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14 STATISTICS Exercise 14.1

Q.1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants 02 24 46 68 810 1012 1214 Number of houses 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why ? Solution.

Number of plants Number of houses( )if

Class mark ( )ix

i if x

02 24 46 68 810 1012 1214

1 2 1 5 6 2 3

1 3 5 7 9 11 13

1 6 5 35 54 22 39

Total = 20if = 162i if x

Mean 162 8.1 plants20

i i

i

f xxf

= = = = We have used direct method because numerical values of ix and if are small.

Q.2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs)

100120 120140 140160 160180 180200

Number of workers

12 14 8 6 10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution. Assumed mean ( ) 150a =Width of the class ( ) 20h =

Daily Wages (in Rs)

Number of workers

( )if

Class mark ( )ix

i id x= a ii

x auh

= i if u



100120 120140 140160 160180 180200

12 14 8 6 10

110 130 150 170 190

40 20 0 20 40

2 1

0 1 2

24 14 0 6 20

Total = 50if = 12i if u

Mean 12150 20 150 4.8 145.250

= = + = + = = i i

i

f ux a h

f

Hence, the mean daily wages of the workers of factory is Rs 145.20. Q.3. The following distribution shows the daily packet allowance of children of a

locality. The mean pocket allowance is Rs 18. Find the missing frequence f.

Daily pocket allowance (in Rs)

1113 1315 1517 1719 1921 2123 2325

Number of children 7 6 9 13 f 5 4 Solution.

Daily pocket allowance

(in Rs)

Number of children

( )if

Class mark ( )ix

i if x

1113 1315 1517 1719 1921 2123 2325

7 6 9 13 f 5 4 12 14 16 18 20 22 24

84 84 144 234 20f 110 96

Total = + 44if f = 20 + 752i if x f

Mean i ii

f xxf

= =

20 7521844

ff+= +

20 752 18( 44) 20 752 18 792 20 18 792 752

2 4040 202

+ = + + = + = = = =

f ff ff f

f

f

Hence, the missing frequency is 20.



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

Number of heart beats per minute

6568 6871 7174 7477 7780 8083 8386

Number of women

2 4 3 8 7 4 2

Solution. Assumed mean ( ) 75.5a = , width of the class ( ) 3h =

Number of heart beats per minute

Number of women

( )if

Class mark ( )ix

i id x= a ii

x auh

= i if u

6568 6871 7174 7477 7780 8083 8386

2 4 3 8 7 4 2

66.5 69.5 72.5 75.5 78.5 81.5 84.5

9 6 3

0 3 6 9

3 2 1

0 1 2 3

6 8 3

0 7 8 6


Mean 475.5 3 75.5 0.4 75.930

i i

i

f ux a hf

= = + = + = + =

Hence, the mean heart beats per minute is 75.9 Q.5. In a retail market, fruit vendors were selling mangoes kept in packing boxes.

These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes 5052 5355 5658 5961 6264 Number of boxes 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose ? Solution. Assumed mean ( ) 57a = Width of class ( ) 3h =

Number of mangoes

Number of boxes ( )if

Class mark ( )ix

i id x= a ii

x auh

= i if u

5052 5355 5658 5961 6264

15 110 135 115 25

51 54 57 60 63

6 3

0 3 6

2 1

0 1 2

30 110 0

115 50




Mean 2557 3 57 0.19 57.19400

i i

i

f ux a hf

= = + = + = + =

Q.6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs) 100150 150200 200250 250300 300350 Number of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method. Solution. Assumed mean ( ) 225a =

Width of class ( ) 50h =

Daily expenditure

(in Rs)

Number of households

( )if

Class mark ( )ix

i id x= a ii

x auh

= i if u

100150 150200 200250 250300 300350

4 5 12 2 2

125 175 225 275 325

100 50 0 50 100

2 1

0 1 2

8 5

0 2 4


Mean 7225 50 225 14 Rs 21125

i i

i

f ux a hf

= = + = + = =

Hence, the mean daily expenditure on food is Rs 211 Q.7. To find out the concentration of in the air (in parts per million, i.e., ppm),

the data was collected for 30 localities in a certain city and is presented below: 2SO

