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Page No 213:

Question 1:

Calculate median of the following data:

145 130 200 210 198 234 159 160 178
257 260 300 345 360 390      

Answer:

Arranging the data in ascending order.

130, 145, 159, 160, 178, 198, 200, 210, 234, 257, 260, 300, 345, 360, 390

  N=15Median=Size of N+12th item           =Size of 15+12th item           =Size of 8th item = 210Therefore, median is 210.

Page No 213:

Question 2:

Find out median of the following information:

Marks : 10, 70, 50, 20, 95, 55, 42, 60, 48, 80

Answer:

Arranging the data in ascending order.

10, 20, 42, 48, 50, 55, 60, 70, 80, 95

Here, the number of observations is even.
Median=Size of N2th item+Size of N2+1th item2      =Size of 102th item+Size of 102+1th item2      =Size of 5th item+Size of 6th item2      =50+552=52.5
Thus, median marks is 52.5.

Page No 213:

Question 3:

We have the following frequency distribution of the size of 51 households . Calculate the arithmetic mean and the median.

Size 2 3 4 5 6 7 Total
Number of households 2 3 9 21 11 5 51

Answer:

Size

(X)

Households

(f)

fX

c.f

 

2

2

4

2

 

3

3

9

5

 

4

9

36

14

 

5

21

105

35

 

6

11

66

46

 

7

5

35

51

 

 

Ʃf = 51

Ʃfx = 255

 

 

X¯=ΣfXΣf=25551= 5Median = Size of N+12th item=51+12th=26th item
Now, we need to look at the column of cumulative frequency. The item just exceeding the 26th item is 35, which corresponds to 5.
Hence, the median is 5.
 

Page No 213:

Question 4:

Find out median:
(a)

Serial No. 1 2 3 4 5 6 7 8 9
Values 2 4 10 8 15 20 12 25 30
 
(b) 
X 5 10 15 20 25
f 2 4 10 8 15
 

Answer:

(a) For the calculation of median the given values are first arranged in ascending order as follows.
 

Serial Number Values Ascending Order
1
2
3
4
5
6
7
8
9
2
4
10
8
15
20
12
25
30
2
4
8
10
12
15
20
25
30
Here, N = 9
Median =Size of N+12thitem              = 9+12th            =5th item            =12
Thus, median is 12.

(b)

X

f

c.f

 

5

2

2

 

10

4

6

 

15

6

12

 

20

8

20

 

25

10

30

 


Median=Size of  N+12thitem           = Size of 30+12th           =Size of 15.5th item
Now, we need to look at the column of cumulative frequency. The item just exceeding the 15.5th item is 20, which corresponds to 20.
Thus, median is 20.

Page No 213:

Question 5:

Find out median , first quartile and third quartile of the following series:

Height (in inches) 58 59 60 61 62 63 64 65 66
No of Persons 2 3 6 15 10 5 4 3 1
 

Answer:

 

Height

(X)

Persons

(f)

c.f

 

58

2

2

 

59

3

5

 

60

6

11

 

61

15

26

 

62

10

36

 

63

5

41

 

64

4

45

 

65

3

48

 

66

1

49

 


(i)Median
Median = Size of N+12thitem            = Size of 49+12th=25th item 
This corresponds to height 61. Thus, the median is 61.

(ii) First Quartile
Q1= Size of N+14thitem    =Size of 49+14th item=12.5th item 
This corresponds to height 61. Thus, first quartile is 61.

(iii) Third Quartile
Q3= Size of 3N+14thitem    =Size of 349+14th=37.5th item 
This corresponds to height 63. Thus, third quartile is 63.
 

Page No 213:

Question 6:

The marks obtained by 68 students in an examination are given below. Compute the median.

Marks Below 20 20-40 40-60 60-80 Above 80
No of students 0 5 22 25 16
 

Answer:

 

Marks

Students

(f)

c.f

 

Below  20

0

0

 

20 – 40

5

5

 

40 – 60

22

 27

 

60 –80

25

52

 

Above 80

16

68

 

Median class is given by the size of the N2th item, i.e. the 682th item, which is the 34th item. This corresponds to the class interval 60−80, so this is the median class.

