RS Agarwal Solution | Class 10th | Chapter-18 | Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive| Edugrown

Exercise Ex. 18A

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 = 28Question 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 = A 7090110	-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 544Question 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 seriesSolution 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

Exercise Ex. 18B

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 19.5Question 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 136.5Question 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	525×187

Solution 7

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

Class	Frequency	C.F
0 – 1010 – 2020 – 3030 – 4040 – 50	525×187	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 yearsQuestion 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 93.50Question 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

Exercise Ex. 18C

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 = 28.5Question 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. 1820Question 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. 7727.27Question 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 yearsQuestion 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 = 85.71Question 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 = 23.28Question 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

Exercise Ex. 18D

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 = 35.81Question 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 = 154.9Question 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 dataSolution 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 = 125.4Question 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 foodSolution 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

Exercise Ex. 18E

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 = 68Question 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 = 78Question 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 = 47Question 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 cmQuestion 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 unitsQuestion 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/haQuestion 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 medianSolution 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 = 590Question 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 medianSolution 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.

Exercise Ex. 18F

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	5	25	?	18	7

Solution 22

Exercise MCQ

Question 1

Which of the following is not a measure of central tendency?

(a) Mean

(b) Mode

(c) Median

(d) RangeSolution 1

Correct option: (d)

Range is not a measure of central tendency.Question 2

Which of the following cannot be determined graphically? (a) Mean

(b) Median

(c) Mode

(d) None of theseSolution 2

Correct option: (a)

Mean cannot be determined graphically.Question 3

Which of the following measures of central tendency is influenced by extreme values?

(a) Mean

(b) Median

(c) Mode

(d) None of theseSolution 3

Correct option: (a)

Since mean is the average of all observations, it is influenced by extreme values.Question 4

The mode of a frequency distribution is obtained graphically from

(a) a frequency curve

(b) a frequency polygon

(c) a histogram

(d) an ogiveSolution 4

Correct option: (c)

Mode can be obtained graphically from a histogram.Question 5

The median of a frequency distribution is found graphically with the help of

(a) a histogram

(b) a frequency curve

(c) a frequency polygon

(d) ogivesSolution 5

Correct option: (d)

Ogives are used to determine the median of a frequency distribution.Question 6

The cumulative frequency table is useful in determining the

(a) mean

(b) median

(c) mode

(d) all of theseSolution 6

Correct option: (b)

The cumulative frequency table is useful in determining the median.Question 7

The abscissa of the point of intersection of the Less Than Type and of the More Than Type cumulative frequency curves of a grouped data gives its

(a) mean

(b) median

(c) mode

(d) none of theseSolution 7

Correct option: (b)

Median is given by the abscissa of the point of intersection of the Less than Type and More than Type cumulative frequency curves.Question 8

(a) 1

(b) 0

(c) -1

(d) 2Solution 8

Question 9

Solution 9

Question 10

(a) Lower limits of the classes

(b) upper limits of the classes

(c) midpoints of the classes

(d) none of theseSolution 10

Correct option: (c)

d_i‘s are the deviations from A of midpoints of the classes.Question 11

While computing the mean of the grouped data, we assume that the frequencies are

(a) evenly distributed over the classes

(b) centred at the class marks of the classes

(c) centred at the lower limits of the classes

(d) centred at the upper limits of the classesSolution 11

Correct option: (b)

While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.Question 12

The relation between mean, mode and median is

(a) mode = (3 ⨯ mean) – (2 ⨯ median)

(b) mode = (3 ⨯ median) – (2 ⨯ mean)

(c) median = (3 ⨯ mean) – (2 ⨯ mode)

(d) mean = (3 ⨯ median) – (2 ⨯ mode)Solution 12

Correct option: (b)

Mode = (3 x median) – (2 x mean)Question 13

If the ‘less than type’ ogive and ‘more than type’ ogive intersect each other at (20.5, 15.5) then the median of the given data is

(a) 5.5

(b) 15.5

(c) 20.5

(d) 36.0Solution 13

Correct option: (c)

Since the abscissa of the point of intersection of both the ogives gives the median, we have median = 20.5Question 14

Consider the frequency distribution of the height of 60 students of a class:

Height (in cm)	No. of Students	Cumulative Frequency
150-155	16	16
155-160	12	28
160-165	9	37
165-170	7	44
170-175	10	54
175-180	6	60

The sum of the lower limit of the modal class and the upper limit of the median class is

(a) 310

(b) 315

(c) 320

(d) 330Solution 14

Question 15

Consider the following frequency distribution :

Class	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	3	9	15	30	18	5

The modal class is

(a) 10-20

(b) 20-30

(c) 30-40

(d) 50-60Solution 15

Correct option: (c)

Class having maximum frequency is the modal class.

