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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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=5, fk=28, fk-1=18 and fk+1=20 Mode, M0=xk+h×fk-fk-12fk-fk-1-fk+1=20.5+5×28-182×28-18-20=20.5+5×1018=20.5+2.78=23.28

Hence, mode=23.28

Page No 879:

Answer:

It is given that the sum of frequencies is 181.

x + 15 + 18 + 30 + 50 + 48 + x = 181
⇒ 2x + 161 = 181
⇒ 2x = 181 − 161
⇒ 2x = 20
x = 10

Thus, x = 10.

Here the maximum class frequency is 50, and the class corresponding to this frequency is 13−15. So, the modal class is 13−15.

Now,

Modal class = 
13−15, lower limit (l) of modal class = 13, class size (h) = 2,

frequency (f1) of the modal class = 50,

frequency (f0) of class preceding the modal class = 30,

frequency (f2) of class succeeding the modal class = 48.

Now, let us substitute these values in the formula:


Mode=l+f1-f02f1-f0-f2×h        =13+50-30100-30-48×2        =13+2022×2        =13+1.82        =14.82

Hence, the mode is 14.82. 



Page No 881:

Answer:

To find the mean let us put the data in the table given below:
 

Class Frequency (fi) Class mark (xi) fixi
0−10 4 5 20
10−20 4 15 60
20−30 7 25 175
30−40 10 35 350
40−50 12 45 540
50−60 8 55 440
60−70 5 65 325
Total fi = 50   fixi =  1910

Mean=ifixiifi        =191050        =38.2

Thus, mean of the given data is 38.2.

Now, to find the median let us put the data in the table given below:
 
Class Frequency (fi) Cumulative frequency (cf)
0−10 4 4
10−20 4 8
20−30 7 15
30−40 10 25
40−50 12 37
50−60 8 45
60−70 5 50
Total N = ∑fi = 50  

Now, N = 50 N2=25.

The cumulative frequency just greater than 25 is 37, and the corresponding class is 40−50.

Thus, the median class is 40−50.

l = 40, h = 10, N = 50, f = 12 and cf = 25.


Now, 

Median=l+N2-cff×h           =40+25-2512×10           =40

Thus, the median is 40.

We know that,
Mode = 3(median) − 2(mean)
          = 3 × 40 − 2 × 38.2
          = 120 − 76.4
          = 43.6

Hence, Mean = 38.2, Median = 40 and Mode = 43.6

Page No 881:

Answer:

To find the mean let us put the data in the table given below:
 

Class Frequency (fi) Class mark (xi) fixi
0−20 6 10 60
20−40 8 30 240
40−60 10 50 500
60−80 12 70 840
80−100 6 90 540
100−120 5 110 550
120−140 3 130 390
Total fi = 50   fixi =  3120

Mean=ifixiifi        =312050        =62.4

Thus, mean of the given data is 62.4.

Now, to find the median let us put the data in the table given below:
 
Class Frequency (fi) Cumulative frequency (cf)
0−20 6 6
20−40 8 14
40−60 10 24
60−80 12 36
80−100 6 42
100−120 5 47
120−140 3 50
Total N = ∑fi = 50  

Now, N = 50 N2=25.

The cumulative frequency just greater than 25 is 36, and the corresponding class is 60−80.

Thus, the median class is 60−80.

∴ l = 60, h = 20, N = 50, f = 12 and cf = 24.


Now, 

Median=l+N2-cff×h           =60+25-2412×20           =60+1.67           =61.67

Thus, the median is 61.67.

We know that,
Mode = 3(median) − 2(mean)
          = 3 × 61.67 − 2 × 62.4
          = 185.01 − 124.8
          = 60.21

Hence, Mean = 62.4, Median = 61.67 and Mode = 60.21.



