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 10 Maths are prepared by experts and are 100% accurate.

Page No 859:

Question 1:

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:

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.

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:


​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.

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:


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.

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:


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.

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:


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.

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:


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.

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:


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.

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:


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.

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:

 

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.

Answer:

The given data is shown as follows:
 

Daily pocket allowance (in â‚¹) 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:

The given data is shown as follows:
 

Daily pocket allowance (in â‚¹) 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.

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 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.

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 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.

Answer:

The given data is shown as follows:
 

Expenditure
(in â‚¹)
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 given data is shown as follows:
 

Expenditure
(in â‚¹)
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.

Answer:

Class

Frequency

fi

Mid values xi

(fi×xi)

0-20

7

10

70

20-40

f

1

30

30 f

1

40-60

12

50

600

60-80

18- f

1

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:

Class

Frequency

fi

Mid values xi

(fi×xi)

0-20

7

10

70

20-40

f

1

30

30 f

1

40-60

12

50

600

60-80

18- f

1

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.

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:


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.

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:

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%.

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, xŻ=A+(fi×di)fi= 25+300100=28xŻ=28

Page No 861:

Question 18:

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, xŻ=A+(fi×di)fi= 25+300100=28xŻ=28

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, xŻ=A+(fi×di)fi= 15030080=1503.75xŻ=146.25

Page No 861:

Question 19:

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, xŻ=A+(fi×di)fi= 15030080=1503.75xŻ=146.25

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, xŻ=A+(fi×di)fi=50+2760220=50+12.55xŻ=62.55



Page No 862:

Question 20:

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, xŻ=A+(fi×di)fi=50+2760220=50+12.55xŻ=62.55

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:

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.

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:

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.

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:

​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.

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:

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.

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:

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

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:

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

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:

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

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:

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.

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:

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.

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:

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.

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:

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.

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:

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

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:

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.

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:

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

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:

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

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:

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

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:

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

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:

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.

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:

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.

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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.

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:

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.

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:

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. 

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:

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

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:

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

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:

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

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:

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

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:

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

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:

Question 10:

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

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:

Question 1:

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. 

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:

Question 2:

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

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:

Question 3:

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.

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:

Question 4:

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

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:

Question 5:

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.

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:

Question 6:

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

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:

Question 7:

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

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:

Question 1:

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

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:

Question 2:

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.

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:

Question 3:

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.

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:

Question 4:

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:

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:

Question 5:

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:

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:

Question 6:

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:

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 x−coordinate is 70.5.


Thus, median of the data is 70.5.

Page No 893:

Question 7:

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 x−coordinate is 70.5.


Thus, median of the data is 70.5.

Answer:

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

Weekly expenditure (in â‚¹)
(upper class limits)
Cumulative frequency (cf)
Less than 200 5
Less than 300 5 + 6 = 11
Less than 400 11 + 11 = 22
Less than 500 22 + 13 = 35
Less than 600 35 + 5 = 40
Less than 700 40 + 4 = 44
Less than 800 44 + 3 = 47
Less than 900 47 + 2 = 49

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



Now,

The frequency distribution table of more than type is as follows:
 
Weekly expenditure (in â‚¹)
(lower class limits)
Cumulative frequency (cf)
More than 100 44 + 5 = 49
More than 200 38 + 6 = 44
More than 300 27 + 11 = 38
More than 400 14 + 13 = 27
More than 500 9 + 5 = 14
More than 600 5 + 4 = 9
More than 700 2 + 3 = 5
More than 800 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 894:

Question 8:

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

Weekly expenditure (in â‚¹)
(upper class limits)
Cumulative frequency (cf)
Less than 200 5
Less than 300 5 + 6 = 11
Less than 400 11 + 11 = 22
Less than 500 22 + 13 = 35
Less than 600 35 + 5 = 40
Less than 700 40 + 4 = 44
Less than 800 44 + 3 = 47
Less than 900 47 + 2 = 49

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



Now,

The frequency distribution table of more than type is as follows:
 
Weekly expenditure (in â‚¹)
(lower class limits)
Cumulative frequency (cf)
More than 100 44 + 5 = 49
More than 200 38 + 6 = 44
More than 300 27 + 11 = 38
More than 400 14 + 13 = 27
More than 500 9 + 5 = 14
More than 600 5 + 4 = 9
More than 700 2 + 3 = 5
More than 800 2

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

Answer:

From the given table, we may prepare the 'more than' frequency table as shown below:

Score

Number of candidates

More than 750

34

More than 700

52

More than 650

79

More than 600

103

More than 550

135

More than 500

175

More than 450

210

More than 400

230

 

We plot the points A(750,34), B(700,52), C(650,79), D(600,103), E(550,135), F(500,175), G(450,210) and H(400,230).
 Join AB, BC, CD, DE, EF, FG, GH and HA with a free hand to get the curve representing the ‘more than type’ series.


Here, N=230
N2=115
From P(0,115), draw PQ meeting the curve at Q. Draw QM meeting at M.
Clearly, OM = 590 units
Hence, median = 590 units
 

Page No 894:

Question 9:

From the given table, we may prepare the 'more than' frequency table as shown below:

Score

Number of candidates

More than 750

34

More than 700

52

More than 650

79

More than 600

103

More than 550

135

More than 500

175

More than 450

210

More than 400

230

 

We plot the points A(750,34), B(700,52), C(650,79), D(600,103), E(550,135), F(500,175), G(450,210) and H(400,230).
 Join AB, BC, CD, DE, EF, FG, GH and HA with a free hand to get the curve representing the ‘more than type’ series.


Here, N=230
N2=115
From P(0,115), draw PQ meeting the curve at Q. Draw QM meeting at M.
Clearly, OM = 590 units
Hence, median = 590 units
 

Answer:

(i) From the given table, we may prepare the 'less than' frequency table as shown below:

Marks

No. of students

Less than 5

2

Less than 10

7

Less than 15

13

Less than 20

21

Less than 25

31

Less than 30

56

Less than 35

76

Less than 40

94

Less than 45

98

Less than 50

100

 

We plot the points A(5,2), B(10,7), C(15,13), D(20,21), E(25,31), F(30,56), G(35,76), H(40,94), I(45,98) and J(50,100).
 Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve representing the ‘less than type’ series.

(ii) More than series:

Marks

No. of student

More than 0

100

More than 5

98

More than 10

93

More than 15

87

More than 20

79

More than 25

69

More than 30

44

More than 35

24

More than 40

6

More than 45

2

Now, on the same graph paper, we plot the points (0,100), (5,98), (10,94), (15,76), (20,56), (25,31), (30,21), (35,13), (40,6) and (45,2).
Join , with a free hand to get the ‘more than type’ series.



The two curves intersect at point L. Draw LMOX cutting the x-axis at M.
Clearly, M = 29.5
Hence, Median = 29.5

 



Page No 895:

Question 10:

(i) From the given table, we may prepare the 'less than' frequency table as shown below:

Marks

No. of students

Less than 5

2

Less than 10

7

Less than 15

13

Less than 20

21

Less than 25

31

Less than 30

56

Less than 35

76

Less than 40

94

Less than 45

98

Less than 50

100

 

We plot the points A(5,2), B(10,7), C(15,13), D(20,21), E(25,31), F(30,56), G(35,76), H(40,94), I(45,98) and J(50,100).
 Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve representing the ‘less than type’ series.