Median = l1+N2-c.ff× i            =60+34-2725×20            =60+14025            = 60 + 5.6            = 65.6

Thus, median is 65.6.



Page No 214:

Question 7:

Calcualte the mean of the following distribution of daily wages of workers in a factory:

Daily Wages (in ₹) : 100-120 140-160 160-180 180-200 Total
No of Workers : 10 30 15 5 80

Also, calculate the median for the distribution of wages given above.

Answer:

Wages

 

Worker

(f)

X

fX

c.f

100 – 120

10

110

1100

10

120-140 20 130 2600 30

140 – 160

30

150

4500

60

160 – 180

15

170

2550

75

180 – 200

5

190

950

80

 

80

 

Ʃ fX = 11700

 

X¯ =ΣfxΣf=1170080=146.25Median class is given by the size of N2th item, i.e size of 802th, which is the 40th item. This corresponds to the class interval 140-160, so this is the median class. M = l1+N2th-c.ff×i     = 140 +40-3030×20=140+6.66     =146.66

Note: In the above calculation we have assumed a class interval (120-140) and its corresponding frequency as 20.

Page No 214:

Question 8:

The following table gives the marks obtained by 65 students in statistics in a certain examination . Calculate the median.

Marks No of Students
More than 70 8
More than 60 18
More than 50 40
More than 40 45
More than 30 50
More than 20 63
More than 10 65
 

Answer:

Marks Students
c.f
More than 70
More than 60
More than 50
More than 40
More than 30
More than 20
More than 10
8
18
40
45
50
63
65

Now the more than cumulative frequency distribution can be converted into class intervals as follows.
 

C.I.

 f

c.f

 

10 – 20

2

2

 

20 – 30

13

15

 

30 – 40

5

20

 

40 – 50

5

25 

 

50 – 60 

22

47

 

60 – 70

10

57

 

70 and above

8

65

 

 

Ʃf = 65

 

 

Median class is given by the size of the N2th item, i.e. the 652th item, which is the 32.5th item. This corresponds to the class interval 50−60, so this is the median class.

M =l1+N2-c.ff×i    =50+32.5-2522×10    =50+7522=50+3.4    =53.4
Thus, the median marks is 53.4.

Page No 214:

Question 9:

Calculate the Arithmetic mean and median of the following frequency distribution:

Class Interval 10-20 20-40 40-70 70-120 120-200 Total
Frequency 4 10 26 8 2 50
 

Answer:

Class
Interval

 

 f

X

(Midpoint)

fX

c.f

10 – 20

4

15

60

4

20 – 40

10

30

300

14

40 – 70

26

55

1430

40

70 – 120

8

95

  760

48

120 – 200

2

160

  320

50

 

Ʃf = 50

 

ƩfX = 2870

 

X¯=ΣfXΣf=287050=57.4
Median class is given by the size of the N2th item, i.e. the 502th item, which is the 25th item. This corresponds to the class interval 40−70, so this is the median class.
Median =l1+N2-c.ff×i            =40+25-1426×30            =40+33026=40+12.69            =52.69
Thus, mean and median are 57.4 and 52.69 respectively.

Page No 214:

Question 10:

An analysis for more efficiency in a factory , indicating the distribution of ages of workers was as follows:

Age (in years) 16-19 20-29 30-39 40-49 50-59 60-64
Frequency 15 46 49 32 28 14
(a) Calculate the mean and median of the above data.
(b) Draw a histogram and indicate mode therein.

Answer:

(a)

Age

Frequency (f)

Midpoint 

(m)

fm

16 – 19

15

17.5

262.5

20 – 29

46

24.5

1127

30 – 39

49

34.5

1690.5

40 – 49

32

44.5

1424

50 – 59

28

54.5

1526

60 – 64

14

62

868

 

f=184

 

fm=6898

Mean

X¯=ΣfmΣf   =6898184   =37.48

Hence, the mean of the given distribution is 37.48

Median

Note that the given distribution is in the form of inclusive class intervals. For the calculation of median, first the class intervals must be converted into exclusive form using the following formula.
Value of Adjustment = Lower limit of one class - Upper limit of the preceeding class2
 Value of lower limit of one class  Value of upper limit of the preceeding class2
The value of adjustment as calculated is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.
Age

