# R S AGGARWAL AND V AGGARWAL Solutions Mathematics Class 10 Chapter 9 Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive

## Chapter 9 – Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 9A

Question 1

If the mean of 5 observation x, x + 2, x + 4, x + 6 and x  + 8 is 11, find the value of x.

Solution 1

Question 2

If
the mean of 25 observations is 27 and each observation is decreased by 7,
what will be the new mean?

Solution 2

Question 3

Compute
the mean of the following data:

 Class 1-3 3-5 5-7 7-9 Frequency 12 22 27 19

Solution 3

Question 4

Find the mean, using direct method:

 Class Frequency 0 – 1010 – 2020- 3030 – 4040 – 5050 – 60 7561282

Solution 4

We have

 Class Frequency Mid Value 0-1010-2020-3030-4040-5050-60 7561282 51525354555 3575150420360110

Mean

Question 5

Find the mean, using direct method:

 Class Frequency 25 – 3535 – 4545 – 5555 – 6565 – 75 6108124

Solution 5

We have

 Class Frequency Mid – value 25 – 3535 – 4545 – 5555 – 6565 – 75 6108124 3040506070 180400400720280

Mean,

Question 6

Find the mean, using direct method:

 Class Frequency 0 – 100100 – 200200 – 300300 – 400400 – 500 6915128

Solution 6

We have

 Class Frequency Mid Value 0 – 100100 – 200200 – 300300 – 400400 – 500 6915128 50150250350450 3001350375042003600 = 50

Mean,

Question 7

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?

Solution 7

Question 8

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

Solution 8

Question 9

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

Solution 9

Question 10

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

Solution 10

Question 11

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

Solution 11

Question 12

The
daily expenditure of 100 families are given below.
Calculate f1 and f2 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 f1 f2 5

Solution 12

Question 13

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 .

Solution 13

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

Question 14

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 minute 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.

Solution 14

Question 15

Find the mean, using assumed mean method:

 Marks No, of students 0 – 1010 – 2020 -3030 – 4040 – 5050 – 60 12182720176

Solution 15

We have, Let A = 25 be the assumed mean

 Marks Frequency Mid value Deviation 0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60 12182720176 51525 = A354555 -20-100102030 -240-1800200340180 = 100 = 300

Hence mean = 28

Question 16

Find the mean, using assumed mean method:

 Class Frequency 100 – 120120 – 140140 – 160160 – 180180 – 200 102030155

Solution 16

Let the assumed mean be 150, h = 20

 Marks Frequency Mid value Deviationdi = – 150 di 100 – 120120 – 140140 – 160160 – 180180 – 200 102030155 110130150=A170190 -40-2002040 -400-4000300200 = 80 di=-300

Hence, Mean = 146.25

Question 17

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

Solution 17

Let A = 50 be the assumed mean, we have

 Marks Frequency Mid value Deviation 0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120 203552443831 103050 = A7090110 -40-200204060 -800-700088015201860 = 220

Question 18

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

Solution 18

Question 19

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

Solution 19

Question 20

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

Solution 20

Question 21

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.

Solution 21

Question 22

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

Solution 22

Question 23

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.

Solution 23

Question 24

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

Solution 24

Question 25

Find the arithmetic mean of each of the following frequency distribution using step-deviation method:

 Class Frequency 500 – 520520 – 540540 – 560560 – 580580 – 600600 – 620 1495435

Solution 25

Let h = 20 and assume mean = 550, we prepare the table given below:

 Age Frequency Mid value 500 – 520520 – 540540 – 560560 – 580580 – 600600 – 620 1495435 510530550 = A570590610 -2-10123 -27-904615 = 40

Thus, A = 550, h = 20, and = 40,

Hence the mean of the frequency distribution is 544

Question 26

Find the mean age from the following frequency distribution:

 Age(in years) No. of persons 25 – 2930 – 3435 – 3940 – 4445 – 4950 – 5455 – 59 4142216653

Hint: change the given series to the exclusive series

Solution 26

The given series is an inclusive series, making it an exclusive series, we have

 Class Frequency Mid value 24.5 – 29.529.5 – 34.534.5 – 39.539.5 – 44.544.5 – 49.549.5 – 54.554.5 – 59.5 4142216653 27323742 = A475257 -3-2-10123 -12-28-2206109 = 70

Thus, A = 42, h = 5, = 70 and

Hence, Mean = 39.36 years

Question 27

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 – 1415 – 2424 – 3435 – 4445 – 5455 – 64 6112123145

Find the average age of the patients.

