Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 18 Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are provided here with simple step-by-step explanations. These solutions for Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are extremely popular among Class 10 students for Maths Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 859:

Question 1:

If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.   [CBSE 2014]

Answer:

Mean of given observations = Sum of given observationsTotal number of observations

 11=x+(x+2)+(x+4)+(x+6)+(x+8)555=5x+205x=55-205x=35x=355x=7

Hence, the value of x is 7.

Page No 859:

Question 2:

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

Answer:


​Mean of given observations = Sum of given observationsTotal number of observations

Mean of 25 observations = 27

∴ Sum of 25 observations = 27 × 25 = 675

If 7 is subtracted from every number, then the sum = 675 − (25 × 7)
                                                                                  = 675 − 175
                                                                                  = 500

Then, new mean = 50025=20

Thus, the new mean will be 20.

Page No 859:

Question 3:

Compute the mean of the following frequency distribution:
 

Class 10 − 30 30 − 50 50 − 70 70 − 90 90 − 110
Frequency 15 18 25 10 2

Answer:


Here, h = 20

Let the assumed mean, A be 60.
 

Class Mid-Values(xi) Frequency (fi) di=xi-60 ui=xi-6020 fiui
10−30 20 15 −40 −2 −30
30−50 40 18 −20 −1 −18
50−70 60 25 0 0 0
70−90 80 10 20 1 10
90−110 100 2 40 2 4
    fi=70     fiui=-34


We know

Mean =A+h×fiuifi

∴ Mean of the given frequency distribution

=60+20×-3470=60-9.71=50.29 Approx

Thus, the mean of the given frequency distribution is approximately 50.29.

Page No 859:

Question 4:

Compute the mean of the following frequency distribution:
 

Class 0 − 20 20 − 40 40 − 60 60 − 80 80 − 100 100 – 120 120 – 140
Frequency 6 8 10 12 6 5 3

Answer:


Here, h = 20

Let the assumed mean, A be 70.
 

Class Mid-Values(xi) Frequency (fi) di=xi-70 ui=xi-7020 fiui
0−20 10 6 −60 −3 −18
20−40 30 8 −40 −2 −16
40−60 50 10 −20 −1 −10
60−80 70 12 0 0 0
80−100 90 6 20 1 6
100−120 110 5 40 2 10
120−140 130 3 60 3 9
    fi=50     fiui=-19

We know

Mean =A+h×fiuifi

∴ Mean of the given frequency distribution

=70+20×-1950=70-7.6=62.4

Thus, the mean of the given frequency distribution is 62.4.

Page No 859:

Question 5:

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

Number of days 0 − 6 6 − 12 12 − 18 18 − 24 24 − 30 30 – 36 36 – 42
Number of students 10 11 7 4 4 3 1

Answer:


Here, h = 6

Let the assumed mean, A be 21.
 

Number of days Mid-Values(xi) Number of Students (fi) di=xi-21 ui=xi-216 fiui
0−6 3 10 −18 −3 −30
6−12 9 11 −12 −2 −22
12−18 15 7 −6 −1 −7
18−24 21 4 0 0 0
24−30 27 4 6 1 4
30−36 33 3 12 2 6
36−42 39 1 18 3 3
    fi=40     fiui=-46

We know

Mean =A+h×fiuifi

∴ Mean number of days a student was absent

=21+6×-4640=21-6.9=14.1

Thus, the mean number of days a student was absent is 14.1.

Page No 859:

Question 6:

The daily expenses of 40 families has been shown by the following frequency distribution:
 

Expenses (in ₹) 500 − 700 700 − 900 900 − 1100 1100 − 1300 1300 − 1500
Number of families 6 8 10 9 7

Find the mean daily expenses.

Answer:


Here, h = 200

Let the assumed mean, A be ₹1000.
 

Expenses (in ₹) Mid-Values(xi) Number of Families(fi) di=xi-1000 ui=xi-1000200 fiui
500−700 600 6 −400 −2 −12
700−900 800 8 −200 −1 −8
900−1100 1000 10 0 0 0
1100−1300 1200 9 200 1 9
1300−1500 1400 7 400 2 14
    fi=40     fiui=3


We know

Mean =A+h×fiuifi

∴ Mean daily expenses

=1000+200×340=1000+15=1015

Thus, the mean daily expenses is ₹1015.



Page No 860:

Question 7:

During a medical checkup, the number of heartbeats per minute of 40 patients were recorded and summarised as follows:
 

Number of heartbeats per minute 64 − 68 68 − 72 72 − 76 76 − 80 80 − 84 84 – 88
Number of patients 6 8 10 12 3 1

Find the mean heartbeats per minute for these patients.

Answer:


Here, h = 4

Let the assumed mean, A be 78.
 

Number of Heartbeats per Minute Mid-Values(xi) Number of Patients (fi) di=xi-78 ui=xi-784 fiui
64−68 66 6 −12 −3 −18
68−72 70 8 −8 −2 −16
72−76 74 10 −4 −1 −10
76−80 78 12 0 0 0
80−84 82 3 4 1 3
84−88 86 1 8 2 2
    fi=40     fiui=-39

We know

Mean =A+h×fiuifi

∴ Mean heartbeats per minute for these patients

=78+4×-3940=78-3.9=74.1

Thus, the mean heartbeats per minute for these patients is 74.1.