(ii) More than series:

Marks

No. of student

More than 0

100

More than 5

98

More than 10

93

More than 15

87

More than 20

79

More than 25

69

More than 30

44

More than 35

24

More than 40

6

More than 45

2

Now, on the same graph paper, we plot the points (0,100), (5,98), (10,94), (15,76), (20,56), (25,31), (30,21), (35,13), (40,6) and (45,2).
Join , with a free hand to get the ‘more than type’ series.



The two curves intersect at point L. Draw LMOX cutting the x-axis at M.
Clearly, M = 29.5
Hence, Median = 29.5

 

Answer:

(i) Less than series:

Mark

s

No. of students

Less than 144

3

Less than 148

12

Less than 152

36

Less than 156

67

Less than 160

109

Less than 164

173

Less than 168

248

Less than 172

330

Less than 176

416

Less than 180

450

 

We plot the points A(144,3), B(148,12), C(152,36), D(156,67), E(160,109) F(164,173), G(168,248), H(172,330), I(176,416) and J(180,450). Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve representing  the ‘less than type’ series.

(ii) More than series:

Marks

No. of students

More than 140

450

More than 144

447

More than 148

438

More than 152

414

More than 156

383

More than 160

341

More than 164

277

More than 168

202

More than 172

120

More than 176

34

 

Now on the same graph paper, we plot the points A1(140,450),  B1(144,447), C1(148,438),  D1(152,414), E1(156,383), F1(160,341), G1(164,277), H1(168,202), I1(172,120) and J1(176,34).
Join A1B1, B1C1, C1D1, D1E1, E1F1, F1G1, G1H1, H1I1 and I1J1
with a free hand to get the ‘more than type’ series.

The two curves intersect at point L. Draw LMOX  cutting the x-axis at M.
Clearly, M = 166 cm
Hence, Median = 166 cm

 



Page No 896:

Question 1:

(i) Less than series:

Mark

s

No. of students

Less than 144

3

Less than 148

12

Less than 152

36

Less than 156

67

Less than 160

109

Less than 164

173

Less than 168

248

Less than 172

330

Less than 176

416

Less than 180

450

 

We plot the points A(144,3), B(148,12), C(152,36), D(156,67), E(160,109) F(164,173), G(168,248), H(172,330), I(176,416) and J(180,450). Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve representing  the ‘less than type’ series.

(ii) More than series:

Marks

No. of students

More than 140

450

More than 144

447

More than 148

438

More than 152

414

More than 156

383

More than 160

341

More than 164

277

More than 168

202

More than 172

120

More than 176

34

 

Now on the same graph paper, we plot the points A1(140,450),  B1(144,447), C1(148,438),  D1(152,414), E1(156,383), F1(160,341), G1(164,277), H1(168,202), I1(172,120) and J1(176,34).
Join A1B1, B1C1, C1D1, D1E1, E1F1, F1G1, G1H1, H1I1 and I1J1
with a free hand to get the ‘more than type’ series.

The two curves intersect at point L. Draw LMOX  cutting the x-axis at M.
Clearly, M = 166 cm
Hence, Median = 166 cm

 

Answer:

 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 8 16
30−40 10 26
40−50 12 38
50−60 8 46
60−70 4 50
Total N = ∑fi = 50  

Now, N = 50 N2=25.

The cumulative frequency just greater than 25 is 26, and the corresponding class is 30−40.

Thus, the median class is 30−40.



Page No 897:

Question 2:

 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 8 16
30−40 10 26
40−50 12 38
50−60 8 46
60−70 4 50
Total N = ∑fi = 50  

Now, N = 50 N2=25.

The cumulative frequency just greater than 25 is 26, and the corresponding class is 30−40.

Thus, the median class is 30−40.

Answer:

Here the maximum class frequency is 27, and the class corresponding to this frequency is 40−50. So, the modal class is 40−50.

Now,

Modal class = 40−50, lower limit (l) of modal class = 40

Thus, lower limit (
l) of modal class is 40.

Page No 897:

Question 3:

Here the maximum class frequency is 27, and the class corresponding to this frequency is 40−50. So, the modal class is 40−50.

Now,

Modal class = 40−50, lower limit (l) of modal class = 40

Thus, lower limit (
l) of modal class is 40.

Answer:

Here the maximum class frequency is 30, and the class corresponding to this frequency is 150−200. So, the modal class is 150−200.

Also, class mark of the modal class is 150+2002=175.

Page No 897:

Question 4:

Here the maximum class frequency is 30, and the class corresponding to this frequency is 150−200. So, the modal class is 150−200.

Also, class mark of the modal class is 150+2002=175.

Answer:

If the number of observations is odd, then the median is n+12th observation.

Thus, 25+12=13th observation represents the median.

Page No 897:

Question 5:

If the number of observations is odd, then the median is n+12th observation.

Thus, 25+12=13th observation represents the median.

Answer:

There is an empirical relationship between the three measures of central tendency:

3Median = Mode + 2Mean

Mean=3Median-Mode2            =31250-10002            =1375

Thus, the mean is 1375.

Page No 897:

Question 6:

There is an empirical relationship between the three measures of central tendency:

3Median = Mode + 2Mean

Mean=3Median-Mode2            =31250-10002            =1375

Thus, the mean is 1375.

Answer:

Here the maximum class frequency is 25, and the class corresponding to this frequency is 40−60.

So, the modal class is 40
−60.

Now, to find the median class let us put the data in the table given below:
 

Marks obtained Number of students (fi) Cumulative frequency (cf)
020 4 4
2040 6 10
4060 25 35
6080 10 45
80100 5 50
Total N = ∑fi = 50  


Now, N = 50 N2=25.

The cumulative frequency just greater than 25 is 35, and the corresponding class is 40−60.

Thus, the median class is 40
60.

Page No 897:

Question 7:

Here the maximum class frequency is 25, and the class corresponding to this frequency is 40−60.

So, the modal class is 40
−60.

Now, to find the median class let us put the data in the table given below:
 

Marks obtained Number of students (fi) Cumulative frequency (cf)
020 4 4
2040 6 10
4060 25 35
6080 10 45
80100 5 50
Total N = ∑fi = 50  


Now, N = 50 N2=25.

The cumulative frequency just greater than 25 is 35, and the corresponding class is 40−60.

Thus, the median class is 40
60.

Answer:

Class mark = Upper limit+Lower limit2

∴ class mark of 10−25 = 10+252
                                    = 17.5

and class mark of 35−55 = 35+552
                                       = 45

Page No 897:

Question 8:

Class mark = Upper limit+Lower limit2

∴ class mark of 10−25 = 10+252
                                    = 17.5

and class mark of 35−55 = 35+552
                                       = 45

Answer:

According to assumed-mean method,

x¯=A+ifidiifi  =25+11050  =25+2.2  =27.2

Thus, mean is 27.2.

Page No 897:

Question 9:

According to assumed-mean method,

x¯=A+ifidiifi  =25+11050  =25+2.2  =27.2

Thus, mean is 27.2.

Answer:

According to the question,

4 = X36 and 3 = Y64

⇒ X = 4 × 36 and Y = 3 × 64

⇒ X = 144 and Y = 192

Now, X + Y = 144 + 192 = 336

and total number of observations = 36 + 64 = 100

Thus, mean = 336100=3.36.