Page No 882:

Answer:

To find the mean let us put the data in the table given below:
 

Class Frequency (fi) Class mark (xi) fixi
0−50 2 25 50
50−100 3 75 225
100−150 5 125 625
150−200 6 175 1050
200−250 5 225 1125
250−300 3 275 825
300−350 1 325 325
Total fi = 25   fixi =  4225

Mean=ifixiifi        =422525        =169

Thus, mean of the given data is 169.

Now, to find the median let us put the data in the table given below:
 
Class Frequency (fi) Cumulative frequency (cf)
0−50 2 2
50−100 3 5
100−150 5 10
150−200 6 16
200−250 5 21
250−300 3 24
300−350 1 25
Total N = ∑fi = 25  

Now, N = 25 N2=12.5.

The cumulative frequency just greater than 12.5 is 16, and the corresponding class is 150−200.

Thus, the median class is 150−200.

∴ l = 150, h = 50, N = 25, f = 6 and cf = 10.


Now, 

Median=l+N2-cff×h           =150+12.5-106×50           =150+20.83           =170.83

Thus, the median is 170.83.

We know that,
Mode = 3(median) − 2(mean)
          = 3 × 170.83 − 2 × 169
          = 512.49 − 338
          = 174.49

Hence, Mean = 169, Median = 170.83 and Mode = 174.49

Page No 882:

Answer:

To find the mean let us put the data in the table given below:
 

Marks obtained Number of students (fi) Class mark (xi) fixi
25−35 7 30 210
35−45 31 40 1240
45−55 33 50 1650
55−65 17 60 1020
65−75 11 70 770
75−85 1 80 80
Total fi = 100   fixi =  4970

Mean=ifixiifi        =4970100        =49.7

Thus, mean of the given data is 49.7.

Now, to find the median let us put the data in the table given below:
 
Class Frequency (fi) Cumulative frequency (cf)
25−35 7 7
35−45 31 38
45−55 33 71
55−65 17 88
65−75 11 99
75−85 1 100
Total N = ∑fi = 100  

Now, N = 100 N2=50.

The cumulative frequency just greater than 50 is 71, and the corresponding class is 45−55.

Thus, the median class is 45−55.

∴ l = 45, h = 10, N = 100, f = 33 and cf = 38.


Now, 

Median=l+N2-cff×h           =45+50-3833×10           =45+3.64           =48.64

Thus, the median is 48.64.

We know that,
Mode = 3(median) − 2(mean)
          = 3 × 48.64 − 2 × 49.70
          = 145.92 − 99.4
          = 46.52

Hence, Mean = 49.70, Median = 48.64 and Mode = 46.52.

Page No 882:

Answer:

We have the following:

Height in cm Mid value
xi
Frequency
fi
Cumulative frequency fi×xi
120-130 125 2 2 250
130-140 135 8 10 1080
140-150 145 12 22 1740
150-160 155 20 42 3100
160-170 165 8 50 1320
    fi=50   fi×xi=7490
Mean, x¯=fi×xifi
=749050

=149.8

Now, N=50N2=25
The cumulative frequency just greater than 25 is 42 and the corresponding class is 150-160.
Thus, the median class is 150-160.
Now, l=150, h=10, f=20, c=cf of preceding class = 22 and N2=25
Median, Me=l+h×N2-cf=150+10×25-2220=150+10×320  

              =151.5
∴ Mode = 3(Median) - 2(Mean)
= 3×151.5-2×149.8
= 154.9

Page No 882:

Answer:

We have the following:

Daily income Mid value
xi
Frequency
fi
Cumulative frequency fi×xi
100-120 110 12 12 1320
120-140 130 14 26 1820
140-160 150 8 34 1200
160-180 170 6 40 1020
180-200 190 10 50 1900
    fi=50   fi×xi=7260

Mean, x¯=fi×xifi
=726050

=145.2

 Here, N=50N2=25
The cumulative frequency just greater than 25 is 26 and the corresponding class is 120-140.
Thus, the median class is 120-140.
Now, l=120, h=20, f=14, c=cf of preceding class = 12 andN2=25
Median, Me=l+h×N2-cf
=120+20×25-1214=120+20×1314