Frequency
(f)

Cumulative
Frequency

(c.f)

15.5 – 19.5
19.5 – 29.5
L129.5 – 39.5
39.5 – 49.5
49.5 – 59.5
59.5 – 64.5
15
46
        49 f1
32
28
14
15
         61c.f
110
142
170
184

Median class is given by the
Size of 
N2th item = 1842th item = 92th item.
This corresponds to the class interval of (29.5 39.5), so this is the median class.
Median=L1+N2-c.ff×iso, Median=29.5+1842-6149×10or, Median=29.5+3149×10 Median= 29.5+6.32 =35.826

Hence, the median is 35.826

(b)
Here the data is in the form of unequal class interval. So, we will first make appropriate adjustment in the frequencies to make the class intervals equal.
Age

Frequency
(f)

Adjusted
Frequency

15.5 – 19.5

19.5 – 29.5
29.5 – 39.5
39.5 – 49.5
49.5 – 59.5
59.5 – 64.5
15

46
49
32
28
14

15×104=37.5
-
-
-
-
14×105=28
 



From the above graph, it can be seen that mode is 30.5. However, mean can not be determined graphically.

Page No 214:

Question 11:

The marks obtained by 100 students  in an examination are given below. Compute the median, 1st and 3rd Quartiles.

Marks 30-35 35-40 40-45 45-50 50-55 55-60 60-65
No of students 14 16 18 23 18 8 3
 

Answer:

 

Marks

 f

c.f

 

   30 – 35

14 

  14

 

 35 – 40

16

  30

 

  40 – 45

18

 48

 

45 – 50

 23

71

 

50 – 55

18

89

 

  55 – 60

8

97

 

  60 – 65

3

100

 

 
N=100
Median class is given by the size of the N2th item, i.e. the 1002th item, which is the 50th item. This corresponds to the class interval 45−50, so this is the median class.
Median =l1+N2-c.ff×i            =45+50-4823×5            =45+223×5            =45+0.43            = 45.43 Median= 45.43

First quartile class is given by the size of the N4th item, i.e. the 1004th item, which is the 25th item. This corresponds to the class interval 35−40, so this is the first quartile class.

Q1=l1+N4-c.ff×i    =35+25-1416×5    =35+5516   =35+3.43   =38.43Q1=38.43

Third quartile class is given by the size of the 3N4th item, i.e. the 31004th item, which is the 75th item. This corresponds to the class interval 50−55, so this is the third quartile class.
Q3=l1+3N4-c.ff×i    =50+75-7118×5    =50+2018    =50+1.11    =51.11Q3=51.11

Page No 214:

Question 12:

Compute mode  from the following series:

Size  of items : 2 3 4 5 6 7
Frequency : 3 8 10 12 16 14
Size of items : 8 9 10 11 12 13
Frequency : 10 8 17 5 4 1
 

Answer:

Size of Items

Frequency

 

2

3

 

3

8

 

4

10

 

5

12

 

6

16

 

7

14

 

8

10

 

9

8

 

10

17

→ Modal class

11

5

 

12

4

 

13

1

 


Here, mode is equal to 10 as it has the highest frequency.



Page No 215:

Question 13:

Calcualte mode for the following data:

No. of persons 1 2 3 4 5 6 7 8 9 10
Families 26 113 120 95 60 42 21 14 5 4
 

Answer:

No. of Persons

(X)

Families

(f)

 

1

26

 

2

113

 

3

120

Modal class

4

 95

 

5

60

 

6

42

 

7

21

 

8

14

 

9

 5

 

10

 4

 

Here, mode is equal to 3 as it has the highest frequency.