Concentraion of (in ppm) 2SO Frequency

0.000.04 0.040.08 0.080.12 0.120.16 0.160.20 0.200.24

4 9 9 2 4 2

Find the mean concentration of in the air. 2SOSolution. Assumed mean ( ) 0.14a =

Width of class ( ) 0.04h =

Concentration of 2SO

Frequency ( )if

Class mark

i id x= a ii

x auh

= i if u



(in ppm) ( )ix 0.000.04 0.040.08 0.080.12 0.120.16 0.160.20 0.200.24

4 9 9 2 4 2

0.02 0.06 0.10 0.14 0.18 0.22

0.12 0.08

0.04 0

0.04 0.08

3 2 1

0 1 2

12 18 9

0 4 4


Mean 310.14 (0.04) 0.14 0.041 0.099 ppm30

i i

i

f ux a hf

= = + = + = =

Therefore, the mean concentration of in the air is 0.099 ppm 2SO

Q.8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days 06 610 1014 1420 2028 2838 3840 Number of students 11 10 7 4 4 3 1 Solution.

Number of days Number of students

( )if

Class mark ( )ix

i if x

06 610

1014 1420 2028 2838 3840

11 10 7 4 4 3 1

3 8

12 17 24 33 39

33 80 84 68 96 99 39

Total = 40if = 499i if x

Mean 499 12.47540

i i

i

f xxf

= = = = Hence, the mean number of days a student was absent is 12.48

Q.9. The following table gives the literacy rate (in percentage) of 35 citites. Find the mean literacy rate.

Literacy rate (in %) 4555 5565 6575 7585 8595 Number of cities 3 10 11 8 3

Solution. Assumed mean ( ) 70a = Width of the class ( ) 10h =



Literacy rate

(in %)

Number of cities ( )if

Class mark ( )ix

i id x= a ii

x auh

= i if u

4555 5565 6575 7585 8595

3 10 11 8 3

50 60 70 80 90

20 10 0 10 20

2 1

0 1 2

6 10 0 8 6


Mean 2 470 10 70 70 0.57 69.4335 7

= = + = + = = = i i

i

f ux a hf

Hence, the mean literacy rate is 69.43%

Exercise 14.2 Q.1. The following table shows the ages of the patients admitted in a hospital during

a year:

Age (in years) 515 1525 2535 3545 4555 5565 Number of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution. Mode: Here, the maximum frequency is 23 and the class corresponding to this frequency is

3545. So, modal class is 3545. Modal class 35 45= 1 0 235, 23, 21, 14 and 10l f f f h= = = = = Mode 1 0

1 0 2

23 2135 102 2 23 21 14

f fl hf f f

= + = +

2 2035 10 35 35 1.8 (approx.) 36.8 years (approx.)11 11

= + = + = + = Mean: Assumed mean ( ) 40a = Width of class ( ) 10h =

Age (in years)

Number of patients

( )if

Class mark ( )ix

i id x= a ii

x auh

= i if u

515 1525 2535 3545

6 11 21 23

10 20 30 40

30 20 10 0

3 2 1

0

18 22 21 0



4555 5565

14 5

50 60

10 20

1 2

14 10


Mean 3740 1080

i i

i

f ux a hf

= = + = +

3740 40 4.63 35.378

= = = Maximum number of patients admitted in the hospital are of the age 36.8 years

(approx.), while on an average the age of patient admitted to the hospital is 35.37 years Q.2. The following data gives the information on the observed lifetimes (in hours) of

225 electrical components:

Lifetimes (in hours) 020 2040 4060 6080 80100 100120 Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components. Solution. The maximum frequency is 61 and the class corresponding to this frequency is 60 80. So, it a modal class.

60=l , 20=h1 61=f , , 0 52=f 2 38=f

Mode 1 01 0 2

61 5260 202 2 61 52 38

= + = + f fl h

f f f

9 180 4560 20 60 60 60 5.625 65.625122 90 32 8

= + = + = + = + = Hence, the modal lifetime of the components is 65.625 hours

Q.3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure (in Rs) Number of families 10001500 15002000 20002500 25003000 30003500 35004000 40004500 45005000

24 40 33 28 30 22 16 7

Solution. Mode:

Since the maximum number of families have their total monthly expenditure (in Rs) in the interval 15002000, the modal class is 15002000.