Page No 897:

Question 10:

According to the question,

4 = X36 and 3 = Y64

⇒ X = 4 × 36 and Y = 3 × 64

⇒ X = 144 and Y = 192

Now, X + Y = 144 + 192 = 336

and total number of observations = 36 + 64 = 100

Thus, mean = 336100=3.36.

Answer:

Uppar class boundary = Lowest class boundary + width × number of classes

                                    = 8.1 + 2.5 × 12

                                    = 8.1 + 30

                                    = 38.1

Thus, upper class boundary of the highest class is 38.1.



Page No 898:

Question 11:

Uppar class boundary = Lowest class boundary + width × number of classes

                                    = 8.1 + 2.5 × 12

                                    = 8.1 + 30

                                    = 38.1

Thus, upper class boundary of the highest class is 38.1.

Answer:

If number of observations is even, then the median will be the average of n2th and the n2+1th observations.

In the given case, n = 10 n2th=5th and n2+1th=6th observation.

Thus, 63=x+x+22
126=2x+2124=2xx=62

Thus, the value of x is 62.

Page No 898:

Question 12:

If number of observations is even, then the median will be the average of n2th and the n2+1th observations.

In the given case, n = 10 n2th=5th and n2+1th=6th observation.

Thus, 63=x+x+22
126=2x+2124=2xx=62

Thus, the value of x is 62.

Answer:

Since, 8 is less than 30 and 32 is more than 30, so the middle value remains unchanged.

Thus, the median of 21 observations taken together is 30.

Page No 898:

Question 13:

Since, 8 is less than 30 and 32 is more than 30, so the middle value remains unchanged.

Thus, the median of 21 observations taken together is 30.

Answer:

Arranging the observations in ascending order, we have

x5,x4,x3,x2,x.

Thus, the median is x3.

x3=8x=3×8x=24

Thus, the value of x is 24.

Page No 898:

Question 14:

Arranging the observations in ascending order, we have

x5,x4,x3,x2,x.

Thus, the median is x3.

x3=8x=3×8x=24

Thus, the value of x is 24.

Answer:

​Here the maximum class frequency is 23, and the class corresponding to this frequency is 12−15.

So, the modal class is 12−15.


Now, to find the cumulative frequency let us put the data in the table given below:
 

Class Frequency (fi) Cumulative frequency (cf)
3−6 7 7
6−9 13 20
9−12 10 30
12−15 23 53
15−18 4 57
18−21 21 78
21−24 16 94
Total N = ∑fi = 94  

Thus, the cumulative frequency of the modal class is 53.

Page No 898:

Question 15:

​Here the maximum class frequency is 23, and the class corresponding to this frequency is 12−15.

So, the modal class is 12−15.


Now, to find the cumulative frequency let us put the data in the table given below:
 

Class Frequency (fi) Cumulative frequency (cf)
3−6 7 7
6−9 13 20
9−12 10 30
12−15 23 53
15−18 4 57
18−21 21 78
21−24 16 94
Total N = ∑fi = 94  

Thus, the cumulative frequency of the modal class is 53.

Answer:

Here the maximum class frequency is 18, 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 = 18,

frequency (f0) of class preceding the modal class = 6,

frequency (f2) of class succeeding the modal class = 10.

Now, let us substitute these values in the formula:

Mode=l+f1-f02f1-f0-f2×h        =40+18-636-6-10×20        =40+1220×20        =40+12        =52


Hence, the mode is 52.

Page No 898:

Question 16:

Here the maximum class frequency is 18, 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 = 18,

frequency (f0) of class preceding the modal class = 6,

frequency (f2) of class succeeding the modal class = 10.

Now, let us substitute these values in the formula:

Mode=l+f1-f02f1-f0-f2×h        =40+18-636-6-10×20        =40+1220×20        =40+12        =52


Hence, the mode is 52.

Answer:

A 'less than type' cumulative frequency distribution table is given below:
 

Age (in years) Cumulative frequency (cf)
 Less than 20 60
Less than 30 102
Less than 40 157
Less than 50 227
Less than 60 280
Less than 70 300

Page No 898:

Question 17:

A 'less than type' cumulative frequency distribution table is given below:
 

Age (in years) Cumulative frequency (cf)
 Less than 20 60
Less than 30 102
Less than 40 157
Less than 50 227
Less than 60 280
Less than 70 300

Answer:

Here, p = 11 + 12 = 23

and 33 + q = 46 ⇒ q = 46 − 33 = 13

Thus, p is 23 and q is 13.

Now,

Here the maximum class frequency is 20, and the class corresponding to this frequency is 500−600.

So, the modal class is 500−600.

Also, ∑ f N = 80 N2=40.


The cumulative frequency just greater than 40 is 46, and the corresponding class is 400−500.

Thus, the median class is 400−500.



Page No 899:

Question 18:

Here, p = 11 + 12 = 23

and 33 + q = 46 ⇒ q = 46 − 33 = 13

Thus, p is 23 and q is 13.

Now,

Here the maximum class frequency is 20, and the class corresponding to this frequency is 500−600.

So, the modal class is 500−600.

Also, ∑ f N = 80 N2=40.


The cumulative frequency just greater than 40 is 46, and the corresponding class is 400−500.

Thus, the median class is 400−500.

Answer:

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

Monthly consumption (in units)
(lower class limits)
Cumulative frequency (cf)
More than 65 60 + 4 = 64
More than 85 55 + 5 = 60
More than 105 42 + 13 = 55
More than 125 22 + 20 = 42
More than 145 8 + 14 = 22
More than 165 8

Page No 899:

Question 19:

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

Monthly consumption (in units)
(lower class limits)
Cumulative frequency (cf)
More than 65 60 + 4 = 64
More than 85 55 + 5 = 60
More than 105 42 + 13 = 55
More than 125 22 + 20 = 42
More than 145 8 + 14 = 22
More than 165 8

Answer:

The frequency distribution is as follows:
 

Life-time
(in days)
Frequency (f)
0−50 7
50−100 14
100−150 31
150−200 27
200−250 12
250−300 9

Page No 899:

Question 20:

The frequency distribution is as follows:
 

Life-time
(in days)
Frequency (f)
0−50 7
50−100 14
100−150 31
150−200 27
200−250 12
250−300 9

Answer:

(a) â€‹â€‹The frequency distribution into the continuous form is as follows:
 

Marks obtained
(in per cent)
Number of students (f)
10.5−20.5 141
20.5−30.5 221
30.5−40.5 439
40.5−50.5 529
50.5−60.5 495
60.5−70.5 322
70.5−80.5 153

(b) Now, to find the median class let us put the data in the table given below:
 
Marks obtained
(in per cent)
Number of students (f) Cumulative frequency (cf)
10.5−20.5 141 141
20.5−30.5 221 362
30.5−40.5 439 801
40.5−50.5 529 1330
50.5−60.5 495 1825
60.5−70.5 322 2147
70.5−80.5 153 2300

Now, N = 2300 N2=1150.

The cumulative frequency just greater than 1150 is 1330, and the corresponding class is 40.5−50.5.

Thus, the median class is 40.5−50.5.


Now, class mark = upper class limit+lower class limit2
                           = 40.5+50.52=912=45.5

Thus, class mark of the median class is 45.5.

(c) Here the maximum class frequency is 529, and the class corresponding to this frequency is 40.5−50.5.