Page No 215:

Question 14:

Find out the Mode from any of the following two distributions:
(a) 

X : 30-40 40-50 50-60 60-70 70-80 80-90 90-100
f : 6 10 16 14 10 5 2

(b) 
Marks : 0-9 10-19 20-29 30-39 40-49
No of candidates : 6 29 87 181 247
Marks : 50-59 60-69 70-79 80-89 90-99
No of candidates : 263 113 49 9 2
 

Answer:

(a)

X

f

 

   30 – 40

 6

 

   40 – 50

 10

 

  50 – 60

 16

Modal class

   60 – 70

 14

 

   70 – 80

10

 

   80 – 90

5

 

   90 – 100

2

 

By inspection, we can spot that (50–60) has the highest frequency, i.e. 16. So, this is the modal class.

Mode=l1+f1-f02f1-f0-f2×i        =50+16-102×16-10-14×10        =50+608        =50+7.5        =57.5

Thus, the mode is 57.5.

(b)
Since this is an inclusive class interval series, to calculate mode, first we need to convert the inclusive class intervals into the exclusive class intervals by using the following formula.

Value of Adjustment=Lower limit of ine class-Upper limit of the preceding class2Valueofadjustment=LowerlimitofoneclassUpperlimitofthepreceedingclass2

The value of adjustment (as calculated) is then added to the upper limit of each class and subtracted from the lower limit of each class. In this manner, we get the following distribution.
 

Marks

f

 

–0.5 – 9.5

  6

 

9.5 – 19.5

 29

 

19.5 – 29.5

 87

 

29.5 – 39.5

181

 

39.5 – 49.5

 247

 

49.5 – 59.5

263

Modal class

59.5 – 69.5

 113

 

69.5 – 79.5

 49

 

79.5 – 89.5

   9

 

89.5 – 99.5

   2

 

By inspection, we can spot that (49.5 – 59.5) has the highest frequency, i.e.263. So, this is the modal class.
Mode=l1+f1-f02f1-f0-f2×i        =49.5+263-2472×263-247-113×10        =49.5+160166       =49.5+0.963      =50.46

Thus, the mode is 50.46.

Page No 215:

Question 15:

Life of electric lamps is given in the following table. Calculate the median and the mode.

Life in  hours Number of lamps
Below 400 4
400-800 12
800-1200 40
1200-1600 41
1600-2000 27
2000-2400 13
2400-2800 9
Above 2800 4
 

Answer:

Class Interval

 f

c.f.

 

Below 400

4

     4

 

400 – 800

12

    16

 

800 – 1200

40

    56

 

1200 – 1600

41

   97

 

1600 – 2000

27

   124

 

2000 – 2400

13

   137

 

2400 – 2800

9

  146

 

above 2800

4

  150

 

Median class is given by the size of the N2th item, i.e. the 1502th item, which is the 75th item. This corresponds to the class interval 1200−1600, so this is the median class.
Median= l1+N2-c.f.f×i           =1200+75-5641×400           =1200+760041           =1200+185.36            =1385.36 Median =1385.36

By inspection, we can spot that (1200–1600) has the highest frequency, i.e. 41. So, this is the modal class.
Mode =1200+41-402×41-40-27×400          =1200+40015          =1200+26.66          =1226.66 Mode =1226.66

Page No 215:

Question 16:

Calculate Mean, Median and Mode fromt he following data:

Marks (mid points) No of Students
59 1
61 2
63 9
65 48
67 131
69 102
71 40
73 17
  Total = 350
 

Answer:

Converting the series into exclusive series
 

C.I.

m

f.

fm

c.f

 

   58 – 60

59

   1

  59

1

 

   60 – 62

61

   2

122

3

 

   62 – 64

63

   9

567

12

 

   64 – 66

65

    48

3120

60

 

  66 – 68

67

   131

8777

191

 

   68 – 70

69

  102

7038

293

 

   70 – 72

71

 40

2840

333

 

   72 – 74

73

 17

1241

350

 

 

 

Ʃf = 350

Ʃfm23764

 

 


Mean:X =ΣfmΣf=23764350=67.89Median:Median class is given by size of N2th item i.e. =3502th=175th item This item lies in the c.f group of 191. So, the median class (66-68).Median=l1+N2-c.ff×i           =66+175-60131×2           =66+230131=66+1.75           =67.75

Mode:Mode =l1+f1-f02f1-f0-f2×i          =66+131-482×131-48-102×2           = 66+166112          =66+1.48          =67.48

Page No 215:

Question 17:

Following is the distribution of marks of 50 students in a class:

Marks (more than) : 0 10 20 30 40 50
No of students : 50 46 40 20 10 3
Calculate the Median Marks . If 60% of students pass this examination , find out the minimum marks obtained by a pass candidate.
 