1 0 21500, 500, 40, 24, 33l h f f f= = = = =

Mode 1 01 0 2

40 241500 5002 2 40 24 33

f fl hf f f

= + = +

16 16 500 80001500 500 1500 150080 57 23 23

= + = + = + 1500 347.83 1847.83.= + =Hence, the modal monthly expenditure of the families is Rs 1847.83 Mean: Take assumed mean ( ) 3250a = Width of the class ( ) 500h =

Expenditure (in Rs)

Number of families

( )if

Class mark ( )ix

i id x= a ii

x auh

= i if u

10001500 15002000 20002500 25003000 30003500 35004000 40004500 45005000

24 40 33 28 30 22 16 7

1250 1750 2250 2750 3250 3750 4250 4750

2000 1500 1000 500 0

500 1000 1500

4 3 2 1

0 1 2 3

96 120 66 28 0 22 32 21


Mean 2353250 500200

i i

i

f ux a hf

= = + = +

235 5 11753250 3250 3250 587.50 2662.502 2= = = =

Hence, the mean monthly expenditure is Rs 2662.50 Q.4. The following distribution gives the statewise teacherstudent ration in higher

secondary schools of India. Find the mode and mean of this data. Interpret, the two measures.

Number of students per teacher Number of states/U.T. 1520 2025 2530 3035 3540 4045 4550 5055

3 8 9 10 3 0 0 2



Solution. Mode: Since the maximum number of states/U.T. is 10, so modal class is 3035.

30=l , , , , 5=h 1 10=f 0 9=f 2 3=f Mode 1 0

1 0 2

10 930 52 2 10 9 3

f fl hf f f

= + = +

530 30 0.62 30.628

= + = + = Hence, the mode of the given data is 30.6. Mean: Assumed mean ( ) 37.5a = Width of class ( ) 5h =

Number of students

per teacher

Number of states/U.T.

( )if

Class mark ( )ix

i id x= a ii

x auh

= i if u

1520 2025 2530 3035 3540 4045 4550 5055

3 8 9 10 3 0 0 2

17.5 22.5 27.5 32.5 37.5 42.5 47.5 52.5

20 15 10 5

0 5 10 15

4 3 2 1

0 1 2 3

12 24 18 10 0 0 0 6

Total = 35if = 58i if u Using deviation method

5837.5 5 37.5 8.3 29.235

i i

i

f ux a hf

= + = + = =

Hence, mean of the given data is 29.2. Hence, most states/U.T. have a student teacher ratio of 30.6 and on an average, this

ratio is 29.2 Q.5. The given distribution shows the number of runs scored by some top batsmen

of the world in oneday international cricket matches.

Runs scored Number of batsmen 30004000 40005000 50006000 60007000 70008000 80009000 900010000

4 18 9 7 6 3 1



1000011000 1

Find the mode of the data. Solution. Since the maximum number of batsmen have their runs scored in the interval

40005000, the modal class is 40005000. 4000=l , , , 1000=h 1 18=f 0 4=f , 2 9=f

Mode 1 01 0 2

18 44000 10002 2 18

f fl hf f f

= + = + 4 9

140004000 4000 608.7 4608.723

= + = + = Hence, the mode of the given data is 4608.7 runs

Q.6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Number of cars 010 1020

2030

3040

4050

5060

6070

7080

Frequence 7 14 13 12 20 11 15 8

Solution. Maximum frequency is 20. So, the modal class is 4050. 1 240, 20, 10, 11, 12l f h f f= = = = =0Mode 1 0

1 0 2

20 1240 102 2 20 12 11

f fl hf f f

= + = +

8040 40 4.7 44.717

= + = + = Hence, the mode of the data is 44.7 cars

Exercise 14.3 Q.1. The following frequency distribution gives the monthly consumption of

electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units) Number of consumers 6585 85105 105125 125145 145165 165185 185205

4 5 13 20 14 8 4

Solution. Median:

Monthly consumption Number of consumers Cumulative frequency



(in units) ( )if C.F.