Here, maximum frequency = 30

Hence, the modal class is 30 – 40.Question 16

Mode =?

Solution 16

Question 17

Median = ?

Solution 17

Question 18

If the mean and median of a set of numbers are 8.9 and 9 respectively then the mode will be

(a) 7.2

(b) 8.2

(c) 9.2

(d) 10.2Solution 18

Correct option: (c)

Mean = 8.9

Median = 9

Mode = 3Median – 2Mean

= 3 x 9 – 2 x 8.9

= 27 – 17.8

= 9.2Question 19

Look at the frequency distribution table given below:

Class interval	35-45	45-55	55-65	65-75
Frequency	8	12	20	10

The median of the above distribution is

(a) 56.5

(b) 57.5

(c) 58.5

(d) 59Solution 19

Question 20

Consider the following table :

Class interval	10-14	14-18	18-22	22-26	26-30
Frequency	5	11	16	25	19

The mode of the above data is

(a) 23.5

(b) 24

(c) 24.4

(d) 25Solution 20

Question 21

The mean and mode of a frequency distribution are 28 and 16 respectively. The median is

(a) 22

(b) 23.5

(c) 24

(d) 24.5Solution 21

Question 22

The median and mode of a frequency distribution are 26 and 29 respectively. The mean is

(a) 27.5

(b) 24.5

(c) 28.4

(d) 25.8Solution 22

Question 23

For a symmetrical frequency distribution, we have

(a) mean < mode < median

(b) mean > mode > median

(c) mean = mode = median

(d) Solution 23

Correct option: (c)

For a symmetrical distribution, we have

Mean = mode = medianQuestion 24

Look at the cumulative frequency distribution table given below :

Monthly income	Number of families
More than Rs.10000	100
More than Rs.14000	85
More than Rs.18000	69
More than Rs.20000	50
More than Rs.25000	37
More than Rs.30000	15

Number of families having income range 20000 to 25000 is

(a) 19

(b) 16

(c) 13

(d) 22Solution 24

Correct option: (c)

Number of families having income more than Rs. 20000 = 50

Number of families having income more than Rs. 25000 = 37

Hence, number of families having income range 20000 to 25000 = 50 – 37 = 13Question 25

The median of first 8 prime numbers is

(a) 7

(b) 9

(c) 11

(d) 13Solution 25

Question 26

The mean of 20 numbers is zero. Of them, at the most, how many may be greater than zero?

(a) 0

(b) 1

(c) 10

(d) 19Solution 26

Correct option: (d)

Mean of 20 numbers = 0

Hence, sum of 20 numbers = 0 x 20 = 0

Now, the mean can be zero if

sum of 10 numbers is (S) and the sum of remaining 10 numbers is (-S),

sum of 11 numbers is (S) and the sum of remaining 9 numbers is (-S),

…….

sum of 19 numbers is (S) and the 20^th number is (-S), then their sum is zero.

So, at the most, 19 numbers can be greater than zero.Question 27

If the median of the data 4, 7, x – 1, x – 3, 16, 25, written in ascending order, is 13 then x is equal to

(a) 13

(b) 14

(c) 15

(d) 16Solution 27

Question 28

The mean of 2, 7, 6 and x is 5 and the mean of 18, 1, 6, x and y is 10.

What is the value of y?