Answer:

Convert the series into exclusive class interval
 

C.I.

more than

c.f.

f

less than

c.f

 

     0 – 10

50

  4

 4

 

   10 – 20

46

  6

     10

 

20 – 30

40

  20

30

→ Median class

   30 – 40

20

10

40

 

   40 – 50

10

 7

47

 

   50 – 60

3

3

50

 

N = 50
Median class is given as size of N2th item i.e. 25th item which falls in 30th cumulative frequency group.Median=l1+N2-c.ff×i           =20+25-1020×10            =20+15020           =20+7.5           =27.5 Median=27.5

If 60% students pass, it means 40% students fail. Thus, we have to find D4.
4th decile class is given by the size of 4N10th item i.e. 20th item which falls in 30 cumulative frequency group.
D4=20+20-1020×10    =20+10020   =20+5   =25 D4=25
 



Page No 216:

Question 18:

The age in completed years of 50 persons is given below:

32 61 52 56 22 49 97 35 30 30 95 67 42 20
31 64 20 10 62 60 27 53 31 90 54 25 43 47
35 21 43 75 45 22 36 13 46 23 51 11 15 39
50 42 77 73 81 40 40 55            

Prepare frequency table taking 10-19, 20-29, ,.......... class intervals and calculate modal age.

Answer:

Age

Tall marks

No. of persons

(f)

Exclusive

Class Interval

 

10 – 19

4

  9.5 – 19.5

 

20 – 29


8

   19.5 – 29.5

 

30 – 39


9

   29.5 – 39.5

 

40 – 49

10

   39.5 – 49.5

Modal class

50 – 59


  7 

    49.5 – 59.5

 

60 – 69

5

    59.5 – 69.5

 

70 – 79


3

    69.5 – 79.5

 

80 – 89

1

    79.5 – 89.5

 

90 – 99


2

    89.5 – 99.5

 

 

 

Ʃf =49

 



 


By inspection, we can spot that (39.5–49.5) has the highest frequency, i.e. 10. So, this is the modal class.
Mode = l1+f1-f02f1-f0-f2×i          =39.5+10-92×10-9-7×10          =39.5+104          =39.5+2.5          =42
Thus, the modal age is 42 years.

Page No 216:

Question 19:

Determine the value of mode for the following data by using the formula:
Mode = 3 Median  − 2   Mean.

Marks No. of students
Less than 10 5
Less than 20 15
Less than 30 98
Less than 40 242
Less than 50 367
Less than 60 405
Less than 70 425
Less than 80 438
Less than 90 439
 

Answer:

 

Marks

c.f.

f

m

fm

 

     0 – 10

5

5

5

   25

 

   10 – 20

15

  10

15

  150

 

   20 – 30

     98

  83

25

2075

 

30 – 40

242

  144

35

5040

 

   40 – 50

367

125

45

5625

 

   50 – 60

405

 38

55

2090

 

   60 – 70

425

20

65

1300

 

   70 – 80

438

13

75

  975

 

   80 – 90

439

 1

85

85

 

 

 

Ʃf  = 439

 

Ʃfm =17365

 

X¯=ΣfmΣf=17365439=39.55
Median class is given by the size of the N2th item, i.e. the 4392th item, which is the 219.5th item. This corresponds to the class interval 30−40, so this is the median class.
Median=l1+N2-c.ff×i            =30+219.5-98144×10            =30+1215144            =30+8.43            =38.43

Now,
Mode = 3 Median – 2 Mean
      = 3 × 38.43 – 2× 39.55
      = 115.29 – 79.1
Mode = 36.19

Page No 216:

Question 20:

For the data given below find graphically the following :
(a) First and Third quartile
(b) The central 50% limit of the age .
(c) The number of workers falling in the age group of 28 to 57 years.