6585 85105 105125 125145 145165 165185 185205

4 5 13 20 14 8 4

4 9 22 42 56 64 68

Total = 68if Here, 68if n = =And, 68 34

2 2n = =

This observation lies in the class 125145. Therefore, 125145 is the median class So, , 125l =

22,20,20

===

cffh

Median 34 222 125 20 125 12 137 units20

n cfl h

f

= + = + = + =

Hence, median is 137 units Mean: Assumed mean 135a= = Width of the class 20h= =

Monthly consumption

(in units)

Number of consumers

( )if

Class mark ( )ix

i id x= a ii

x auh

= i if u

6585 85105 105125 125145 145165 165185 185205

4 5 13 20 14 8 4

75 95 115 135 155 175 195

60 40 20 0 20 40 60

3 2 1

0 1 2 3

12 10 13 0 14 16 12


Mean 7135 2068

i i

i

f ux a hf

= = + = + 35135 135 2.05 137.0517

= + = + = Hence, mean is 137.05 units



Mode: Maximum frequency 20= 1 0 220, 20, 20, 13, 14l h f f f= = = = = Mode 1 0

1 0 2

20 13125 202 2 20 13 14

f fl hf f f

= + = +

7 140125 20 125 125 10.76 135.76 units13 13

= + = + = + = Hence, mode is 135.76 units The three measures are approximately the same in this case.

Q.2. If the median of the distribution given below is 28.5, find the values of x and y.

Class interval Frequency 010 1020 2030 3040 4050 5060

5 x 20 15 y 5

Total 60 Solution.

Class Interval Frequency ( )if

Cumulative frequency C.F.

010 1020 2030 3040 4050 5060

5 x

20 15 y 5

5 5 x+ 25 x+ 40 x+

40 x y+ + 45 x y+ +

Total = 60if 60if n = =

45 60 60 45 15 ...(i)

+ + = + = + =

x yx yx y

The median is 28.5, which lies in the class 2030 So, 20, 20, 5 , 10l f cf x h= = = + =

Median 2n cf

l hf

= +



60 (5 ) 252 28.5 20 10 2020 2

25 28.5 20 8.52

25 8.5 2 17 25 17 8 (ii)

+ = + = + = =

= = = =

x x

x

xx

From equation (i) and equation (ii), we have

8 15

15 8 7+ =

= =yy

Hence, the values of x and y are 8 and 7 respectively. Q.3. A life insurance agent found the following data for distribution of ages of 100

policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 year.

Age (in years) Number of policy holders Below 20 Below 25 Below 30 Below 35 Below 40 Below 45 Below 50 Below 55 Below 60

2 6 24 45 78 89 92 98 100

Solution.

Class Intervals Frequency ( )if

Cumulative frequency

Below 20 2025 2530 3035 3540 4045 4550 5055 5560

2 4 18 21 33 11 3 6 2

2 6 24 45 78 89 92 98 100

Total = 100if



100if n = = 100 50

2 2n = =

This observation lies in the class 3540 So, 3540 is the median class. So, 35, 5, 45, 33= = = =l h cf f

Median 50 452 35 533

n cfl h

f

= + = +

2535 35 0.75 35.75 years33

= + = + = Hence, the median age is 35.75 years

Q.4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represent in the following table:

Length (in mm) Number of leaves 118126 127135 136144 145153 154162 163171 172180

3 5 9 12 5 4 2

Find the median length of the leaves.

Solution.

Length (in mm) Number of leaves ( )if


117.5126.5 126.5135.5 135.5144.5 144.5153.5 153.5162.5 162.5171.5 171.5180.5

3 5 9 12 5 4 2

3 8 17 29 34 38 40

Total = 40if



Where, 40in f= = 40 20

2 2n = =

This observation lies in the class 144.5153.5. So, 144.5153.5 is the median class 144.5, 9, 17, 12= = = =l h cf f

Median 20 172 144.5 9 144.5 2.25 146.75 mm12

n cfl h

f

= + = + = + =

Hence, the median length of the leaves is 146.75 mm

Q.5. The following table gives the distribution of the life time of 400 neon lamps :

Life time (in hours) Number of lamps

15002000

20002500

25003000

30003500

35004000

40004500

45005000

14

56

60

86

74

62

48

Find the median life time of a lamp. Solution.

Life time (in hours) Number of lamps ( )if


15002000 20002500 25003000 30003500 35004000 40004500 45005000

14 56 60 86 74 62 48

14 70 130 216 290 352 400

Total = 400if 400if n = = 400 200

2 2n = = (Where n is total number of lamps)

This observation lies in the class 30003500



So, 30003500 is the median class 3000, 500, 130, 86l h cf f= = = =

Median 200 1302 3000 50086

n cfl h

f

= + = +

3000 406.98 3406.98 hours= + =Hence, the median life of a lamp is 3406.98 hours

Q.6. 100 surname were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters 14 47 710 1013 1316 1619 Number of surnames

6 30 40 16 4 4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

Solution. Median

Number of letters Number of surnames ( )if

Cumulative frequency C.F.

