 | 6 13 25 28 30 |
N = 30 | _{} |
Let assumed mean = 225 and h = 50
(i)Mean = _{}
(ii)_{}
Cumulative frequency just after 15 is 25
_{}corresponding class interval is 200 – 250
_{}Median class is 200 – 250
Cumulative frequency c just before this class = 13
_{}
_{}
Hence, Mean = 205 and Median = 208.33
Chapter 9 – Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 9E
Find
the median of the following data by making a ‘less than ogive‘.
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
Number of students | 5 | 3 | 4 | 3 | 3 | 4 | 7 | 9 | 7 | 8 |
We plot the points (10, 5), (20, 8), (30, 12), (40, 15), (50, 18),
(60, 22), (70, 29), (80, 38), (90, 45) and (100, 53) to get the ‘less than
type’ ogive as follows:
At
y = 26.5, affix A.
Through
A, draw a horizontal line meeting the curve at P.
Through
P, a vertical line is drawn which meets OX at M.
OM
= 68 units
Hence,
median marks = 68
The
given distribution shows the number of wickets taken by the bowlers in one
-day international cricket matches:
Number of wickets | Less than 15 | Less than 30 | Less than 45 | Less than 60 | Less than 75 | Less than 90 | Less than 105 | Less than 120 |
Number of bowlers | 2 | 5 | 9 | 17 | 39 | 54 | 70 | 80 |
Draw
a ‘less type’ ogive from the above data. Find the
median.
Number of wickets | Less than 15 | Less than 30 | Less than 45 | Less than 60 | Less than 75 | Less than 90 | Less than 105 | Less than 120 |
Number of bowlers | 2 | 5 | 9 | 17 | 39 | 54 | 70 | 80 |
We
plot the points (15, 2), (30, 5), (45, 9), (60, 17), (75, 39), (90, 54),
(105, 70) and (120, 80) to get the ‘less than type’ ogive
as follows:
At y = 40, affix A.
Through
A, draw a horizontal line meeting the curve at P.
Through
P, a vertical line is drawn which meets OX at M.
OM
= 78 units
Hence,
median number of wickets = 78
Draw
a ‘more than’ ogive for the data given below which
gives the marks of 100 students.
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Number of students | 4 | 6 | 10 | 10 | 25 | 22 | 18 | 5 |
We plot the points (0, 100), (10, 96), (20, 90), (30, 80), (40, 70),
(50, 45), (60, 23) and (70, 5) to get the ‘more than type’ ogive as follows:
At y = 50, affix A.
Through
A, draw a horizontal line meeting the curve at P.
Through
P, a vertical line is drawn which meets OX at M.
OM
= 47 units
Hence, median marks
= 47
The
height of 50 girls of Class X of a school are recorded as follows:
Height (in cm) | 135-140 | 140-145 | 145-150 | 150-155 | 155-160 | 160-165 |
Number of girls | 5 | 8 | 9 | 12 | 14 | 2 |
Draw
a ‘more than type’ ogive for the above data.
We plot the points (135, 50), (140, 45), (145, 37), (150, 28), (155,
16) and (160, 2) to get the ‘more than type’ ogive
as follows:
At
y = 25, affix A.
Through
A, draw a horizontal line meeting the curve at P.
Through
P, a vertical line is drawn which meets OX at M.
OM
= 151 units
Hence, median
height = 151 cm
The
monthly consumption of electricity (in units) of some families of a locality
is given in the following frequency distribution:
Monthly consumption (in units) | 140-160 | 160-180 | 180-200 | 200-220 | 220-240 | 240-260 | 260-280 |
Number of families | 3 | 8 | 15 | 40 | 50 | 30 | 10 |
Prepare
a ‘more than type’ ogive for the given frequency
distribution.
We plot the points (140, 156), (160, 153), (180, 145), (200, 130),
(220, 90), (240, 40) and (260, 10) to get the ‘more than type’ ogive as follows:
At
y = 78, affix A.
Through
A, draw a horizontal line meeting the curve at P.
Through
P, a vertical line is drawn which meets OX at M.
OM
= 226 units
Hence, median
consumption of electricity = 226 units
The
following table gives the production yield per hectare of wheat of 100 farms
of a village.
Production yield (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 its ogive. Using ogive, find the
median of the given data.
We plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54)
and (75, 16) to get the ‘more than type’ ogive as
follows:
At
y = 50, affix A.
Through
A, draw a horizontal line meeting the curve at P.