(a) 5

(b) 10

(c) 20

(d) 30

Note: Question modifiedSolution 28

Question 29

Match the following columns :

	Column I		Column II
(a)	The most frequent value in a data is known as …….	(p)	Standard deviation
(b)	Which of the following cannot be determined graphically out of mean, mode and median?	(q)	Median
(c)	An ogive is used to determine…	(r)	Mean
(d)	Out of mean, mode, median and standard deviation, which is not a measure of central tendency?	(s)	Mode

Solution 29

(a) – (s)

The most frequent value in a data is known as mode.

(b) – (r)

Mean cannot be determined graphically.

(c) – (q)

An ogive is used to determine median.

(d) – (p)

Standard deviation is not a measure of central tendency.Question 30

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

(a) Both Assertion (A) and Reason (R) are true and Reason (R.) is a correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).

(c) Assertion (A) is true and Reason (R) is false.

(d)Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
If the median and mode of a frequency distribution are 150 and 154 respectively, then its mean is 148.	Mean, median and mode of a frequency distribution are related as:mode = 3 median – 2 mean.

The correct answer is: (a)/(b)/(c)/(d).Solution 30

Question 31

Assertion (A)	Reason (R)
Consider the following frequency distribution :Class interval3-66-99-1212-1515-1818-21Frequency2521231012 The mode of the above data is 12.4	The value of the variable which occurs most often is the mode.

The correct answer is: (a)/(b)/(c)/(d).Solution 31

Exercise FA

Question 1

Which one of the following measures is determined only after the construction of cumulative frequency distribution?

(a) Mean

(b) Median

(c) Mode

(d) None of theseSolution 1

Correct option: (b)

The cumulative frequency table is useful in determining the median.Question 2

If the mean of a data is 27 and its median is 33 then the mode is

(a) 30

(b) 43

(c) 45

(d) 47Solution 2

Correct option: (c)

Mean = 27

Median = 33

Mode = 3Median – 2Mean

= 3 x 33 – 2 x 27

= 99 – 54

= 45Question 3

Consider the following distribution :

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

The sum of the lower limits of the median class and the modal class is

(a) 15

(b) 25

(c) 30

(d) 35Solution 3

Question 4

Consider the following frequency distribution :

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

The upper limit of the median class is

(a) 16.5

(b) 18.5

(c) 18

(d) 17.5Solution 4

Question 5

If the mean and mode of a frequency distribution be 53.4 and 55.2 respectively, find the median.Solution 5

Question 6

In the table given below, the times taken by 120 athletes to run a 100-m-hurdle race are given.

Class	13.8-14	14-14.2	14.2-14.4	14.4-14.6	14.6-14.8	14.8-15
Frequency	2	4	15	54	25	20

Find the number of athletes who completed the race in less than 14.6 seconds.Solution 6

Number of athletes who completed the race in less than 14.6 seconds

= 2 + 4 + 15 + 54

= 75Question 7

Consider the following frequency distribution :

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

Find the upper limit of the median class.Solution 7

Question 8

The annual profits earned by 30 shops of a shopping complex in a locality are recorded in the table shown below :

Profit (in lakhs Rs.)	Number of shops
More than or equal to 5	30
More than or equal to 10	28
More than or equal to 15	16
More than or equal to 20	14
More than or equal to 25	10
More than or equal to 30	7
More than or equal to 35	3

If we draw the frequency distribution table for the above data, find the frequency corresponding to the class 20-25Solution 8

The frequency table is as follows:

ClassesProfit (in lakhs Rs.)	FrequencyNumber of shops
5 – 10	2
10 – 15	12
15 – 20	2
20 – 25	4
25 – 30	3
30 – 35	4
35 – 40	3

The frequency corresponding to the class 20 – 25 is 4.Question 9

Find the mean of the following frequency distribution :

Class	1-3	3-5	5-7	7-9
Frequency	9	22	27	18

Solution 9

Question 10

The maximum bowling speeds (in km/hr) of 33 players at a cricket coaching centre are given below :

Speed in km/hr	85-100	100-115	115-130	130-145
No. of players	10	4	7	9

Calculate the median bowling speed.Solution 10

Question 11

The arithmetic mean of the following frequency distribution is 25.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	16	p	30	32	14

Find the value of p.