 =138.57
Mode = 3(Median) - 2(mean)
=3×138.57-2×145.2=125.31

Page No 882:

Answer:

We have the following:

Daily income Mid value Frequencyfi Cumulative frequency fi×xi
100-150 125 6 6 750
150-200 175 7 13 1225
200-250 225 12 25 2700
250-300 275 3 28 825
300-350 325 2 30 650
    fi=30   fi×xi=6150

Mean, x¯=fi×xifi
=615030

=205

Now, N=30N2=15

The cumulative frequency just greater than 15 is 25 and the corresponding class is 200-250.
Thus, the median class is 200-250.
Now, l=200, h=50, f=12, c=cf of preceding class = 13 and N2=15
Median, Me=l+h×N2-cf
=200+50×15-1312=200+50×212
=200 + 8.33
=208.33



Page No 892:

Answer:

The frequency distribution table of less than type is given as follows:
 

Marks (upper class limits) Cumulative frequency (cf)
Less than 10 5
Less than 20 5 + 3 = 8
Less than 30 8 + 4 = 12
Less than 40 12 + 3 = 15
Less than 50 15 + 3 = 18
Less than 60 18 + 4 = 22
Less than 70 22 + 7 = 29
Less than 80 29 + 9 = 38
Less than 90 38 + 7 = 45
Less than 100 45 + 8 = 53

Taking upper class limits of class intervals on x−axis and their respective frequencies on y−axis, its ogive can be drawn as follows:



Here, N = 53 N2=26.5.

Mark the point A whose ordinate is 26.5 and its x−coordinate is 66.4.



Thus, median of the data is 66.4.

Page No 892:

Answer:

Taking upper class limits of class intervals on x−axis and their respective frequencies on y−axis, its ogive can be drawn as follows:



Here, N = 80 N2=40.

Mark the point A whose ordinate is 40 and its x−coordinate is 76.



Thus, median of the data is 76.



Page No 893:

Answer:

The frequency distribution table of more than type is as follows:
 

Marks (lower class limits) Cumulative frequency (cf)
More  than 0 96 + 4 = 100
More  than 10 90 + 6 = 96
More  than 20 80 + 10 = 90
More  than 30  70 + 10 = 80
More  than 40 45 + 25 = 70
More  than 50 23 + 22 = 45
More  than 60 18 + 5 = 23
More  than 70 5

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows:

Page No 893:

Answer:

The frequency distribution table of more than type is as follows:
 

Height (in cm) (lower class limits) Cumulative frequency (cf)
More  than 135 5 + 45 = 50
More  than 140 8 + 37 = 45
More  than 145 9 + 28 = 37
More  than 150  12 + 16 = 28
More  than 155 14 + 2 = 16
More  than 160 2

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows:

Page No 893:

Answer:

The frequency distribution table of more than type is as follows:
 

Height (in cm) (lower class limits) Cumulative frequency (cf)
More  than 140 3 + 153 = 156
More  than 160 8 + 145 = 153
More  than 180 15 + 130 = 145
More  than 200  40 + 90 = 130
More  than 220 50 + 40 = 90
More  than 240 30 + 10 = 40
More  than 260 10

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows:

Page No 893:

Answer:

The frequency distribution table of more than type is as follows:
 

Production yield (kg/ha)
(lower class limits)
Cumulative frequency (cf)
More than 50 2 + 98 = 100
More than 55 8 + 90 = 98
More than 60 12 + 78 = 90
More than 65 24 + 54 = 78
More than 70 38 + 16 = 54
More than 75 16

Taking the lower class limits on x−axis and their respective cumulative frequencies on y−axis, its ogive can be obtained as follows:


Here, N = 100 N2=50.

Mark the point A whose ordinate is 50 and its