Page No 860:

Question 8:

The table given below shows the age distribution of 1000 persons who visited a marketing centre on a Sunday.
 

Age (in years) 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 – 60 60 – 70
Number of persons 105 222 220 138 102 113 100

Find the mean age of the persons visiting the marketing centre on that day.

Answer:


Here, h = 10

Let the assumed mean, A be 35.
 

Age (in years) Mid-Values(xi) Number of persons (fi) di=xi-35 ui=xi-3510 fiui
0−10 5 105 −30 −3 −315
10−20 15 222 −20 −2 −444
20−30 25 220 −10 −1 −220
30−40 35 138 0 0 0
40−50 45 102 10 1 102
50−60 55 113 20 2 226
60−70 65 100 30 3 300
    fi=1000     fiui=-351

We know

Mean =A+h×fiuifi

∴ Mean age of the persons visiting the marketing centre on that day

=35+10×-3511000=35-3.51=31.49

Thus, the mean age of the persons visiting the marketing centre on that day 31.49 years.

Page No 860:

Question 9:

The arithmetic mean of the following frequency distribution is 53. Find the value of x.
 

Class 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100
Frequency 12 15 32 x 13

Answer:

 

Class Mid-Values(xi) Frequency (fi) fi xi
0−20 10 12 120
20−40 30 15 450
40−60 50 32 1600
60−80 70 x 70x
80−100 90 13 1170
    fi=72+x fixi=3340+70x

Mean of the given frequency distribution = 53

We know

Mean =fixifi

3340+70x72+x=533340+70x=3816+53x70x-53x=3816-3340
17x=476x=47617=28

Thus, the value of x is 28.

Page No 860:

Question 10:

The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is ₹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

Answer:

The given data is shown as follows:
 

Daily pocket allowance (in ₹) Number of children (fi) Class mark (xi) fixi
11−13 7 12 84
13−15 6 14 84
15−17 9 16 144
17−19 13 18 234
19−21 f 20 20f
21−23 5 22 110
23−25 4 24 96
Total ∑ fi = 44 + f   ∑ fixi = 752 + 20f

The mean of given data is given by

x¯=ifixiifi18=752+20f44+f18(44+f)=752+20f792+18f=752+20f20f-18f=792-7522f=40f=20

​Hence, the value of f is 20.

Page No 860:

Question 11:

If the mean of the following frequency distribution is 54, find the value of p.   [CBSE 2006C]
 

Class 0−20 20−40 40−60 60−80 80−100
Frequency 7 p 10 9 13

Answer:

The given data is shown as follows:
 

Class Frequency (fi) Class mark (xi) fixi
0−20 7 10 70
20−40 p 30 30p
40−60 10 50 500
60−80 9 70 630
80−100 13 90 1170
Total ∑ fi = 39 + p   ∑ fixi = 2370 + 30p

The mean of given data is given by

x¯=ifixiifi54=2370+30p39+p54(39+p)=2370+30p2106+54p=2370+30p54p-30p=2370-210624p=264p=11

​Thus, the value of is 11.

Page No 860:

Question 12:

The mean of the following data is 42, find the missing frequencies and y if the sum of the 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

Answer:

The given data is shown as follows:
 

Class interval Frequency (fi) Class mark (xi) fixi
0−10 7 5 35
10−20 10 15 150
20−30 x 25 25x
30−40 13 35 455
40−50 y 45 45y
50−60 10 55 550
60−70 14 65 910
70−80 9 75 675
Total ∑ fi = 63 + y   ∑ fixi = 2775 + 25x + 45y

Sum of the frequencies = 100

ifi=10063+x+y=100x+y=100-63x+y=37y=37-x    ....(1)

Now, The mean of given data is given by

x¯=ifixiifi42=2775+25x+45y1004200=2775+25x+45y4200-2775=25x+45y1425=25x+45(37-x)      from (1)1425=25x+1665-45x20x=1665-142520x=240x=12

If x = 12, then y = 37 − 12 = 25

​Thus, the value of is 12 and y is 25.



Page No 861:

Question 13:

The daily expenditure of 100 families are given below. Calculate f1 and f2 if the mean daily expenditure is ₹188.
 

Expenditure
(in ₹)
140−160 160−180 180−200 200−220 220−240
Number of families 5 25 f1 f2 5

Answer:

The given data is shown as follows:
 

Expenditure
(in ₹)
Number of families
(fi)
Class mark (xi) fixi
140−160 5 150 750
160−180 25 170 4250
180−200 f1 190 190f1
200−220 f2 210 210f2
220−240 5 230 1150
Total ∑ fi = 35 + f1 f2   ∑ fixi = 6150 + 190f1 + 210f2

Sum of the frequencies = 100

ifi=10035+f1+f2=100f1+f2=100-35f1+f2=65f2=65-f1    ....(1)

Now, The mean of given data is given by

x¯=ifixiifi188=6150+190f1+210f210018800=6150+190f1+210f218800-6150=190f1+210f212650=190f1+21065-f1            from (1)12650=190f1-210f1+1365020f1=13650-1265020f1=1000f1=50

If f1 = 50, then f2 = 65 − 50 = 15

​Thus, the value of f1 is 50 and f2 is 15.