Age in years : 20-24 25-29 30-34 35-39 40-44 45-49
No.of workers : 5 10 15 25 65 40
 
Age in years : 50-54 55-59 60-64 65-69
No.of workers : 23 10 5 2
 

Answer:

(a)
In order to find quartiles graphically, we first convert the inclusive series into exclusive series and then convert the frequency distribution into cumulative frequency distribution as follows:
 

Age Age Frequency
(f)

 
Cumulative Frequency
(c.f.)
19.5 – 24.5
24.5 – 29.5
29.5 – 34.5
34.5 – 39.5
39.5 – 44.5
44.5 – 49.5
49.5 – 54.5
54.5 – 59.5
59.5 – 64.5
64.5 – 69.5
Less than 24.5
Less than 29.5
Less than 34.5
Less than 39.5
Less than 44.5
Less than 49.5
Less than 54.5
Less than 59.5
Less than 64.5
Less than 69.5
5
10
15
25
65
40
23
10
5
2
5
15
30
55
120
160
183
193
198
200

 

Thus, the first and third quartiles are 38.5 and 48.25 respectively.

(b) The two limits (i.e. Q1 and Q3) with which central 50% of items lie are 38.5 and 48.25.

(c)


     From the graph it is clear that total Number of workers of age 57 years = 190
 Total Number of workers of age 28 years = 11
 Number of workers falling in the age group of 28 to 57 years = 190 -11=179
 

Page No 216:

Question 21:

Draw a 'less than' ogive from the following data and hence find out the value of Median.
 

Class 20-25 25-30 30-35 35-40 40-45 45-50 50-55 55-60
Frequency 6 9 13 23 19 15 9 6

Answer:

For constructing a less than ogive, first the given frequency distribution is converted into a less than cumulative frequency distribution as follows:
 

Class Frequency
(f)
Cumulative Frequency
(
c.f.)
Less than 25
Less than 30
Less than 35
Less than 40
Less than 45
Less than 50
Less than 55
Less than 60
6
9
13
23
19
15
9
6
6
15
28
51
70
85
94
100

We now plot the cumulative frequencies against the upper limit of the class intervals. The curve obtained on joining the points so plotted is known as the less than ogive.


Hence, median is 39.78.

Page No 216:

Question 22:

The following table gives the distribution of the wages of 65 employees in a factory.

Wages (in ₹ )     : 50 60 70 80 90 100 110 120
(Equal to more than)
Number of employees
: 65 57 47 31 17 7 2 0
Draw a 'less than' curve from the above data and estimate the number of employees earning atleast ₹ 63 but less than ₹ 75.

Answer:

Class
Interval

 
More than Cumulative Frequency Frequency Wages Less than Cumulative Frequency
50-60
60-70
70-80
80-90
90-100
100-110
110-120
120-130
65
57
47
31
17
7
2
0
65-57= 8
57-47=10
47-31=16
31-17=14
17-7= 10
7-2=5
2-0=2
0
Less than 60
Less than 70
Less than 80
Less than 90
Less than 100
Less than 110
Less than 120
Less than 130
8
18
34
48
58
63
65
65




Number of employees earning Rs 75 = 24
Number of employees earning atleast Rs 63 = 10
Number of employees earning between Rs 63 and Rs 75 = 24 -10 =14

Therefore, 14 employees are earning between Rs 63 and Rs 75.



Page No 217:

Question 23:

Draw the histogram and estimate the value of mode from the following data:

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
No of students 0 2 3 7 13 11 9 2 1

Answer:



Hence, mode is 47.5.

Page No 217:

Question 24:

Represent the following data by means of a histogram  and find out mode.

Weekly wages : 10-15 15-20 20-25 25-30 30-35 35-40 40-45
No of Workers : 7 19 27 15 12 12 8
 

Answer:




Hence, mode is 22.

Page No 217:

Question 25:

For a particular set of data in individual incomes the 'less than type' Ogive and the 'more than type' Ogive were found to intersect at ₹ 715. Discuss the significance of '₹ 715' for the given set of data.

Answer:

In a graphical presentation of a data, the place where the 'less than type' ogive and 'more than type' ogive intersect is the middle most point of the data. Thus, we may say that the point at which the two ogives intersect represents the median of the given data.