So, the modal class is 40.5−50.5 and its cumulative frequency is 1330.

Page No 899:

Question 21:

(a) â€‹â€‹The frequency distribution into the continuous form is as follows:
 

Marks obtained
(in per cent)
Number of students (f)
10.5−20.5 141
20.5−30.5 221
30.5−40.5 439
40.5−50.5 529
50.5−60.5 495
60.5−70.5 322
70.5−80.5 153

(b) Now, to find the median class let us put the data in the table given below:
 
Marks obtained
(in per cent)
Number of students (f) Cumulative frequency (cf)
10.5−20.5 141 141
20.5−30.5 221 362
30.5−40.5 439 801
40.5−50.5 529 1330
50.5−60.5 495 1825
60.5−70.5 322 2147
70.5−80.5 153 2300

Now, N = 2300 N2=1150.

The cumulative frequency just greater than 1150 is 1330, and the corresponding class is 40.5−50.5.

Thus, the median class is 40.5−50.5.


Now, class mark = upper class limit+lower class limit2
                           = 40.5+50.52=912=45.5

Thus, class mark of the median class is 45.5.

(c) Here the maximum class frequency is 529, and the class corresponding to this frequency is 40.5−50.5.

So, the modal class is 40.5−50.5 and its cumulative frequency is 1330.

Answer:

The given data is shown as follows:
 

Class Frequency (fi) Class mark (xi) fixi
0−10 8 5 40
10−20 p 15 15p
20−30 12 25 300
30−40 13 35 455
40−50 10 45 450
Total ∑ f= 43 + p   ∑ fix= 1245 + 15p

The mean of given data is given by

x¯=ifixiifi27=1245+15p43+p1161+27p=1245+15p27p-15p=1245-116112p=84p=7

Thus, the value of p is 7.

Page No 899:

Question 22:

The given data is shown as follows:
 

Class Frequency (fi) Class mark (xi) fixi
0−10 8 5 40
10−20 p 15 15p
20−30 12 25 300
30−40 13 35 455
40−50 10 45 450
Total ∑ f= 43 + p   ∑ fix= 1245 + 15p

The mean of given data is given by

x¯=ifixiifi27=1245+15p43+p1161+27p=1245+15p27p-15p=1245-116112p=84p=7

Thus, the value of p is 7.

Answer:

Let the missing frequency be x.

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

Age (in years) Number of persons (f) Cumulative frequency (cf)
0−10 5 5
10−20 25 30
20−30 x 30 + x
30−40 18 48 + x
40−50 7 55 + x

The given median is 24,

∴ the median class is 20−30.

∴ 
l = 20, h = 10, N = 55 + xf = x and cf = 30

Median=l+N2-cff×h24=20+55+x2-30x×1024-20=55+x-602x×104=x-52x×108x=10x-502x=50x=25

Thus, the missing frequency is 25.



Page No 900:

Question 1:

Let the missing frequency be x.

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

Age (in years) Number of persons (f) Cumulative frequency (cf)
0−10 5 5
10−20 25 30
20−30 x 30 + x
30−40 18 48 + x
40−50 7 55 + x

The given median is 24,

∴ the median class is 20−30.

∴ 
l = 20, h = 10, N = 55 + xf = x and cf = 30

Median=l+N2-cff×h24=20+55+x2-30x×1024-20=55+x-602x×104=x-52x×108x=10x-502x=50x=25

Thus, the missing frequency is 25.

Answer:

(d) Standard deviation

The standard deviation is a measure of dispersion. It is the action or process of distributing things over a wide area (nothing about central location).

Page No 900:

Question 2:

(d) Standard deviation

The standard deviation is a measure of dispersion. It is the action or process of distributing things over a wide area (nothing about central location).

Answer:

(a) Mean

The mean can not be determined graphically because the values cannot be summed.

Page No 900:

Question 3:

(a) Mean

The mean can not be determined graphically because the values cannot be summed.

Answer:

Mean is influenced by extreme values.

Hence, the correct answer is option (a).

Page No 900:

Question 4:

Mean is influenced by extreme values.

Hence, the correct answer is option (a).

Answer:

The correct option is (c).

The mode of a frequency distribution can be obtained graphically from a histogram.

Page No 900:

Question 5:

The correct option is (c).

The mode of a frequency distribution can be obtained graphically from a histogram.

Answer:

(d) ogives

​This is because median of a frequency distribution is found graphically with the help of ogives.

Page No 900:

Question 6:

(d) ogives

​This is because median of a frequency distribution is found graphically with the help of ogives.

Answer:

The cumulative frequency table is useful in determining the (b) median.

Page No 900:

Question 7:

The cumulative frequency table is useful in determining the (b) median.

Answer:

The abscissa of the point of intersection of the 'less than type' and that of the 'more than type' cumulative frequency curves of a grouped data gives its (b) median.

Page No 900:

Question 8:

The abscissa of the point of intersection of the 'less than type' and that of the 'more than type' cumulative frequency curves of a grouped data gives its (b) median.

Answer:


We know that x¯=fixifix¯fi=fixi      ...(i)Now, fi(xix¯)=fixix¯fifi(xix¯)=fixifixi        [Using (i)]fi(xix¯)=0Hence, the correct option is (b).

Page No 900:

Question 9:


We know that x¯=fixifix¯fi=fixi      ...(i)Now, fi(xix¯)=fixix¯fifi(xix¯)=fixifixi        [Using (i)]fi(xix¯)=0Hence, the correct option is (b).

Answer:

(b) ui=(xi-A)h



Page No 901:

Question 10:

(b) ui=(xi-A)h

Answer:

The di's are the deviations from A of (c) midpoints of the classes.

Page No 901:

Question 11:

The di's are the deviations from A of (c) midpoints of the classes.

Answer:

While computing the mean of the group data, we assume that the frequencies are (b) centred at the class marks of the classes.

Page No 901:

Question 12:

While computing the mean of the group data, we assume that the frequencies are (b) centred at the class marks of the classes.

Answer:

(b) mode=3×median-2×mean

Page No 901:

Question 13:

(b) mode=3×median-2×mean

Answer:

The x−coordinate represents the median of the given data.

Thus, median of the given data is 20.5.

Hence, the correct answer is option (c).

Page No 901:

Question 14:

The x−coordinate represents the median of the given data.

Thus, median of the given data is 20.5.

Hence, the correct answer is option (c).

Answer:

(b) 315
The class having the maximum frequency is the modal class.

So, the modal class is 150-155 and its lower limit is 150.Also, N=60N2=30 The cumulative freequency just more than 30 is 37 and its class is 160-165, whose upper limit is 165.Required sum=150+165=315



Page No 902:

Question 15:

(b) 315
The class having the maximum frequency is the modal class.

So, the modal class is 150-155 and its lower limit is 150.Also, N=60N2=30 The cumulative freequency just more than 30 is 37 and its class is 160-165, whose upper limit is 165.Required sum=150+165=315

Answer:

(c) 30-40. The class 30-40 has the maximum freequency, i.e., 30. So, the modal class is 30-40.

Page No 902:

Question 16:

(c) 30-40. The class 30-40 has the maximum freequency, i.e., 30. So, the modal class is 30-40.