Page No 861:

Question 14:

The mean of the following frequency distribution is 57.6 and the sum of the observations is 50

Class 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 7 f1 12 f2 8 5

Answer:

Class

Frequency fi

Mid values xi

(fi×xi)

0-20

7

10

70

20-40

f1

30

30 f1

40-60

12

50

600

60-80

18- f1

70

1260-70 f1

80-100

8

90

720

100-120

5

110

550

 

fi=50

 

(fi×xi)=3200-40f1

  We have: 7+f1+12+f2+8+5=50f1+f2=18f2=18f1 Mean, x¯=(fi×xi)fi57.6=320040f150        [Mean=57.6]40f1=320f1=8And f2=188f2=10The missing frequencies are f1=8 and f2 =10.

Page No 861:

Question 15:

If the mean of the following frequency distribution is 188, find the missing frequencies x and y, if the sum of all frequencies is 100.
 

Class 0 – 80 80 – 160 160 – 240 240 – 320 320 – 400
Frequency 20 25 x y 10

Answer:


Here, h = 80

Let the assumed mean, A be 200.
 

Class Mid-Values(xi) Frequency(fi) di=xi-200 ui=xi-20080 fiui
0−80 40 20 −160 −2 −40
80−160 120 25 −80 −1 −25
160−240 200 x 0 0 0
240−320 280 y 80 1 y
320−400 360 10 160 2 20
    fi=x+y+55     fiui=-45+y

fi=x+y+55=100         (Given)

x+y=45         .....1

We know

Mean =A+h×fiuifi

200+80×-45+y100=188          (Given)

-45+y100=188-20080=-0.15-45+y=-15y=45-15=30

Substituting the value of y in (1), we get

x + 30 = 45

x = 45 − 30 = 15

Thus, the missing frequencies x and y are 15 and 30, respectively.

Page No 861:

Question 16:

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

Answer:

Using Direct method, the given data is shown as follows:
 

Literacy rate (%) Number of cities
(fi)
Class mark (xi) fixi
45−55 4 50 200
55−65 11 60 660
65−75 12 70 840
75−85 9 80 720
85−95 4 90 360
Total ∑ fi = 40   ∑ fixi = 2780

The mean of given data is given by

x¯=ifixiifi  =278040  =69.5

​Thus, the mean literacy rate is 69.5%.

Page No 861:

Question 17:

Find the mean, using assumed-mean method:

Marks 0-10 10-20 20-30 30-40 40-50 50-60
Number of students 12 18 27 20 17 6

Answer:

Class

Frequency fi

Mid valuesxi

Deviation di
di=xi-25

(fi×di)

0-10

12

5

-20

-240

10-20

18

15

-10

-180

20-30

27

25=A

0

0

30-40

20

35

10

200

40-50

17

45

20

340

50-60

6

55

30

180

 

fi=100

 

 

(fi×di)=300

 Let A=25 be the assumed mean. Then we have:Mean, x¯=A+(fi×di)fi= 25+300100=28x¯=28

Page No 861:

Question 18:

Find the mean, using assumed-mean method:

Class 100-120 120-140 140-160 160-180 180-200
Frequency 10 20 30 15 5

Answer:

Class

Frequency fi

Mid valuesxi

Deviation di
di=xi-150

(fi×di)

100-120

10

110

-40

-400

120-140

20

130

-20

-400

140-160

30

150=A

0

0

160-180

15

170

20

300

180-200

5

190

40

200

 

fi=80

 

 

(fi×di)=300

 Let A=150 be the assumed mean. Then we have:Mean, x¯=A+(fi×di)fi= 15030080=1503.75x¯=146.25

Page No 861:

Question 19:

Find the mean, using assumed-mean method:

Class 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 20 35 52 44 38 31

Answer:

Class

Frequency fi

Mid Valuesxi

Deviation di
di=xi-50

(fi×di)

0-20

20

10

-40

-800

20-40

35

30

-20

-700

40-60

52

50=A

0

0

60-80

44

70

20

880

80-100

38

90

40

1520

100-120

31

110

60

1860

 

fi=220

 

 

(fi×di)=2760

 Let A=50 be the assumed mean. Then we have:Now, mean, x¯=A+(fi×di)fi=50+2760220=50+12.55x¯=62.55



Page No 862:

Question 20:

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

Answer:

Let us choose a = 25, h = 10, then di = xi − 25 and uixi-2510.

Using Step-deviation method, the given data is shown as follows:
 

Class Frequency
(fi)
Class mark (xi) di xi − 25 uixi-2510 fiui
0−10 7 5 −20 −2 −14
10−20 10 15 −10 −1 −10
20−30 15 25 0 0 0
30−40 8 35 10 1 8
40−50 10 45 20 2 20
Total ∑ fi = 50       ∑ fiui = 4

The mean of given data is given by

x¯=a+ifiuiifi×h  =25+450×10  =25+45  =125+45  =1295  =25.8

​Thus, the mean is 25.8.