Solution 27

The given series is an inclusive series making it an exclusive series,we get

 class Frequency Mid value 4.5 – 14.514.5 – 24.524.5 – 34.534.5 – 44.544.5 – 54.554.5 – 64.5 6112123145 9.519.529.5=A39.549.559.5 -2-10123 -12-110232815 = 80

Thus, A = 29.5, h = 10, = 80 and

Hence, Mean = 34.87 years

Question 28

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.

Solution 28

Question 29

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.

Solution 29

## Chapter 9 – Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 9B

Question 1

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

Solution 1

Question 2

Compute the median from the following data:

 Marks No. of students 0 – 77 – 1414 – 2121 – 2828 – 3535 – 4242 – 49 347110169

Solution 2

We prepare the frequency table, given below

 Marks No. of students C.F. 0 – 77 – 1414 – 2121 – 2828 – 3535 – 4242 – 49 347110169 371425254150 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.

Question 3

The following table shows the daily wages of workers in a factory:

 Daily wages No. of workers 0 – 100100 – 200200 – 300300 – 400400 – 500 403248228

Find the median daily wage income of the workers.

Solution 3

We prepare the frequency table given below:

 Daily wages Frequency C.F. 0 – 100100 – 200200 – 300300 – 400400 – 500 403248228 4072120142150 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.

Hence the median is 28.
Question 4

Calculate the median from the following frequency distribution:

 Class Frequency 5 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535- 4040 – 45 5615105422

Solution 4

We prepare the frequency table, given below:

 Class Frequency C.F 5 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535- 4040 – 45 5615105422 511263641454749 = 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

Question 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 – 8585 – 105105 – 125125 – 145145 – 165165 – 185185 – 205 4513201474

Calculate the median.

Solution 5

We prepare the cumulative frequency table as given below:

 Consumption Frequency C.F 65 – 8585 – 105105 – 125125 – 145145 – 165165 – 185185 – 205 4513201474 492242566367 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

Question 6

Calculate the median from the following data:

 Height(in cm) No. of boys 135 – 140140 – 145145 – 150150 – 155155 – 160160 – 165165 – 170170 – 175 6101822201563

Solution 6

Frequency table is given below:

 Height Frequency C.F 135 – 140140 – 145145 – 150150 – 155155 – 160160 – 165165 – 170170 – 175 6101822201563 6163456769197100 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

Question 7

Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

 Class Frequency 0 – 1010 – 2020 – 3030 – 4040 – 50 525x187
Solution 7

The frequency table is given below. Let the missing frequency be x.

 Class Frequency C.F 0 – 1010 – 2020 – 3030 – 4040 – 50 525x187 53030 + x48 + x55 + 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.

Question 8

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

Solution 8

Question 9

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

Solution 9

Question 10

If the median of the following frequency distribution is 32.5, find the value of .

Solution 10

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

Question 11

Calculate the median for the following data:

 Age(in years) Frequency 19 – 2526 – 3233 – 3940 – 4647 – 5354 – 60 359668102354

Solution 11

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 – 25.525.5 – 32.532.5 – 39.539.5 – 46.546.5 – 53.553.5 – 60.5 359668102354 35131199301336340 fi = 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

Question 12

Find the median wages for the following frequencies distribution:

 Wages per day(in Rs) Frequency 61 – 7071 – 8081 – 9091 – 100101 – 110111 – 120 5152030208

Solution 12

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 – 70.570.5 – 80.580.5 – 90.590.5 – 100.5100.5 – 110.5110.5 – 120.5 5152030208 52040709098 fi = 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