Answer:

(b) xk+hfk-fk-12fk-fk-1-fk+1

Page No 902:

Question 17:

(b) xk+hfk-fk-12fk-fk-1-fk+1

Answer:

(a) l+h×N2-cff

Page No 902:

Question 18:

(a) l+h×N2-cff

Answer:

(c) 9.2It is given that the mean and median are 8.9 and 9, respectively.Mode=3×Median-2×MeanMode=3×9-2×8.9=27-17.8=9.2

Page No 902:

Question 19:

(c) 9.2It is given that the mean and median are 8.9 and 9, respectively.Mode=3×Median-2×MeanMode=3×9-2×8.9=27-17.8=9.2

Answer:

(b) 57.5

Class interval 35-45 45-55 55-65 65-75
Frequency 8 12 20 10
Cumulative frequency 8 20 40 50

Here, N=50N2=25, which lies in the class interval of 55-65.Now, cf=55, f = 20 and l= 50Median=l+h×N2-cff=50+65-5520×25-20=57.5

Page No 902:

Question 20:

(b) 57.5

Class interval 35-45 45-55 55-65 65-75
Frequency 8 12 20 10
Cumulative frequency 8 20 40 50

Here, N=50N2=25, which lies in the class interval of 55-65.Now, cf=55, f = 20 and l= 50Median=l+h×N2-cff=50+65-5520×25-20=57.5

Answer:

(c) 24.4
The maximum frequency is 25 and the modal class is 22-26.

Now, xk=22, fk=25, fk-1=16, fk+1=19 and h=4Mode=xk+h×fk-fk-12fk-fk-1-fk+!=22+4×25-162×25-16-19=22+4×25-1650-16-19=22+4×915=22+125=22+2.4=24.4



Page No 903:

Question 21:

(c) 24.4
The maximum frequency is 25 and the modal class is 22-26.

Now, xk=22, fk=25, fk-1=16, fk+1=19 and h=4Mode=xk+h×fk-fk-12fk-fk-1-fk+!=22+4×25-162×25-16-19=22+4×25-1650-16-19=22+4×915=22+125=22+2.4=24.4

Answer:

(c) 24
Mode=3×median-2×mean3×Median=mode+2mean3×Median=16+563×Median=72Median=723Median=24

Page No 903:

Question 22:

(c) 24
Mode=3×median-2×mean3×Median=mode+2mean3×Median=16+563×Median=72Median=723Median=24

Answer:

(b) 24.5
Mode=3×median-2×mean2×Mean=3×median-mode2×Mean=3×26-292×Mean=49Mean=492 Mean=24.5

Page No 903:

Question 23:

(b) 24.5
Mode=3×median-2×mean2×Mean=3×median-mode2×Mean=3×26-292×Mean=49Mean=492 Mean=24.5

Answer:

(c) mean=mode=median
A symmetric distribution is one where the left and right hand sides of the distribution are roughly equally balanced around the mean.

Page No 903:

Question 24:

(c) mean=mode=median
A symmetric distribution is one where the left and right hand sides of the distribution are roughly equally balanced around the mean.

Answer:

Converting the given data into a frequency table, we get:
 

Monthly income

No. of families

Frequency

30,000 and Above

15

15

25,000-30,000

37

(37 − 15) = 22

20,000-25,000

50

(50 − 37) = 13

18,000-20,000

69

(69 − 50) = 19

14,000-18,000

85

(85 − 69) = 16

10,000-14,000

100

(100 − 85) = 15

 













Hence, the number of families having an income range of Rs 20,000-Rs 25,000 is 13.
​The correct option is (c).

Page No 903:

Question 25:

Converting the given data into a frequency table, we get:
 

Monthly income

No. of families

Frequency

30,000 and Above

15

15

25,000-30,000

37

(37 − 15) = 22

20,000-25,000

50

(50 − 37) = 13

18,000-20,000

69

(69 − 50) = 19

14,000-18,000

85

(85 − 69) = 16

10,000-14,000

100

(100 − 85) = 15

 













Hence, the number of families having an income range of Rs 20,000-Rs 25,000 is 13.
​The correct option is (c).

Answer:

First 8 prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19.

Median of 8 numbers is average of 4th and 5th terms.

i.e. average of 7 and 11

Thus, the median is 9.

​Hence, the correct answer is option (b).

Page No 903:

Question 26:

First 8 prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19.

Median of 8 numbers is average of 4th and 5th terms.

i.e. average of 7 and 11

Thus, the median is 9.

​Hence, the correct answer is option (b).

Answer:

It is given that mean of 20 numbers is zero.

i.e. average of 20 numbers is zero.

i.e. sum of 20 numbers is zero.

Thus, at most, there can be 19 positive numbers.
(such that if sum of 19 positive numbers is x, 20th number will be −x)

Hence, the correct answer is option (d).

Page No 903:

Question 27:

It is given that mean of 20 numbers is zero.

i.e. average of 20 numbers is zero.

i.e. sum of 20 numbers is zero.

Thus, at most, there can be 19 positive numbers.
(such that if sum of 19 positive numbers is x, 20th number will be −x)

Hence, the correct answer is option (d).

Answer:

​Median of 6 numbers is the average of 3rd and 4th term.

13=x-1+x-3226=2x-42x=30x=15

Thus, x is equal to 15.

Hence, the correct answer is option (c).

Page No 903:

Question 28:

​Median of 6 numbers is the average of 3rd and 4th term.

13=x-1+x-3226=2x-42x=30x=15

Thus, x is equal to 15.

Hence, the correct answer is option (c).

Answer:

Mean=Sum of observationsNumber of observations15=2+7+6+x460=15+xx=45  ....(1)  

Now,

Mean=Sum of observationsNumber of observations10=18+1+6+x+y550=25+x+yy=25-xy=25-45    From 1y=-20  

Disclaimer : There is a misprinting in the question. If the mean of 2, 7, 6 and x is 15 then the value of y be −20 and if the mean of 2, 7, 6 and x is 5 then the value of y be 20. So, the correct answer is option (c).



Page No 904:

Question 29:

Mean=Sum of observationsNumber of observations15=2+7+6+x460=15+xx=45  ....(1)  

Now,

Mean=Sum of observationsNumber of observations10=18+1+6+x+y550=25+x+yy=25-xy=25-45    From 1y=-20  

Disclaimer : There is a misprinting in the question. If the mean of 2, 7, 6 and x is 15 then the value of y be −20 and if the mean of 2, 7, 6 and x is 5 then the value of y be 20. So, the correct answer is option (c).

Answer:

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

Page No 904:

Question 30:

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

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
Clearly, reason (R) is true.
Using the relation given in reason (R), we have:
2mean=3×median-mode=3×150-154=450-154=296Mean=148, which is true.This assertion A and reasonR are both true and reason Ris the correct explanation of assertion A.

Page No 904:

Question 31:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
Clearly, reason (R) is true.
Using the relation given in reason (R), we have:
2mean=3×median-mode=3×150-154=450-154=296Mean=148, which is true.This assertion A and reasonR are both true and reason Ris the correct explanation of assertion A.

Answer:

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is a not a correct explanation of Assertion (A).
Clearly, reason (R) is true.
The maximum frequency is 23 and the modal class is 12-15.
Now, xk=12, fk=23, fk-1=21, fk+1=23 and h=3Mode=12+3×23-212×23-21-10=12+3×215=12+0.4=12Assertion A is true.However, reason R is not a correct explanation of assertion A.