Through
P, a vertical line is drawn which meets OX at M.
OM
= 70.5 units
Hence, median
production yield = 70.5 kg/ha
The
table given below shows the weekly expenditures on food of some households in
a locality.
Weekly expenditure (in Rs.) | Number of households |
100-200 | 5 |
200-300 | 6 |
300-400 | 11 |
400-500 | 13 |
500-600 | 5 |
600-700 | 4 |
700-800 | 3 |
800-900 | 2 |
Draw
a ‘less tha type’ and a ‘more than type’ ogive for this distribution.
Less Than Series:
Class interval | Cumulative Frequency |
Less than 200 | 5 |
Less than 300 | 11 |
Less than 400 | 22 |
Less than 500 | 35 |
Less than 600 | 40 |
Less than 700 | 44 |
Less than 800 | 47 |
Less than 900 | 49 |
We
plot the points (200, 5), (300, 11), (400, 22), (500, 35), (600, 40), (700,
44), (800, 47) and (900, 49) to get ‘less than type’ ogive.
More Than Series:
Class interval | Frequency |
More than 100 | 49 |
More than 200 | 44 |
More than 300 | 38 |
More than 400 | 27 |
More than 500 | 14 |
More than 600 | 9 |
More than 700 | 5 |
More than 800 | 2 |
We
plot the points (100, 49), (200, 44), (300, 38), (400, 27), (500, 14), (600,
9), (700, 5) and (800, 2) to get more than ogive.
From the following frequency distribution, prepare the ‘More than Ogive’
Score | No. of candidates |
400 – 450 450 – 500 500 – 550 550 – 600 600- 650 650 – 700 700 – 750 750 – 800 | 20 35 40 32 24 27 18 34 |
Total | 230 |
Also find the median
More than series
We plot the points (400, 230), (450, 210), (500, 175), (550, 135), (600, 103), (650, 79), (700, 52), (750, 34)
Hence,_{}
Take a point A(0, 115) on the y-axis and draw AP||x-axis meeting the curve at P, Draw PM _{}x-axis intersecting x-axis at M
Then,OM = 590
Hence median = 590
The marks obtained by 100 students of a class in an examination are given below:
Marks | No. of students |
0 – 5 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 | 2 5 6 8 10 25 20 18 4 2 |
Draw cumulative frequency curves by using (1), less than series and (2) more than series
Hence, find the median
(i) Less than series:
Marks | No. of students |
Less than 5 Less than 10 Less than 15 Less than 20 Less than 25 Less than 30 Less than 35 Less than 40 Less than 45 Less than 50 | 2 7 13 21 31 56 76 94 98 100 |
Plot the points (5, 2), (10, 7), (15, 13), (20, 21), (25, 31), (30, 56), (35, 76), (40, 94), (45, 98) and (50, 100)
Join these points free hand to get the curve representing “less than” cumulative curve.
(ii)From the given table we may prepare the ‘more than’ series as shown below
Marks | No. of students |
More than 45 More than 40 More than 35 More than 30 More than 25 More than 20 More than 15 More than 10 More than 5 More than 0 | 2 6 24 44 69 79 87 93 98 100 |
Now, on the same graph paper as above, we plot the point (0, 100), (5, 98), (10, 93), (15, 87), (20, 79), (25, 69), (30, 44), (35, 24) and (40, 6) and (45, 2)
Join these points free hand to get required curve
Here _{}
Two curves intersect at point P(28, 50)
Hence, the median = 28
From the following data, draw the two types of cumulative frequency curves and determine the median:
Height (in cm) | Frequency |
140 – 144 144 – 148 148 – 152 152 – 156 156 – 160 160 – 164 164 – 168 168 – 172 172 – 176 176 – 180 | 3 9 24 31 42 64 75 82 86 34 |
We may prepare less than series and more than series
(i)Less than series
Height in (cm) | Frequency |
Less than 140 Less than 144 Less than 148 Less than 152 Less than 156 Less than 160 Less than 164 Less than 168 Less than 172 Less than 176 Less than 180 | 0 3 12 36 67 109 173 248 330 416 450 |
Now on graph paper plot the points (140, 0), (144, 3), (148, 12), (152, 36), (156, 67), (160, 109), (164, 173), (168, 248), (172, 330), (176, 416), (180, 450)
(ii)More than series
Height in cm | C.F. |
More than 140 More than 144 More than 148 More than 152 More than 156 More than 160 More than 164 More than 168 More than 172 More than 176 More than 180 | 450 447 438 414 383 341 277 202 120 34 0 |
Now on the same graph plot the points (140, 450), (144, 447), (148, 438), (152, 414), (156, 383), (160, 341), (164, 277), (168, 202), (172, 120), (176, 34), (180, 0)
The curves intersect at (167, 225).