Page No 862:

Question 21:

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

Answer:

Let us choose a = 40, h = 10, then di = xi − 40 and ui = xi-4010.

Using Step-deviation method, the given data is shown as follows:
 

Class Frequency
(fi)
Class mark (xi) di xi − 40 ui = xi-4010 fiui
5−15 6 10 −30 −3 −18
15−25 10 20 −20 −2 −20
25−35 16 30 −10 −1 −16
35−45 15 40 0 0 0
45−55 24 50 10 1 24
55−65 8 60 20 2 16
65−75 7 70 30 3 21
Total ∑ fi = 86       ∑ fiui = 7

The mean of given data is given by

x¯=a+ifiuiifi×h  =40+786×10  =40+7086  =40+0.81  =40.81

​Thus, the mean is 40.81.

Page No 862:

Question 22:

​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.                     [CBSE 2013]

Answer:

​Let us choose a = 202.5, h = 1, then di = xi − 202.5 and ui = xi-202.51.

Using Step-deviation method, the given data is shown as follows:
 

Weight
(in grams)
Number of packets (fi) Class mark (xi) di xi − 202.5 ui = xi-202.51 fiui
200−201 13 200.5 −2 −2 −26
201−202 27 201.5 −1 −1 −27
202−203 18 202.5 0 0 0
203−204 10 203.5 1 1 10
204−205 1 204.5 2 2 2
205−206 1 205.5 3 3 3
Total ∑ fi = 70       ∑ fiui = −38

The mean of given data is given by

x¯=a+ifiuiifi×h  =202.5+-3870×1  =202.5-0.542  =201.96

​Hence, the mean is 201.96 g.

Page No 862:

Question 23:

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

Answer:

Let us choose a = 45, h = 10, then di = xi − 45 and ui = xi-4510.

Using Step-deviation method, the given data is shown as follows:
 

Class Frequency
(fi)
Class mark (xi) di xi − 45 ui = xi-4510 fiui
20−30 25 25 −20 −2 −50
30−40 40 35 −10 −1 −40
40−50 42 45 0 0 0
50−60 33 55 10 1 33
60−70 10 65 20 2 20
Total ∑ fi = 150       ∑ fiui = −37

The mean of given data is given by

x¯=a+ifiuiifi×h  =45-37150×10  =45-3715  =45-2.466  =42.534

​Thus, the mean is 42.534.

Page No 862:

Question 24:

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

Age (in years) 18-24 24-30 30-36 36-42 42-48 48-54
Number of workers 6 8 12 8 4 2

Answer:

Class

Frequency fi

Mid valuesxi

ui=(xiA)h=(xi33)6

(fi×ui)

18-24

6

21

2

12

24-30

8

27

1

8

30-36

12

33 = A

0

0

36-42

8

39

1

8

42-48

4

45

2

8

48-54

2

51

3

6

 

fi=40

 

 

(fi×ui)=2

 Now, A=33, h=6, fi=40 and (fi×ui)=2 Mean, x¯=A+{h×(fi×ui)fi}=33+{6×240}=33+0.3=33.3x=33.3 years

Page No 862:

Question 25:

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

Class 500-520 520-540 540-560 560-580 580-600 600-620
Frequency 14 9 5 4 3 5

Answer:

Class

Frequencyfi

Mid valuesxi

ui=(xiA)h=(xi550)20

(fi×ui)

500-520

14

510

2

28

520-540

9

530

1

9

540-560

5

550 = A

0

0

560-580

4

570

1

4

580-600

3

590

2

6

600-620

5

610

3

15

 

fi=40

 

 

(fi×ui)=12

 Now, A=550, h=20, fi=40 and (fi×ui)=12 Mean, x¯=A+{h×(fi×ui)fi}=550+{20×(12)40}=550-6=544x¯=544

Page No 862:

Question 26:

Find the mean age from the following frequency distribution:

Age (in years) 25-29 30-34 35-39 40-44 45-49 50-54 55-59
No. of person 4 14 22 16 6 5 3

Answer:

Converting the series into exclusive form, we get:


Class

Frequencyfi

Mid valuesxi

ui=(xiA)h=(xi42)5

(fi×ui)

24.5-29.5

4

27

3

12

29.5-34.5

14

32

2

28

34.5-39.5

22

37

1

22

39.5-44.5

16

42 = A

0

0

44.5-49.5

6

47

1

6

49.5-54.5

5

52

2

10

54.5-59.5

3

57

3

9

 

fi=70

 

 

(fi×ui)=37

 Now, A=42, h=5, fi=70 and (fi×ui)=37 Mean, x¯=A+{h×(fi×ui)fi}=42+{5×(37)70}=42-2.64=39.36x¯=39.36Mean age=39.36 years



Page No 863:

Question 27:

The following table shows the age distribution of patients of malaria in a village during a particular month.

Age (in years) 5-14 15-24 25-34 35-44 45-54 55-64
No. of cases 6 11 21 23 14 5

Find the average age of the patients.