Page No 907:

Question 1:

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is a not a correct explanation of Assertion (A).
Clearly, reason (R) is true.
The maximum frequency is 23 and the modal class is 12-15.
Now, xk=12, fk=23, fk-1=21, fk+1=23 and h=3Mode=12+3×23-212×23-21-10=12+3×215=12+0.4=12Assertion A is true.However, reason R is not a correct explanation of assertion A.

Answer:

The cumulative frequency table is useful in determining the median.

Hence, the correct option is (b).

Page No 907:

Question 2:

The cumulative frequency table is useful in determining the median.

Hence, the correct option is (b).

Answer:

(c) 45
Here, mean=27 and mode=33 Mode=3Median-2Mean=3×33-2×27=99-54=45

Page No 907:

Question 3:

(c) 45
Here, mean=27 and mode=33 Mode=3Median-2Mean=3×33-2×27=99-54=45

Answer:

(b) 25

Class

Frequency

Cumulative frequency

0-5

10

10

5-10

15

25

10-15

12

37

15-20

20

57

20-25

9

63

 

Now, N=63
N2=32.5
 The cumulative frequency just greater than 32.5 is 37 and the corresponding class is 10-15.
Median class=10-15
Here, the highest frequency is 20 and its corresponding class is 15-20.
Modal Class=15-20
Sum of the lower limits of the median class and modal class = 10 + 15 = 25

Page No 907:

Question 4:

(b) 25

Class

Frequency

Cumulative frequency

0-5

10

10

5-10

15

25

10-15

12

37

15-20

20

57

20-25

9

63

 

Now, N=63
N2=32.5
 The cumulative frequency just greater than 32.5 is 37 and the corresponding class is 10-15.
Median class=10-15
Here, the highest frequency is 20 and its corresponding class is 15-20.
Modal Class=15-20
Sum of the lower limits of the median class and modal class = 10 + 15 = 25

Answer:

(d) 17.5
Converting the given series into continuous series, we get:

 

Class

Frequency

Cumulative frequency

0.5-5.5

13

13

5.5-11.5

10

23

11.5-17.5

15

38

17.5-23.5

8

46

23.5-29.5

11

57

 

Now, N =57
N2=28.5

The cumulative frequency just greater than 28.5 is 38 and its corresponding class is 11.5-17.5.
∴ The median class is 11.5-17.5 and the corresponding upper limit is 17.5.

Page No 907:

Question 5:

(d) 17.5
Converting the given series into continuous series, we get:

 

Class

Frequency

Cumulative frequency

0.5-5.5

13

13

5.5-11.5

10

23

11.5-17.5

15

38

17.5-23.5

8

46

23.5-29.5

11

57

 

Now, N =57
N2=28.5

The cumulative frequency just greater than 28.5 is 38 and its corresponding class is 11.5-17.5.
∴ The median class is 11.5-17.5 and the corresponding upper limit is 17.5.

Answer:

Given:
Mean=53.4Mode=55.2We know that mode=3median-2mean55.2=3median-2×53.43Median=55.2+106.8Median=1623=54

Page No 907:

Question 6:

Given:
Mean=53.4Mode=55.2We know that mode=3median-2mean55.2=3median-2×53.43Median=55.2+106.8Median=1623=54

Answer:

Given distribution table can be written as following:

Class

Frequency

Cumulative frequency

Less than 14

2

2

Less than 14.2

4

6

Less than 14.4

15

21

Less than 14.6

54

75

Less than 14.8

25

100

Less than 15

20

120

Number of athletes who completed the race in less than 14.6 sec = 75



Page No 908:

Question 7:

Given distribution table can be written as following:

Class

Frequency

Cumulative frequency

Less than 14

2

2

Less than 14.2

4

6

Less than 14.4

15

21

Less than 14.6

54

75

Less than 14.8

25

100

Less than 15

20

120

Number of athletes who completed the race in less than 14.6 sec = 75

Answer:

Class

Frequency

Cumulative frequency

0.5-5.5

13

13

5.5-11.5

10

23

11.5-17.5

15

38

17.5-23.5

8

46

23.5-29.5

11

57

 N=57
N2=28.5


The cumulative frequency just greater than 28.5 is 38 and its corresponding class is 11.5-17.5.

Now, we have:Median class=11.5-17.5Upper limit =17.5

Page No 908:

Question 8:

Class

Frequency

Cumulative frequency

0.5-5.5

13

13

5.5-11.5

10

23

11.5-17.5

15

38

17.5-23.5

8

46

23.5-29.5

11

57

 N=57
N2=28.5


The cumulative frequency just greater than 28.5 is 38 and its corresponding class is 11.5-17.5.

Now, we have:Median class=11.5-17.5Upper limit =17.5

Answer:

The frequency distribution table of the given data is as follows:
 

Profit (in lakhs â‚¹) Number of shops
5−10 30 − 28 = 2
10−15 28 − 16 = 12
15−20 16 − 14 = 2
20−25 14 − 10 = 4
25−30 10 − 7 = 3
30−35 7 − 3 = 4
35−40 3

Thus, the frequency corresponding to the class 20−25 is 4.

Page No 908:

Question 9:

The frequency distribution table of the given data is as follows:
 

Profit (in lakhs â‚¹) Number of shops
5−10 30 − 28 = 2
10−15 28 − 16 = 12
15−20 16 − 14 = 2
20−25 14 − 10 = 4
25−30 10 − 7 = 3
30−35 7 − 3 = 4
35−40 3

Thus, the frequency corresponding to the class 20−25 is 4.

Answer:

We have the following table:

Class

Mid value xi

Frequency

fi

 fixi

1-3

2

9

18

3-5

4

22

88

5-7

6

27

162

7-9

8

18

144

 

 

fi=76

fixi=412

  Mean, x¯=fixifi              =41276              =5.42

Page No 908:

Question 10:

We have the following table:

Class

Mid value xi

Frequency

fi

 fixi

1-3

2

9

18

3-5

4

22

88

5-7

6

27

162

7-9

8

18

144

 

 

fi=76

fixi=412

  Mean, x¯=fixifi              =41276              =5.42

Answer:

Speed (km/h)

No. of players fi

c.f

85-100

10

10

100-115

4

14

115-130

7

21

130-145

9

30

 

N =30
N2=15
The cumulative frequency just greater than 15 is 21.
Median Class=115-130
i.e., xk=115, h=15, f=7, c=c.f of the preceeding class=14, N2=15
Me=xk+h×N2-cf=115+15×15-147=115+2.1=117.1
Hence, the required speed is 117.1 km/h.

Page No 908:

Question 11:

Speed (km/h)

No. of players fi

c.f

85-100

10

10

100-115

4

14

115-130

7

21

130-145

9

30

 

N =30
N2=15
The cumulative frequency just greater than 15 is 21.
Median Class=115-130
i.e., xk=115, h=15, f=7, c=c.f of the preceeding class=14, N2=15
Me=xk+h×N2-cf=115+15×15-147=115+2.1=117.1
Hence, the required speed is 117.1 km/h.

Answer:

Class

Mid value xi

Frequency fi

xi×fi

0-10

5

16

80

10-20

15

p

15p

20-30

25

30

750

30-40

35

32

1120

40-50

45

14

630

 

 

fi=p+92

fixi=15p+2580

 Mean, xŻ=fixifi=15p+2580p+92Now,15p+2580p+92=5050p+4600=15p+258035p=-2020p=-4047 frequency cannot be negative



So, there is an error in question.
 