In the given question, the 'less than' and the 'more than' ogives intersect at Rs715.
Therefore, 715 is the median of the given data. It divides the whole data into two  equal parts.

Page No 217:

Question 26:

Calculate missing frequencies, if median value is 28 and N =44
 

Class 0-10 10-20 20-30 30-40 40-50 50-60
No of students 3 ? 15 9 ? 4

Answer:

Class

Frequency

(f)

m

c.f

0 – 10

3

5

3

10 – ​20 ? (f1) 15 3 + f1

20 – 30

15

25

18 + f1

30 – 40 9 35 27 + f1

40 – 50

? (f2)

45

27 + f1 + f2

50 – 60

4

55

44

 

44

 

 

We know, total frequency = 44
So, 31 + f1 f2 = 44
or, f1 + f2 = 13

Median is given as 28.Median class is given by the size of N2th item, i.e size of 442th, which is the 22nd item. This corresponds to the class interval 20-30, so this is the median class. M = l1+N2th-c.ff×i28= 20 +22-(3+f1)15×1028-20=22-(3+f1)15÷108×1.5=22-3-f1f1=7

As, we know f1 + f2 = 13, and f1 is 7, so, f2 becomes (13-7 =) 6 .
 

Page No 217:

Question 27:

Calculate Median from the data given below:

Class Interval 0-10 10-20 20-40 40-60 60-80 80-100
Frequency 10 8 22 10 20 10

Answer:

Class

Frequency

(f)

c.f

0 – 10

10

10

 

10 – ​20 8 18

20 – 40

22

40

40 – 60 10 50

60 – 80

20

70

80 – 100

10

80

 

80

 


Median class is given by the size of N2th item, i.e size of 802th, which is the 40th item. This corresponds to the class interval 20-40, so this is the median class. M = l1+N2th-c.ff×iM= 20 +40-1822×20M=40

 

Page No 217:

Question 28:

Calculate mode from  the following data:

Marks (more than) 0-9 10-19 20-29 30-39 40-49 50-59
No of students 2 5 15 25 10 3

Answer:

 

X

f

 

   0 – 9.5

 2

 

9.5 – 19.5 5  

  19.5 – 29.5 

 15

 

   29.5 – 39.5

 25

Modal class

  39.5 – 49.5

 10

 

    49.5 – 59.5

3

 


By inspection, we can spot that (29.5–39.5) has the highest frequency, i.e. 25. So, this is the modal class.

Mode=l1+f1-f02f1-f0-f2×i        =29.5+25-152×25-15-10×10        =29.5+1025×10        =33.5

Thus, the mode is 33.5.

Page No 217:

Question 29:

Calculate mode from the following data:

Class Interval 0-10 10-20 20-30 30-40 40-50 50-70
Frequency 7 10 8 20 32 28

Answer:

 

X

f

 

0 – 10

 7

 

10 – 20 10  
20 – 30 8  

30 – 40 

20

 

40 – 50

32

Modal class

50 – 70

28

 


By inspection, we can spot that (40 – 50) has the highest frequency, i.e. 32. So, this is the modal class.

Mode=l1+f1-f02f1-f0-f2×i        =40+32-202×32-20-14×10        =40+1230×10        =44

Thus, the mode is 44.
Note, f2 will be 14 as the adjusted frequency for the class 50 - 70 is 14.

Page No 217:

Question 30:

Calculate mode from following data:

Class Interval 5-14 15-24 25-34 35-44 44-54 55-64
Frequency 8 10 15 25 40 20

 

Answer:

 

X

f

 

4.5 – 14.5

8

 

14.5 – 24.5 10  
24.5 – 34.5 15  

 34.5 – 44.5 

25

 

44.5 – 54.5

40

Modal class

54.5 – 64.5

20

 


By inspection, we can spot that (44.5–54.5) has the highest frequency, i.e. 40. So, this is the modal class.

Mode=l1+f1-f02f1-f0-f2×i        =44.5+40-252×40-25-20×10        =44.5+1535×10        =48.78

Thus, the mode is 48.78.



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