Answer:

Converting the series into exclusive form, we get:
 

Class

Frequency fi

Mid valuesxi

ui=(xiA)h=(xi29.5)10

(fi×ui)

4.5-14.5

6

9.5

2

12

14.5-24.5

11

19.5

1

11

24.5-34.5

21

29.5 = A

0

0

34.5-44.5

23

39.5

1

23

44.5-54.5

14

49.5

2

28

54.5-64.5

5

59.5

3

15

 

fi=80

 

 

(fi×ui)=43

 Now, A=29.5, h=10,fi=80 and (fi×ui)=43 Mean, x¯=A+{h×(fi×ui)fi}=29.5+{10×4380}=29.5+5.375=34.875x¯=34.875The average age of the patients is 34.87 years.

Page No 863:

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.

Answer:

Let us choose a = 92, h = 5, then di = xi − 92 and ui = xi-925.

Using Step-deviation method, the given data is shown as follows:
 

Weight
(in grams)
Number of eggs (fi) Class mark (xi) di xi − 92 ui = xi-925 fiui
74.5−79.5 4 77 −15 −3 −12
79.5−84.5 9 82 −10 −2 −18
84.5−89.5 13 87 −5 −1 −13
89.5−94.5 17 92 0 0 0
94.5−99.5 12 97 5 1 12
99.5−104.5 3 102 10 2 6
104.5−109.5 2 107 15 3 6
Total ∑ fi = 60       ∑ fiui = −19

The mean of given data is given by

x¯=a+ifiuiifi×h  =92+-1960×5  =92-1.58  =90.42  90

​Thus, the mean weight to the nearest gram is 90 g.

Page No 863:

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.

Answer:

Let us choose a = 17.5, h = 5, then di = xi − 17.5 and ui = xi-17.55.

Using Step-deviation method, the given data is shown as follows:
 

Marks Number of students (cf) Frequency (fi) Class mark (xi) di xi − 17.5 ui = xi-17.55 fiui
0−5 3 3 2.5 −15 −3 −9
5−10 10 7 7.5 −10 −2 −14
10−15 25 15 12.5 −5 −1 −15
15−20 49 24 17.5 0 0 0
20−25 65 16 22.5 5 1 16
25−30 73 8 27.5 10 2 16
30−35 78 5 32.5 15 3 15
35−40 80 2 37.5 20 4 8
Total   ∑ fi = 80       ∑ fiui = 17

The mean of given data is given by

x¯=a+ifiuiifi×h  =17.5+1780×5  =17.5+1.06  =18.56

​Thus, the mean marks correct to 2 decimal places is 18.56.



Page No 870:

Question 1:

In a hospital, the ages of diabetic patients were recorded as follows. Find the median age.    [CBSE 2014]
 

Age (in years) 0−15 15−30 30−45 45−60 60−75
Number of patients 5 20 40 50 25 

Answer:

We prepare the cumulative frequency table, as shown below:
 

Age (in years) Number of patients (fi) Cumulative Frequency (cf)
0−15 5 5
15−30 20 25
30−45 40 65
45−60 50 115
60−75 25 140
Total N∑ fi = 140  

Now, N = 140 N2=70.

The cumulative frequency just greater than 70 is 115 and the corresponding class is 45−60.

Thus, the median class is 45−60.

∴ l = 45, h = 15, f = 50, N = 140 and cf = 65.

Now,

Median=l+N2-cff×h          =45+1402-6550×15          =45+70-6550×15          =45+1.5          =46.5 

Hence, the median age is 46.5 years.

Page No 870:

Question 2:

Compute the median from the following data:

Marks 0-7 7-14 14-21 21-28 28-35 35-42 42-49
Number of students 3 4 7 11 0 16 9

Answer:

Class

Frequency (f)

Cumulative

frequency

0-7

3

3

7-14

4

7

14-21

7

14

21-28

11

25

28-35

0

25

35-42

16

41

42-49

9

50

 

N=f=50

 

 N=50N2=25The cumulative frequency just greater than 25 is 41 and the corresponding class is 35-42.Thus, the median class is 35-42.l=35, h=7, f=16, cf=c.f. of preceding class = 25 and N2=25Median=l+N2-c.ff×h=35+7×(2525)16=35+0=35

Page No 870:

Question 3:

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

Daily wages (in Rs) 0-100 100-200 200-300 300-400 400-500
Number of workers 40 32 48 22 8

Find the median daily wage income of the workers.

Answer:

Class

Frequency(f)

Cumulative

frequency

0-100

 

40

40

100-200

32

72

200-300

48

120

300-400

22

142

400-500

8

150

 

N=f=150

 

 N=150N2=75The cumulative frequency just greater than 75 is 120 and the corresponding class is 200-300.Thus, the median class is 200-300.l=200, h=100, f=48, cf=c.f. of preceding class=72 and N2=75Median, M=l+h×N2cff=200+100×(7572)48=200+6.25=206.25Hence, the median daily wage income of the workers is Rs 206.25.