Page No 908:

Question 12:

Class

Mid value xi

Frequency fi

xi×fi

0-10

5

16

80

10-20

15

p

15p

20-30

25

30

750

30-40

35

32

1120

40-50

45

14

630

 

 

fi=p+92

fixi=15p+2580

 Mean, xŻ=fixifi=15p+2580p+92Now,15p+2580p+92=5050p+4600=15p+258035p=-2020p=-4047 frequency cannot be negative



So, there is an error in question.
 

Answer:

As the class 20-30 has the maximum frequency, it is the modal class.

xk=20, h=10, fk=30, fk-1=16 and fk+1=9Mode, M0=xk+h×fk-fk-12fk-fk-1-fk+1=20+10×30-162×30-16-9=20+10×1435=20+4=24



Page No 909:

Question 13:

As the class 20-30 has the maximum frequency, it is the modal class.

xk=20, h=10, fk=30, fk-1=16 and fk+1=9Mode, M0=xk+h×fk-fk-12fk-fk-1-fk+1=20+10×30-162×30-16-9=20+10×1435=20+4=24

Answer:

 Let us plot the points A(10,3), B(20,11), C(30,28), D(40,48) and E(50,70).
 Now, let us join AB, BC, CD and DE with a free hand to get the curve representing the ‘less than type’ series.

Page No 909:

Question 14:

 Let us plot the points A(10,3), B(20,11), C(30,28), D(40,48) and E(50,70).
 Now, let us join AB, BC, CD and DE with a free hand to get the curve representing the ‘less than type’ series.

Answer:

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

Class interval Frequency (fi) Cumulative frequency (cf)
0−10 8 8
10−20 16 24
20−30 36 60
30−40 34 94
40−50 6 100
Total N = ∑ fi = 100  

Now, N = 100 N2=50.

The cumulative frequency just greater than 50 is 60, and the corresponding class is 20−30.

Thus, the median class is 20
−30.

∴ l = 20, h = 10, N = 100, f = 36 and cf = 24.


Now, 

Median=l+N2-cff×h          =20+50-2436×10          =20+26036          =27.22

Thus, the median is 27.22.

Page No 909:

Question 15:

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

Class interval Frequency (fi) Cumulative frequency (cf)
0−10 8 8
10−20 16 24
20−30 36 60
30−40 34 94
40−50 6 100
Total N = ∑ fi = 100  

Now, N = 100 N2=50.

The cumulative frequency just greater than 50 is 60, and the corresponding class is 20−30.

Thus, the median class is 20
−30.

∴ l = 20, h = 10, N = 100, f = 36 and cf = 24.


Now, 

Median=l+N2-cff×h          =20+50-2436×10          =20+26036          =27.22

Thus, the median is 27.22.

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 = 100

Mark the point A whose ordinate is 50 and its x−coordinate is 55.



Thus, the median of the data is 55.

Page No 909:

Question 16:

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 = 100

Mark the point A whose ordinate is 50 and its x−coordinate is 55.



Thus, the median of the data is 55.

Answer:

Let, f1 and f2be the frequencies of the class intervals 20-30 and 40-50, respectively.
Then 10+20+f1+40+f2+25+15=170 f1+f2=60

The median is 35 which lies in the class of 30-40. So, the median class is 30-40.
Now, l=30, h=10, f=40, N=170 and cf=10+20+f1=f1+30Median, M=l+h×N2-cff30+10×85-(f1+30)40=3530+55-f14=3555-f1=20f1=35Now, f2=60-35=25Hence, f1=35 and f2=25

Page No 909:

Question 17:

Let, f1 and f2be the frequencies of the class intervals 20-30 and 40-50, respectively.
Then 10+20+f1+40+f2+25+15=170 f1+f2=60

The median is 35 which lies in the class of 30-40. So, the median class is 30-40.
Now, l=30, h=10, f=40, N=170 and cf=10+20+f1=f1+30Median, M=l+h×N2-cff30+10×85-(f1+30)40=3530+55-f14=3555-f1=20f1=35Now, f2=60-35=25Hence, f1=35 and f2=25

Answer:

Class

Frequency

fi

Mid values xi

(fi×xi)

0-20

17

10

170

20-40

f1

30

30 f1

40-60

32

50

1600

60-80

52- f1

70

3640-70 f1

80-100

19

90

1710

 

fi=120

 

(fi×xi)=7120-40f1

  We have:17+f1+32+f2+19=120f1+f2=52f2=52f1 Mean, xŻ=(fi×xi)fi50=712040f1120               [Mean=50]40f1=1120f1=28And f2=5228f2=24The missing frequencies are f1=28 and f2 =24.

Page No 909:

Question 18:

Class

Frequency

fi

Mid values xi

(fi×xi)

0-20

17

10

170

20-40

f1

30

30 f1

40-60

32

50

1600

60-80

52- f1

70

3640-70 f1

80-100

19

90

1710

 

fi=120

 

(fi×xi)=7120-40f1

  We have:17+f1+32+f2+19=120f1+f2=52f2=52f1 Mean, xŻ=(fi×xi)fi50=712040f1120               [Mean=50]40f1=1120f1=28And f2=5228f2=24The missing frequencies are f1=28 and f2 =24.

Answer:

Let us choose = 105, h = 6, then di = xi − 105 and uxi-1056.

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

Class Frequency (fi) Class mark (xi) di = xi − 105 uxi-1056 fiui
84−90 15 87 −18 −3 −45
90−96 22 93 −12 −2 −44
96−102 20 99 −6 −1 −20
102−108 18 105 0 0 0
108−114 20 111 6 1 20
114−120 25 117 12 2 50
Total ∑ fi = 120       ∑ fiui = −39

The mean of the given data is given by

x¯=a+ifiuiifi×h  =105+-39120×6  =105-1.95  =103.05

Thus, mean of the given data is 103.05.

Page No 909:

Question 19:

Let us choose = 105, h = 6, then di = xi − 105 and uxi-1056.

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

Class Frequency (fi) Class mark (xi) di = xi − 105 uxi-1056 fiui
84−90 15 87 −18 −3 −45
90−96 22 93 −12 −2 −44
96−102 20 99 −6 −1 −20
102−108 18 105 0 0 0
108−114 20 111 6 1 20
114−120 25 117 12 2 50
Total ∑ fi = 120       ∑ fiui = −39

The mean of the given data is given by

x¯=a+ifiuiifi×h  =105+-39120×6  =105-1.95  =103.05

Thus, mean of the given data is 103.05.

Answer:

Here the maximum class frequency is 15, 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 = 15,

frequency (f0) of class preceding the modal class = 10,

frequency (f2) of class succeeding the modal class = 5.

Now, let us substitute these values in the formula:


Mode=l+f1-f02f1-f0-f2×h        =30+15-1030-10-5×10        =30+5015        =33.33

Hence, the mode is 33.33.

Now, to find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (xi) fixi
0−10 6 5 30
10−20 8 15 120
20−30 10 25 250
30−40 15 35 525
40−50 5 45 225
50−60 4 55 220
60−70 2 65 130
Total ∑ fi = 50   ∑ fixi = 1500

Mean=ifixiifi        =150050        =30

Thus, mean of the given data is 30.