Note: Question modifiedSolution 11

Question 12

Find the median of the following frequency distribution :

Marks	0-10	10-20	20-30	30-40	40-50
Number of students	6	16	30	9	4

Solution 12

Question 13

Following is the distribution of marks of 70 students in a periodical test :

Marks	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50
Number of students	3	11	28	48	70

Draw a cumulative frequency curve for the above data.Solution 13

Marks	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50
Number of students	3	11	28	48	70

We plot the points (10, 3), (20, 11), (30, 28), (40, 48) and (50, 70) to get the cumulative frequency curve as follows:

Question 14

Find the median of the following data.

Class interval	0-10	10-20	20-30	30-40	40-50	Total
Frequency	8	16	36	34	6	100

Solution 14

Question 15

For the following distribution draw a ‘less than type’ ogive and from the curve find the median.

Marks obtained	Less than 20	Less than 30	Less than 40	Less than 50	Less than 60	Less than 70	Less than 80	Less than 90	Less than 100
Number of students	2	7	17	40	60	82	85	90	100

Solution 15

Marks obtained	Less than 20	Less than 30	Less than 40	Less than 50	Less than 60	Less than 70	Less than 80	Less than 90	Less than 100
Number of students	2	7	17	40	60	82	85	90	100

We plot the points (20, 2), (30, 7), (40, 17), (50, 40), (60, 60), (70, 82), (80, 85), (90, 90) and (100, 100) to get the cumulative frequency curve 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 = 56.

Hence, median = 56  Question 16

The median value for the following frequency distribution is 35 and the sum of all the frequencies is 170. Using the formula for median, find the missing frequencies.

Class	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	10	20	?	40	?	25	15

Solution 16

Question 17

Find the missing frequencies f₁ and f₂ in the table given below, it being given that the mean of the given frequency distribution is 50.

Class	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	f1	32	f2	19	120

Solution 17

Question 18

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

Class	84-90	90-96	96-102	102-108	108-114	114-120
Frequency	15	22	20	18	20	25

Solution 18

Question 19

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

Class	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	6	8	10	15	5	4	2

Solution 19

Question 20

Draw ‘less than ogive’ and more than ogive’ on a single graph paper and hence find the median of the following data :

Class interval	5-10	10-15	15-20	20-25	25-30	30-35	35-40
Frequency	2	12	2	4	3	4	3

Solution 20

Less Than Series:

Class interval	Frequency
Less than 10	2
Less than 15	14
Less than 20	16
Less than 25	20
Less than 30	23
Less than 35	27
Less than 40	30

We plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27) and (40, 30) to get less than ogive.

More Than Series:

Class interval	Frequency
More than 5	30
More than 10	28
More than 15	16
More than 20	14
More than 25	10
More than 30	7
More than 35	3

We plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) to get more than ogive.

The two curves intersect at L. Draw LM ⊥ OX.

Thus, median = OM = 16 Question 21

The production yield per hectare of wheat of some farms of a village are given in the following table :

Production yield (in kg/ha)	40-45	45-50	50-55	55-60	60-65	65-70	70-75	75-80	80-85
Number of farms	1	9	15	18	40	26	16	14	10

Draw a less than type ogive and a more than type ogive for this dataSolution 21

Less Than Series:

Class interval	Frequency
Less than 45	1
Less than 50	10
Less than 55	25
Less than 60	43
Less than 65	83
Less than 70	109
Less than 75	125
Less than 80	139
Less than 85	149

We plot the points (45, 1), (50, 10), (55, 25), (60, 43), (65, 83), (35, 27), (70, 109), (75, 125), (80, 139) and (85, 149) to get less than ogive.

More Than Series:

Class interval	Frequency
More than 40	149
More than 45	148
More than 50	139
More than 55	124
More than 60	106
More than 65	66
More than 70	40
More than 75	24
More than 80	10

We plot the points (40, 149), (45, 148), (50, 139), (55, 124), (60, 106), (65, 66), (70, 40), (75, 24) and (80, 10) to get more than ogive.

Question 22

The following table gives the marks obtained by 50 students in a class test :

Marks	11-15	16-20	21-25	26-30	31-35	36-40	41-45	46-50
Number of students	2	3	6	7	14	12	4	2

Calculate the mean, median and mode for the above data.Solution 22