Page No 870:

Question 4:

Calculate the median from the following frequency distribution:

Class 5-10 10-15 15-20 20-25 20-30 30-35 35-40 40-45
Frequency 5 6 15 10 5 4 2 2

Answer:

Class

Frequency(f)

Cumulative

frequency

5-10

5

5

10-15

6

11

15-20

15

26

20-25

10

36

25-30

5

41

30-35

4

45

35-40

2

47

40-45

2

49

 

N=f=49

 

 N=49N2=24.5The cumulative frequency just greater than 24.5 is 26 and the corresponding class is 15-20.Thus, the median class is 15-20.Now, l=15, h=5, f=15, cf=c.f. of preceding class = 11 and N2=24.5Median, M=l+h×N2cff=15+5×(24.511)15=15+4.5=19.5 Hence, median=19.5

Page No 870:

Question 5:

Given below is the number of units of electricity consumed in a week in a certain locality:

Consumption
(in units)
65-85 85-105 105-125 125-145 145-165 165-185 195-205
Number of
consumers
4 5 13 20 14 7 4

Calculate the median

Answer:

Class

Frequency(f)

Cumulative

frequency

65-85

4

4

85-105

5

9

105-125

13

22

125-145

20

42

145-165

14

56

165-185

7

63

185-205

4

67

 

N=f=67

 

 N=67N2=33.5The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125-145.Thus, the median class is 125-145.Now, l=125, h=20, f=20, cf=c.f. of preceding class = 22 and N2=33.5Median, M=l+h×N2cff=125+20×(33.522)20=125+11.5=136.5Hence, median=136.5

Page No 870:

Question 6:

Calculate the median from the following data:

Height
(in cm)
135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175
No. of
boys
6 10 18 22 20 15 6 3

Answer:

Class

Frequency(f)

Cumulative

frequency

135-140

6

6

140=145

10

16

145-150

18

34

150-155

22

56

155-160

20

76

160-165

15

91

165-170

6

97

170-175

3

100

 

N=f=100

 

 N=100N2=50The cumulative frequency just greater than 50 is 56 and the corresponding class is 150-155.Thus, the median class is 150-155.Now, l=150, h=5, f=22, cf=c.f. of preceding class=34 and N2=50Median, M=l+h×N2cff=150+5×(5034)22=150+3.64=153.64Hence, median=153.64

Page No 870:

Question 7:

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

Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 25 ? 18 7

Answer:

Class Frequency (fi) c.f
0-10 5 5
10-20 25 30
20-30 x x+30
30-40 18 x+48
40-50 7 x+55

Median is 24 which lies in 20-30Median Class=20-30Let the unknown frequency be xHere, l=20,n2=x+552, c.f of the preceding class=c.f=30,f=x, h=10Median=l+n2-c.ff×h24=20+x+552-30x×1024=20+x+55-602x×1024=20+x-52x×1024=20+5x-25x24=20x+5x-25x24x=25x-25-x=-25x=25Hence, the unknown frequency is 25



Page No 871:

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
Frequency 12 a 12 15 b 6 6 4

Answer:

We prepare the cumulative frequency table, as shown below:
 

Class Frequency (fi) Cumulative frequency (cf)
0−5 12 12
5−10 a 12 + a
10−15 12 24 + a
15−20 15 39 + a
20−25 b 39 + a + b
25−30 6 45 + a + b
30−35 6 51 + a + b
35−40 4 55 + a + b
Total N = ∑fi = 70  
 
Let a and b be the missing frequencies of class intervals 5−10 and 20−25 respectively. Then,

55 a + = 70 ⇒ a + = 15     ...(1)

Median is 16, which lies in 15−20. So, the median class is 15−20.

l = 15, h = 5, N = 70, f = 15 and cf = 24 + a


Now,

Median=l+N2-cff×h16=15+702-24+a15×516=15+35-24-a316=15+11-a316-15=11-a31×3=11-aa=11-3a=8

b = 15 a    [From (1)]
b = 15 − 8
b = 7

Hence, a = 8 and b = 7.

Page No 871:

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

Answer:

We prepare the cumulative frequency table, as shown below:
 

Runs scored Number of batsmen (fi) Cumulative frequency (cf)
2500−3500 5 5
3500−4500 x 5 + x
4500−5500 y 5 + y
5500−6500 12 17 y
6500−7500 6 23 y
7500−8500 2 25 y
Total N = ∑fi = 60  
 
Let x and y be the missing frequencies of class intervals 3500−4500 and 4500−5500 respectively. Then,

25 x + = 60 ⇒ x + = 35     ...(1)

Median is 5000, which lies in 4500−5500. So, the median class is 4500−5500.

∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x


Now,

Median=l+N2-cff×h5000=4500+602-5+xy×10005000-4500=30-5-xy×1000500=25-xy×1000y=50-2x35-x=50-2x     From 12x-x=50-35x=15

∴ y = 35 − x    [From (1)]
⇒ y = 35 − 15
⇒ y = 20

Hence, x = 15 and y = 20.

Page No 871:

Question 10:

If the median of the following frequency distribution is 32.5, find the values of f1 and f2.