14 47 710 1013 1316 1619

6 30 40 16 4 4

6 36 76 92 96 100

Total = 100if Now, 100n = 100 50

2 2n = =

This observation lies in the class 710. So, 710 is the median class. 7, 3, 40, 36= = = =l h f cf

Median 50 362 7 340

n cfl h

f

= + = +

427 7 1.05 8.0540

= + = + =

Hence, the median number of letters in the surnames is 8.05.

Mean : Assume mean ( ) 8.5a =Width of class ( ) 3h =



Number of letters

Number of surnames

( )if

Class mark ( )ix

=i id x a = ii x au h i if u

14 47 710 1013 1316 1619

6 30 40 16 4 4

2.5 5.5 8.5 11.5 14.5 17.5

6 3

0 3 6 9

2 1

0 1 2 3

12 30 0 16 8 12

Total = 100if = 6i if u 68.5 3 8.5 0.18 8.32

100 = + = + = =

i i

i

f ux a hf

Hence, the mean number of letters in the surnames is 8.32 Mode: Here, modal class is 710. So, 1 0 27, 3, 40, 30 and 16l h f f f= = = = = Mode 1 0

1 0 2

40 307 32 2 40 30 16

f fl hf f f

= + = +

307 7 0.88 7.8834

= + = + = Hence, the modal size of the surnames is 7.88

Q.7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg) 4045 4550 5055 5560 6065 6570 7075 Number of students 2 3 8 6 6 3 2

Solution.

Weight (in kg)

Number of students ( )if


4045 4550 5055 5560 6065 6570 7075

2 3 8 6 6 3 2

2 5 13 19 25 28 30

Total = 30if Here, 30if n = = 30 15

2 2n = =

This observation lies in the class 5560, so, 5560 is the median class.



55, 5, 6, 13= = = =l h f cf Median

15 13 10 52 55 5 55 55 55 1.67 56.676 6 3

n cfl h

f

= + = + = + = + = + =

Hence, the median weight of the student is 56.67 kg.

Exercise 14.4 Q.1. The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs) 100120 120140 140160 160180 180200 Number of workers 12 14 8 6 10

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

Solution.

Daily income (in Rs)

Number of workers ( )if

Cumulative Frequency

100120 120140 140160 160180 180200

12 14 8 6 10

12 26 34 40 50

Total = 50if



Drawing points are: (120, 12), (140, 26), (160, 34), (180, 40), (200, 50)

Q.2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg) Number of students Less than 38 Less than 40 Less than 42 Less than 44 Less than 46 Less than 48 Less than 50 Less than 52

0 3 5 9 14 28 32 35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution.



Here, 35 17.5

2 2= =n

Locate 17.5 on the y-axis. From this point, draw a line parallel to the x-axis cutting the curve at a point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis determines the medium of the given data as 46.5 kg.

The ogive plotting points are:

(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35)

Grouped Frequency Distribution

Weight (in Kg) Number of students ( )if


038 3840 4042 4244 4446

0 3 2 4 5

0 3 5 9 14



4648 4850 5052

14 4 3

28 32 35

Total = 35if 35if n = = 35 17.5

2 2n = = (Where n is number of students)

This observation lies in the class 4648.

So, 4648 is the median class.

46, 35, 2, 14, 14l n h f cf= = = = =

Median 17.5 14 12 46 2 46 46.5 kg14 2

= + = + = + =

n cfl h

f

Hence, median weight of student is 46.5 kg.

We find that the median weight obtained from the graph is the same as the median weight obtained by using the formula.

Q.3. The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (in kg/ha)

5055 5560 6065 6570 7075 7580

Number of farms 2 8 12 24 38 16

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

Production yields (in kg/ha)

Number of farms ( )if

Cumulative Frequency

5055 5560 6065 6570 7075 7580

2 8 12 24 38 16

100 98 90 78 54 16

Total = 100if We draw the ogive by plotting the points:



(50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16).

Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4

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