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Question-1

If the arithmetic mean of the data given below is 28, find
(a) The missing frequency, and
(b) The median of the series:
Profit per retail shop (in Rs)  0-10  10-20  20-30  30-40  40-50  50-60
Number of retail shops             12      18        27       NIL     17        6

Solution:
(a)

Profit

(in Rs)

No of shops

Frequency (f)

Mid value

x

fx

0-10

10-20

20-30

30-40

40-50

50-60

12

18

27

P

17

6

5

15

25

35

45

55

60

270

675

35x

765

330

80+p

2100+35p

Given that Arithmetic mean of the data is 28.

Therefore,

⇒ 2100+35p = 2240+28p

(i.e) 7p = 140

⇒ p = 20

Thus the missing frequency is 20.

(b)

Profit (in Rs)

No of shops f

Cumulative frequency

0-10

10-20

20-30

30-40

40-50

50-60

12

18

27

20

17

6

12

30

57

77

94

100

Median =

= 20 + = = 27.407

Therefore Median = 27.41

Question-2

The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.
 
Workers A B C  D E F G H I J
Daily Income (in Rs) 120 150 180  200 250 300 220 350 370 260

                                                 

Solution:
Total Daily income for ten workers = = 2400

Therefore Arithmetic mean = = 240

Question-3


Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.


Solution:

Income

No of families f

Mid class x

fx

75 - 85

85 - 95

95-105

105-115

115-125

125-135

135-145

145-155

10

25

20

25

10

20

15

25

80

90

100

110

120

130

140

150

800

2250

2000

2750

1200

2600

2100

3750

150

17450

Arithmetic Mean = = 116.33

Question-4


4.  The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

Size of Land Holdings

(in acres)

Number of families

Less than 100

100-200

200-300

300-400

400 and above

40

89

148

64

39


Solution:

Size of land

(in acres)

No of families f

Mid value x

c.f

0 – 100

100-200

200-300

300-400

400-500

40

89

148

64

39

50

150

250

350

450

40

129

277

341

380

380

Median =

             = 200 + = 200 + 41.217 = 241.22

Thus, the median size of land holdings = 241.22 acres.

Question-5


5.  The following series relates to the daily income of workers employed in a firm. Compute

(a) Highest income of lowest 50% workers

(b) Minimum income earned by the top 25% workers and

(c) Maximum income earned by lowest 25% workers.

Daily Income (in Rs) 10–14 15–19 20–24 25–29 30–34 35–39

Number of workers        5        10        15        20       10         5

(Hint: compute median, lower quartile and upper quartile.)

[Ans. (a) Rs 25.11 (b) Rs 19.92 (c) Rs 29.19]


Solution:

Daily income (in Rs)

No of workers f

Mid- value x

fx

Cumulative

Frequency (c.f)

  9.5-14.5

14.5-19.5

19.5-24.5

24.5-29.5

29.5-34.5

34.5-39.5

  5

10

15 f1

20 f2

10 f3

5

12

17

22

27

32

37

60

170

330

540

320

185

5

15

30

50

60

65

65

1605

 

= 24.5 + = = 25.12

Lower Quartile Q1 =

=

= 19.5 + = 19.5 + 0.42 = 19.92

Upper Quartile Q3 =

=

= 29.5 + = 29.5 -

= = 28.87

Thus,

Highest income of lowest 50% workers = 25.12

Minimum income earned by the top 25% workers = 19.92

Maximum income earned by lowest 25% workers = 28.87

Question-6

The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode production yield.

Production yield        50–53 53–56 56–59 59–62 62–65 65–68 68–71 71–74 74–77

(Kg. per hectare)

Number of farms             3          8          14        30         36        28        16         10         5


Solution:

Production yield

(1 kg/hectare)

No of farms f

Mid class x

fx

Cumulative frequency c.f

50-53

53-56

56-59

59-62

62-65

65-68

68-71

71-74

74-77

3

8

14

30 f0

36 f1

28 f2

16

10

5

51.5

54.5

57.5

60.5

63.5

66.5

69.5

72.5

75.5

154.5

436

805

1815

2286

1862

1112

725

377.5

3

11

25

55

91

119

135

145

150

150

9573

Mean = = 63.82

Median =

              = 62 + = = 63.667

Mode =

        =

       = 62 + = = 63.286

Thus, Mean = 63.82 kg/hectare,

Median = 63.667 kg/hectare and Mode = 63.29 kg/hectare.

Question-7

What are the different types of mean and Explain them:

Solution:
Arithmetic mean, Assumed mean and Weighted mean.

Arithmetic mean:

Arithmetic mean is the most commonly used measure of central tendency. It is defined as the sum of the values of all observations divided by the number of observations and is usually denoted by x . In general, if there are N observations as X1, X2, X3, ..., XN, then the Arithmetic Mean is given by

     =

Where, = sum of all observations and N = total number of observations.

Assumed mean:

As in case of individual series the calculations can be simplified by using assumed mean method, as described earlier, with a simple modification. Since frequency (f) of each item is given here, we multiply each deviation (d) by the frequency to get fd. Then we get S fd. The next step is to get the total of all frequencies i.e. S f. Then find out S fd/S f. Finally the arithmetic mean is calculated by

using assumed mean method.

Weighted mean:

Sometimes it is important to assign weights to various items according to their importance, when you calculate the arithmetic mean. For example, there are two commodities, mangoes and potatoes. You are interested in finding the average price of mangoes (p1) and potatoes (p2).

The arithmetic mean will be. However, you might want to give more importance to the rise in price of potatoes (p2). To do this, you may use as ‘weights’ the quantity of mangoes (q1) and the quantity of potatoes (q2). Now the arithmetic mean weighted by the quantities would be.

Question-8

Write the relation among Mean, Median and Mode:

Solution:
Suppose we express, Arithmetic Mean = Me Median = Mi Mode = Mo so that e, i and o are the suffixes. The relative magnitude of the three are Me>Mi>Mo or Me<Mi<Mo (suffixes occurring in alphabetical order). The median is always between the arithmetic mean and the mode.

Question-9

What are the uses of central tendency?

Solution:
Measures of central tendency or averages are used to summarise the data with a single value. Which can represent the entire data. It specifies a single most representative value to describe the data set. Arithmetic mean is the most commonly used average. It is simple to calculate and is based on all the observations. But it is unduly affected by the presence of extreme items. Median is a better summary for such data. Mode is generally used to describe the qualitative data. Median and mode can be easily computed.

Question-10

What do you mean by quartiles?

Solution:
Quartiles are the measures which divide the data into four equal parts, each portion contains equal number of observations. Thus, there are three quartiles.
  The first Quartile (denoted by Q1) or lower quartile has 25% of the items of the distribution below it and 75% of the items are greater than it.
  The second Quartile (denoted by Q2) or median has 50% of items below it and 50% of the observations above it.
  The third Quartile (denoted by Q3) or upper Quartile has 75% of the items of the distribution below it and 25% of the items above it.
  Thus, Q1 and Q3 denote the two limits within which central 50% of the data lies.

Question-12

Which is affected by the extreme values in the data?

Solution:
Arithmetic mean.

Question-13

‘Percentiles’ which divides the data into how many parts?

Solution:
Hundred equal parts.

Question-14

Which measure is repeated the highest number of times in the series?

Solution:
Mode.

Question-15

Write empirical relation among mean, median and mode:

Solution:
2median = mode – 3mean.

Question-16

Which is the central value of the distribution?

Solution:
Median.




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