Class
interval
0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total
Frequency f1 5 9 12 f2 3 2 40

Answer:

Class

Frequency(f)

Cumulative

frequency

0-10

f1

f1

10-20

5

f1+5

20-30

9

f1+14

30-40

12

f1+26

40-50

f2

f1+f2+26

50-60

3

f1+f2+29

60-70

2

f1+f2+31

 

N=f=40

 

 Now, f1+f2+31=40 f1+f2=9 f2=9 f1       ...(i)The median is 32.5 which lies in 30-40.Hence, median class=3040Here, l=30, N2=402=20, f=12 and c.f=14+f1Now, median=32.5l+N2c.ff×h=32.530+20(14+f1)12×10=32.56f112×10=2.56010f112=2.56010f1=3010f1=30f1=3From equation (i), we have:f2=93f2=6

Page No 871:

Question 11:

Calculate the median for the following data:

Age (in years) 19-25 26-32 33-39 40-46 47-53 54-60
Frequency 35 96 68 102 35 4

Answer:

First, we will convert the data into exclusive form.

Class

Frequency(f)

Cumulative

frequency

18.5-25.5

35

35

25.5-32.5

96

131

32.5-39.5

68

199

39.5-46.5

102

301

46.5-53.5

35

336

53.5-60.5

4

340

 

N=f=340

 

 N=340=>N2=170The cumulative frequency just greater than 170 is 199 and the corresponding class is 32.5-39.5.Thus, the median class is 32.5-39.5.l=32.5, h=7, f=68, cf=c.f. of preceding class = 131 and N2=170Median, M=l+h×N2cff= 32.5+7×(170131)68= 32.5+4.01= 36.51Hence, median = 36.51

Page No 871:

Question 12:

Find the median wages for the following frequencies distribution:

Wages per day
(in Rs)
61-70 71-80 81-90 91-100 101-110 111-120
No. of women
workers
5 15 20 30 20 8

Answer:

Converting the given data into exclusive form, we get:

Class

Frequency(f)

Cumulative

frequency

60.5-70.5

5

5

70.5-80.5

15

20

80.5-90.5

20

40

90.5-100.5

30

70

100.5-110.5

20

90

110.5-120.5

8

98

 

N=f=98

 

 N=98N2=49The cumulative frequency just greater than 49 is 70 and the corresponding class is 90.5-100.5.Thus, the median class is 90.5-100.5.Now, l=90.5, h=10, f=30, cf=c.f. of preceding class = 40 and N2=49Median, M=l+{h×(N2cf)f}=90.5+{10×(4940)30}=90.5+3=93.5Hence, median wages = Rs 93.50

Page No 871:

Question 13:

Find the median from the following data:

Class 1-5 6-10 11-15 16-20 21-25 26-30 31-35 35-40 41-45
Frequency 7 10 16 32 24 16 11 5 2

Answer:

Converting into exclusive form, we get:

Class

Frequency(f)

Cumulative

frequency

0.5-5.5

7

7

5.5-10.5

10

17

10.5-15.5

16

33

15.5-20.5

32

65

20.5-25.5

24

89

25.5-30.5

16

105

30.5-35.5

11

116

35.5-40.5

5

121

40.5-45.5

2

123

 

N=f=123

 

 N=123N2=61.5The cumulative frequency just greater than 61.5 is 65 and the corresponding class is 15.5-20.5.Thus, the median class is 15.5-20.5.l=15.5, h=5, f=32, cf=c.f. of preceding class = 33 and N2=61.5Median, M=l+h×N2cff=15.5+5×(61.533)32=15.5+4.45=19.95Hence, median=19.95

Page No 871:

Question 14:

Find the median from the following data:

Marks No. of students
Below 10 12
Below 20 32
Below 30 57
Below 40 80
Below 50 92
Below 60 116
Below 70 164
Below 80 200

Answer:

Class

Cumulative frequency

Frequency (f)

0-10

12

12

10-20

32

20

20-30

57

25

30-40

80

23

40-50

92

12

50-60

116

24

60-70

164

48

70-80

200

36

 

 

N=f=200
 

 N=200N2=100The 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, cf=c.f. of preceding class = 92 and N2=100Median, M=l+h×N2cff= 50 + 10×(10092)24= 50 + 3.33= 53.33Hence, median = 53.33



Page No 877:

Question 1:

Find the mode of the following frequency distribution:         [CBSE 2014]
 

Marks 10−20 20−30 30−40 40−50 50−60
Frequency 12 35 45 25 13

Answer:

Here the maximum class frequency is 45, and the class corresponding to this frequency is 30−40. So, the modal class is 30−40.

Now, 

modal class = 30−40, lower limit (l) of modal class = 30, class size (h) = 10,

frequency (f1) of the modal class = 45,

frequency (f0) of class preceding the modal class = 35,

frequency (
f2) of class succeeding the modal class = 25

Now, let us substitute these values in the formula:

Mode=l+f1-f02f1-f0-f2×h        =30+45-3590-35-25×10        =30+1030×10        =30+3.33        =33.33

Hence, the mode is 33.33.

Page No 877:

Question 2:

Compute the mode of the following data:              [CBSE 2013]

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

Answer:

Here the maximum class frequency is 28, and the class corresponding to this frequency is 40−60. So, the modal class is 40−60.

Now,

Modal class = 40−60, lower limit (l) of modal class = 40, class size (h) = 20,

frequency (f1) of the modal class = 28,

frequency (f0) of class preceding the modal class = 16,

frequency (f2
) of class succeeding the modal class = 20.