Question 13

Find the median from the following table:

 Class Frequency 1 – 56 – 1011 – 1516 – 2021 – 2526 – 3031 – 3535 – 4040 – 45 710163224161152

Solution 13

The given series is converted from inclusive to exclusive form and preparing the cumulative frequency table, we get

 Marks Frequency C.F 0.5 – 5.55.5 – 10.510.5 – 15.515.5 – 20.520.5 – 25.525.5 – 30.530.5 – 35.535.5 – 40.540.5 – 45.5 710163224161152 717336589105116121123 fi = 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

Question 14

Find the median from the following data:

 Marks No. of students Below 10Below 20Below 30Below 40Below 50Below 60Below 70Below 80 1232578092116164200

Solution 14

 Marks Frequency C.F 0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80 1220252312244836 1232578092116164200 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

Question 1

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

Solution 1

Question 2

Compute
the mode of the following data:

 Class 0-20 20-40 40-60 60-80 80-100 Frequency 25 16 28 20 5

Solution 2

Question 3

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.

Solution 3

Question 4

Find the mode of the following distribution:

 Class interval Frequency 10 – 1414 – 1818 – 2222 – 2626 – 3030 – 3434 – 3838 – 42 8611202522104

Solution 4

As the class 26 – 30 has maximum frequency so it is modal class

Hence, mode = 28.5

Question 5

Given below is the distribution of total household expenditure of 200 manual workers in a city:

 Expenditure No. of manual workers 1000 – 15001500 – 20002000 – 25002500 – 30003000 – 35003500 – 40004000 – 45004500 – 5000 244031283223175

Find the average expenditure done by maximum number of manual workers.

Solution 5

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

Question 6

Calculate the mode from the following data:

 Monthly salary(in Rs) No. of employees 0 – 50005000- 1000010000 – 1500015000 – 2000020000 – 2500025000 – 30000 90150100807010
Solution 6

As the class 5000 – 10000 has maximum frequency, so it is modal class

Hence, mode = Rs. 7727.27

Question 7

Compute the mode from the following data:

 Age (in years) No. of patients 0 – 55 – 1010 – 1515 – 2020 – 2525 – 3030 – 35 611182417135
Solution 7

As the class 15 – 20 has maximum frequency so it is modal class.

Hence mode = 17.3 years

Question 8

Compute the mode from the following series:

 Size Frequency 45 – 5555 – 6565 – 7575 – 8585 – 9595 – 105105 – 115 712173032610
Solution 8

As the class 85 – 95 has the maximum frequency it is modal class

Hence, mode = 85.71

Question 9

Compute the mode of the following data:

 Class Interval Frequency 1 – 56 – 1011 – 1516 – 2021 – 2526 – 3031 – 3536 – 4041 – 4546 – 50 381318282013864

Solution 9

The given series is converted from inclusive to exclusive form and on preparing the frequency table, we get

 Class Frequency 0.5 – 5.55.5- 10.510.5 – 15. 515.5 – 20.520.5 – 25. 525.5 – 30.530.5 – 35.535.5 – 40.540.5 – 45.545.5 – 50.5 381318282013863

As the class 20.5 – 25.5 has maximum frequency, so it is modal class

Hence, mode = 23.28

Question 10

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.

Solution 10

## Chapter 9 – Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Exercise Ex. 9D

Question 1

Find the mean, mode and median of the following data:

 Class Frequency 0 – 1010 – 2020 – 3030 – 4040 – 5050 – 6060 – 70 510183020125

Solution 1

Let assumed mean be 35, h = 10, now we have

 Class Frequency Mid-value C.F 0-1010-2020-3030-4040-5050-6060-70 510183020125 5152535 = A455565 -3-2-10123 51533638395100 -15-20-180202415 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

Question 2

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

Solution 2

Question 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

Solution 3

Question 4

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

Solution 4

Question 5

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 – 130130 – 140140 – 150150 – 160160 – 170 2812208 Total 50

Find the mean, Median and mode of the above data.

Solution 5

Let the assumed mean A be 145.Class interval h = 10.