Hence, 167 is the median.
Chapter 9 – Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 9F
Write
the median class of the following distribution:
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 4 | 4 | 8 | 10 | 12 | 8 | 4 |
What
is the lower limit of the modal class of the following frequency
distribution?
Age (in years) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Number of patients | 16 | 13 | 6 | 11 | 27 | 18 |
Class
having maximum frequency is the modal class.
Here,
maximum frequency = 27
Hence, the modal
class is 40 – 50.
Thus, the lower
limit of the modal class is 40.
The
monthly pocket money of 50 students of a class are given in the following
distribution:
Monthly pocket money (in Rs.) | 0-50 | 50-100 | 100-150 | 150-200 | 200-250 | 250-300 |
Number of students | 2 | 7 | 8 | 30 | 12 | 1 |
Find
the modal class and also give class mark of the modal class.
A
data has 25 observations arranged in a descending order. Which observation
represents the median?
For
a cetain distribution, mode and median were found
to be 1000 and 1250 respectively. Find mean for this distribution using an
empirical relation.
In
a class test, 50 students obtained marks as follows:
Marks obtained | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
Number of students | 4 | 6 | 25 | 10 | 5 |
Find
the modal class and the median class.
Find
the class marks of classes 10-25 and 35-55.
While
calculating the mean of a given data by the assumed-mean method, the following
values were obtained:
Find
the mean.
The
distributions X and Y with total number of observations 36 and 64, and mean 4
and 3 respectively are combined. What is the mean of the resulting
distribution X + Y?
In
a frequency distribution table with 12 classes, the class-width is 2.5 and
the lowest class boundary is 8.1, then what is the upper class boundary of
the highest class?
The
observations 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95 are arranged in
ascending order. What is the value of x if the median of the data is 63?
The median of 19
observations is 30. Two more observations are made and the values of these
are 8 and 32. Find the median of the 21 observations taken together.
What
is the cumulative frequency of the modal class of the following distribution?
Class | 3-6 | 6-9 | 9-12 | 12-15 | 15-18 | 18-21 | 21-24 |
Frequency | 7 | 13 | 10 | 23 | 4 | 21 | 16 |
Find
the mode of the given data:
Class interval | 0-20 | 20-40 | 40-60 | 60-80 |
Frequency | 15 | 6 | 18 | 10 |
The
following are the ages of 300 patients getting medical treatment in a
hospital on a particular day:
Age (in year) | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Number of patients | 60 | 42 | 55 | 70 | 53 | 20 |
Form
a ‘less than type’ cumulative frequency distribution.
In
the following data, find the value of p and q. Also, find the median class
and modal class.
Class | Frequency (f) | Cumulative frequency (cf) |
100-200 | 11 | 11 |
200-300 | 12 | p |
300-400 | 10 | 33 |
400-500 | q | 46 |
500-600 | 20 | 66 |
600-700 | 14 | 80 |
The
following frequency distribution gives the monthly consumption of electricity
of 64 consumers of a locality.
Monthly consumption (in units) | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 |
Number of consumers | 4 | 5 | 13 | 20 | 14 | 8 |
Form
a ‘more than type’ cumulative frequency distribution.
The
following table gives the life-time (in days) of 100 electric bulbs of a
certain brand.
Life-time (in days) | Less than 50 | Less than 100 | Less than 150 | Less than 200 | Less than 250 | Less than 300 |
Number of bulbs | 7 | 21 | 52 | 79 | 91 | 100 |
From
this table, construct the frequency distribution table.
The
following table gives the frequency distribution of the percentage of marks
obtained by 2300 students in a competitive examination.
Marks obtained (in percent) | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 |
Number of students | 141 | 221 | 439 | 529 | 495 | 322 | 153 |
(a) Convert the given frequency
distribution into the continuous form.
(b) Find the median class and write
its class mark
(c) Find the modal class and write
its cumulative frequency.
If
the mean of the following distribution is 27, find the value of p.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 8 | p | 12 | 13 | 10 |
Calculate
the missing frequency from the following distribution, it being given that
the median of the distribution is 24.
Age (in years) | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Number of persons |