Now, let us substitute these values in the formula:


Mode=l+f1-f02f1-f0-f2×h        =40+28-1656-16-20×20        =40+1220×20        =40+12        =52

Hence, the mode is 52.



Page No 878:

Question 3:

Heights of students of Class X are given in the foloowing 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.    [CBSE 2014]

Answer:

Here the maximum class frequency is 20, and the class corresponding to this frequency is 160−165. So, the modal class is 160−165.

Now,

Modal class = 
160−165, lower limit (l) of modal class = 160, class size (h) = 5,

frequency (f1) of the modal class = 20,

frequency (f0) of class preceding the modal class = 8,

frequency (f2) of class succeeding the modal class = 12.

Now, let us substitute these values in the formula:


Mode=l+f1-f02f1-f0-f2×h        =160+20-840-8-12×5        =160+1220×5        =160+3        =163

Hence, the mode is 163.

It represents that the height of maximum number of students is 163 cm.

Now, to find the mean let us put the data in the table given below:
 
Height (in cm) Number of students (fi) Class mark (xi) fixi
150−155 15 152.5 2287.5
155−160 8 157.5 1260
160−165 20 162.5 3250
165−170 12 167.5 2010
170−175 5 172.5 862.5
Total fi = 60   fixi =  9670

Mean=ifixiifi        =967060        =161.17

Thus, mean of the given data is 161.17.

It represents that on an average, the height of a student is 161.17 cm. 

Page No 878:

Question 4:

Find the mode of the following distribution:

Class interval 10-14 14-18 18-22 22-26 26-30 30-34 34-38 38-42
Frequency 8 6 11 20 25 22 10 4

Answer:

As the class 26-30 has the maximum frequency, it is the modal class.

Now, xk=26,h=4,fk=25,fk-1=20 ,fk+1=22 Mode, M0=xk+h×fk-fk-12fk-fk-1-fk+1=26+4×25-202×25-20-22=26+4×58=26+2.5=28.5

Page No 878:

Question 5:

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

Expenditure (in Rs) No. of manual workers
1000-1500 24
1500-2000 40
2000-2500 31
2500-3000 28
3000-3500 32
3500-4000 23
4000-4500 17
4500-5000 5
Find the expenditure done by maximum number of manual workers.

Answer:

As the class 1500-2000 has the maximum frequency, it is the modal class.
Now, xk=1500, h=500, fk=40, fk-1=24 and fk+1=31 Mode, M0=xk+h×fk-fk-12fk-fk-1-fk+1=1500+500×40-242×40-24-31=1500+500×1625=1500+320=1820

Hence, mode = Rs 1820

Page No 878:

Question 6:

Calculate the mode from the following data:

Monthly salary (in Rs) No. of employees
0-5000 90
5000-1000 150
10000-15000 100
15000-20000 80
20000-25000 70
25000-30000 10

Answer:

As the class 5000-10000 has the maximum frequency, it is the modal class.

Now, xk=5000, h=5000, fk=150, fk-1=90 and fk+1=100 Mode, M0=xk+h×fk-fk-12fk-fk-1-fk+1=5000+5000×150-902×150-90-100=5000+5000×60110=5000+2727.27=7727.27


Hence, mode = Rs 7727.27



Page No 879:

Question 7:

Compute the mode from the following data:

Age (in years) 0-5 5-10 10-15 15-20 20-25 25-30 30-35
Number of patients 6 11 18 24 17 13 5

Answer:

As the class 15-20 has the maximum frequency, it is the modal class.
Now, xk=15, h=5, fk=24, fk-1=18 and fk+1=17 Mode, M0=xk+h×fk-fk-12fk-fk-1-fk+1=15+5×24-182×24-18-17=15+5×613=15+2.3=17.3

Hence, mode=17.3 years

Page No 879:

Question 8:

Compute the mode from the following series:

Size 45-55 55-65 65-75 75-85 85-95 95-105 105-115
Frequency 7 12 17 30 32 6 10

Answer:

As the class 85-95 has the maximum frequency, it is the modal class.

Now, xk=85, h=10, fk=32, fk-1=30 and fk+1=6 Mode, M0=xk+h×fk-fk-12fk-fk-1-fk+1=85+10×32-302×32-30-6=85+10×228=85+.71=85.71


Hence, mode=85.71

Page No 879:

Question 9:

Compute the mode from the following data:

Class
interval
1-5 6-10 11-15 16-20 21-25 26-30 31-35 36-40 41-45 46-50
Frequency 3 8 13 18 28 20 13 8 6 4

Answer:

Clearly, we have to find the mode of the data. The given data is an inclusive series. So, we will convert it to an exclusive form as given below:
 

Class interval 0.5-5.5 5.5-10.5 10.5-15.5 15.5-20.5 20.5-25.5 25.5-30.5 30.5-35.5 35.5-40.5 40.5-45.5 45.5-50.5
Frequency 3 8 13 18 28 20 13 8 6 4

As the class 20.5-25.5 has the maximum frequency, it is the modal class.

Now, xk=20.5, h=