 Class Frequency Mid-Value C.F. 120-130130-140140-150150-160160-170 2812208 125135145=A155165 -2-1012 -4-802016 210224250 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

Question 6

The following table gives the daily income of 50 workers of a factory:

 Daily income(in Rs) No. of workers 100 – 120120 – 140140 – 160160 – 180180 – 200 12148610

Find the mean, mode and median of the above data

Solution 6
 Class Frequency Mid-value C.F. 100-120120-140140-160160-180180-200 12148610 110130150= A170190 -2-1012 -24-140620 1226344050 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

Question 7

The table below shows the daily expenditure on food of 30 households in a locality:

 Daily expenditure No. of households 100 – 150150 – 200200 – 250250 – 300300 – 350 671232

Find the mean and median daily expenditure on food

Solution 7

 Class Frequency Mid-value C.F. 100-150150-200200-250250-300300-350 671232 125175225275325 -2-1012 -12-7034 613252830 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

Question 1

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

Solution 1

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

Question 2

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.

Solution 2

 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

Question 3

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

Solution 3

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

Question 4

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.

Solution 4

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

Question 5

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.

Solution 5

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

Question 6

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.

Solution 6

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

Question 7

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.

Solution 7

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.

Question 8

From the following frequency distribution, prepare the ‘More than Ogive’

 Score No. of candidates 400 – 450450 – 500500 – 550550 – 600600- 650650 – 700700 – 750750 – 800 2035403224271834 Total 230

Also find the median

Solution 8

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

Question 9

The marks obtained by 100 students of a class in an examination are given below:

 Marks No. of students 0 – 55 – 1010 – 1515 – 2020 – 2525 – 3030 – 3535 – 4040 – 4545 – 50 25681025201842

Draw cumulative frequency curves by using (1), less than series and (2) more than series

Hence, find the median

Solution 9

(i) Less than series:

 Marks No. of students Less than 5Less than 10Less than 15Less than 20Less than 25Less than 30Less than 35Less than 40Less than 45Less than 50 2713213156769498100

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 45More than 40More than 35More than 30More than 25More than 20More than 15More than 10More than 5More than 0 2624446979879398100

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

Question 10

From the following data, draw the two types of cumulative frequency curves and determine the median:

 Height (in cm) Frequency 140 – 144144 – 148148 – 152152 – 156156 – 160160 – 164164 – 168168 – 172172 – 176176 – 180 392431426475828634

Solution 10

We may prepare less than series and more than series

(i)Less than series

 Height in (cm) Frequency Less than 140Less than 144Less than 148Less than 152Less than 156Less than 160Less than 164Less than 168Less than 172Less than 176Less than 180 03123667109173248330416450

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 140More than 144More than 148More than 152More than 156More than 160More than 164More than 168More than 172More than 176More than 180 450447438414383341277202120340

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

Question 1

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

Solution 1

Question 2

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

Solution 2

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.

Question 3

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.

Solution 3

Question 4

A
data has 25 observations arranged in a descending order. Which observation
represents the median?

Solution 4

Question 5

For
a cetain distribution, mode and median were found
to be 1000 and 1250 respectively. Find mean for this distribution using an
empirical relation.

Solution 5

Question 6

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.

Solution 6

Question 7

Find
the class marks of classes 10-25 and 35-55.

Solution 7

Question 8

While
calculating the mean of a given data by the assumed-mean method, the following
values were obtained:

Find
the mean.

Solution 8

Question 9

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?

Solution 9

Question 10

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?

Solution 10

Question 11

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?

Solution 11

Question 12

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.

Solution 12

Question 13

Solution 13

Question 14

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

Solution 14

Question 15

Find
the mode of the given data:

 Class interval 0-20 20-40 40-60 60-80 Frequency 15 6 18 10

Solution 15

Question 16

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.

Solution 16

Question 17

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

Solution 17

Question 18

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.

Solution 18

Question 19

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.

Solution 19

Question 20

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.

Solution 20

Question 21

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

Solution 21

Question 22

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