Now, to find the median let us put the data in the table given below:
 
Class Frequency (fi) Cumulative frequency (cf)
0−10 6 6
10−20 8 14
20−30 10 24
30−40 15 39
40−50 5 44
50−60 4 48
60−70 2 50
Total N = ∑ fi = 50  

Now, N = 50 N2=25.

The cumulative frequency just greater than 25 is 39, and the corresponding class is 30−40.

Thus, the median class is 30−40.

∴ l = 30, h = 10, N = 50, f = 15 and cf = 24.


Now, 

Median=l+N2-cff×h          =30+25-2415×10          =30+1015          =30.67

Thus, the median is 30.67.



Page No 910:

Question 20:

Here the maximum class frequency is 15, 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 = 15,

frequency (f0) of class preceding the modal class = 10,

frequency (f2) of class succeeding the modal class = 5.

Now, let us substitute these values in the formula:


Mode=l+f1-f02f1-f0-f2×h        =30+15-1030-10-5×10        =30+5015        =33.33

Hence, the mode is 33.33.

Now, to find the mean let us put the data in the table given below:
Class Frequency (fi) Class mark (xi) fixi
0−10 6 5 30
10−20 8 15 120
20−30 10 25 250
30−40 15 35 525
40−50 5 45 225
50−60 4 55 220
60−70 2 65 130
Total ∑ fi = 50   ∑ fixi = 1500

Mean=ifixiifi        =150050        =30

Thus, mean of the given data is 30.

Now, to find the median let us put the data in the table given below:
 
Class Frequency (fi) Cumulative frequency (cf)
0−10 6 6
10−20 8 14
20−30 10 24
30−40 15 39
40−50 5 44
50−60 4 48
60−70 2 50
Total N = ∑ fi = 50  

Now, N = 50 N2=25.

The cumulative frequency just greater than 25 is 39, and the corresponding class is 30−40.

Thus, the median class is 30−40.

∴ l = 30, h = 10, N = 50, f = 15 and cf = 24.


Now, 

Median=l+N2-cff×h          =30+25-2415×10          =30+1015          =30.67

Thus, the median is 30.67.

Answer:

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

Class interval
(upper class limits)
Cumulative frequency (cf)
Less than 10 2
Less than 15 2 + 12 = 14
Less than 20 14 + 2 = 16
Less than 25 16 + 4 = 20
Less than 30 20 + 3 = 23
Less than 35 23 + 4 = 27
Less than 40 27 + 3 = 30

and the frequency distribution table of more than type is as follows:
 
Class interval
(lower class limits)
Cumulative frequency (cf)
More than 5 28 + 2 = 30
More than 10 16 + 12 = 28
More than 15 14 + 2 = 16
More than 20 10 + 4 = 14
More than 25 7 + 3 = 10
More than 30 3 + 4 = 7
More than 35 3

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

​

Here both ogives intersect at point A whose x−coordinate is 17.5.



Thus, the median of the data is 17.5.

Page No 910:

Question 21:

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

Class interval
(upper class limits)
Cumulative frequency (cf)
Less than 10 2
Less than 15 2 + 12 = 14
Less than 20 14 + 2 = 16
Less than 25 16 + 4 = 20
Less than 30 20 + 3 = 23
Less than 35 23 + 4 = 27
Less than 40 27 + 3 = 30

and the frequency distribution table of more than type is as follows:
 
Class interval
(lower class limits)
Cumulative frequency (cf)
More than 5 28 + 2 = 30
More than 10 16 + 12 = 28
More than 15 14 + 2 = 16
More than 20 10 + 4 = 14
More than 25 7 + 3 = 10
More than 30 3 + 4 = 7
More than 35 3

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

​

Here both ogives intersect at point A whose x−coordinate is 17.5.



Thus, the median of the data is 17.5.

Answer:

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

Production yield (in kg/ha)
(upper class limits)
Cumulative frequency (cf)
Less than 45 1
Less than 50 1 + 9 = 10
Less than 55 10 + 15 = 25
Less than 60 25 + 18 = 43
Less than 65 43 + 40 = 83
Less than 70 83 + 26 = 109
Less than 75 109 + 16 = 125
Less than 80 125 + 14 = 139
Less than 85 139 + 10 = 149

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

​

The frequency distribution table of more than type is as follows:
 
Production yield (in kg/ha)
(lower class limits)
Cumulative frequency (cf)
More than 40 148 + 1 = 149
More than 45 139 + 9 = 148
More than 50 124 + 15 = 139
More than 55 106 + 18 = 124
More than 60 66 + 40 = 106
More than 65 40 + 26 = 66
More than 70 24 + 16 = 40
More than 75 10 + 14 = 24
More than 80 10

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

Page No 910:

Question 22:

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

Production yield (in kg/ha)
(upper class limits)
Cumulative frequency (cf)
Less than 45 1
Less than 50 1 + 9 = 10
Less than 55 10 + 15 = 25
Less than 60 25 + 18 = 43
Less than 65 43 + 40 = 83
Less than 70 83 + 26 = 109
Less than 75 109 + 16 = 125
Less than 80 125 + 14 = 139
Less than 85 139 + 10 = 149

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

​

The frequency distribution table of more than type is as follows:
 
Production yield (in kg/ha)
(lower class limits)
Cumulative frequency (cf)
More than 40 148 + 1 = 149
More than 45 139 + 9 = 148
More than 50 124 + 15 = 139
More than 55 106 + 18 = 124
More than 60 66 + 40 = 106
More than 65 40 + 26 = 66
More than 70 24 + 16 = 40
More than 75 10 + 14 = 24
More than 80 10

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

Answer:

The frequency distribution into the continuous form is as follows:

Marks 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
Number of students 2 3 6 7 14 12 4 2

Here the maximum class frequency is 14, and the class corresponding to this frequency is 30.5−35.5. So, the modal class is 30.5−35.5.

Now,

Modal class = 30.5−35.5, lower limit (l) of modal class = 30.5, class size (h) = 5,

frequency (f1) of the modal class = 14,

frequency (f0) of class preceding the modal class = 7,

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        =30.5+14-728-7-12×5        =30.5+359        =34.38

Hence, the mode is 34.38.

Now, to find the mean let us put the data in the table given below:
 
Class Frequency (fi) Class mark (xi) fixi
10.5−15.5 2 13 26
15.5−20.5 3 18 54
20.5−25.5 6 23 138
25.5−30.5 7 28 196
30.5−35.5 14 33 462
35.5−40.5 12 38 456
40.5−45.5 4 43 172
45.5−50.5 2 48 96
Total ∑ fi = 50   ∑ fixi = 1600

Mean=ifixiifi        =160050        =32

Thus, mean of the given data is 32.

Now, to find the median let us put the data in the table given below:
 
Class Frequency (fi) Cumulative frequency (cf)
10.5−15.5 2 2
15.5−20.5 3 5
20.5−25.5 6 11
25.5−30.5 7 18
30.5−35.5 14 32
35.5−40.5 12 44
40.5−45.5 4 48
45.5−50.5 2 50
Total N = ∑ fi = 50  

Now, N = 50 N2=25.

The cumulative frequency just greater than 25 is 32, and the corresponding class is 30.5−35.5.

Thus, the median class is 30.5−35.5.

∴ l = 30.5, h = 5, N = 50, f = 14 and cf = 18.


Now, 

Median=l+N2-cff×h          =30.5+25-1814×5          =30.5+3514          =33

Thus, the median is 33.



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