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EXERCISES

1. Which average would be suitable in the following cases?

(i)

Average size of readymade garments.

(ii)

Average intelligence of students in a class.

(iii)

Average production in a factory per shift.

(iv)

Average wage in an industrial concern.

(v)

When the sum of absolute deviations from average is least.

(vi)

When quantities of the variable are in ratios.

(vii)

In case of open-ended frequency distribution.

(i)

Average Size of Readymade Garments: Mode: The most common size (the size that occurs most frequently) is the best measure here. It helps identify the size that is most in demand.

(ii)

Average Intelligence of Students in a Class: Median: Given that intelligence scores can vary widely and may not be symmetrically distributed, the median provides a more accurate central value, as it is not skewed by extremely high or low scores.

(iii)

Average Production in a Factory per Shift: Arithmetic Mean: It provides an overall average production rate, factoring in the total production across all shifts and dividing by the number of shifts.

(iv)

Average Wage in an Industrial Concern: Arithmetic Mean: Useful for calculating the average wage across all employees, taking into account the total sum of wages and dividing by the number of employees.

(v)

When the Sum of Absolute Deviations from Average is Least: Arithmetic Mean: This is the point at which the sum of the deviations of all data points from the average is minimized, especially when considering squared deviations.

(vi)

When Quantities of the Variable are in Ratios: Median: Suitable for ratio variables, especially when there’s a concern about skewed distributions or extreme values. It ensures that the measure is not disproportionately influenced by outlier values.

(vii)

In Case of Open-Ended Frequency Distribution: Median: This is a practical choice in open-ended distributions, as it doesn’t require knowledge of the extreme values. It divides the dataset into two equal halves and is not affected by the lack of boundaries in the distribution.

These choices are based on the statistical properties of each measure of central tendency and how they apply to the nature of the data in each scenario.

2. Indicate the most appropriate alternative from the multiple choices provided against each question.

(i)

The most suitable average for qualitative measurement is

(a)

arithmetic mean

(b)

median ✔

(c)

mode

(d)

geometric mean

(e)

none of the above

(ii)

Which average is affected most by the presence of extreme items?

(a)

median

(b)

mode

(c)

arithmetic mean ✔

(d)

none of the above

(iii)

The algebraic sum of deviation of a set of {n} values from {A.M.} is

(a)

{n}

(b)

0 ✔

(c)

1

(d)

none of the above

(i)

The most suitable average for qualitative measurement is

●

(b) median

●

Explanation: The Median is most suitable for qualitative measurements because it divides the series into two equal parts. This makes it particularly effective for ordinal data, where the data can be ranked but not necessarily quantified, as it is not influenced by extreme values.

(ii)

Which average is affected most by the presence of extreme items?

●

(c) arithmetic mean

●

Explanation: The Arithmetic Mean is highly sensitive to extreme values in the data set. Any significantly high or low values can skew the mean, making it not representative of the central tendency if these outliers are present.

(iii)

The algebraic sum of deviation of a set of {n} values from {A.M.} is

●

(b) 0

●

Explanation: A fundamental property of the Arithmetic Mean is that the sum of the deviations of each value from the mean adds up to zero. This characteristic is a defining feature of the Arithmetic Mean in a data set.

3. Comment whether the following statements are true or false.

(i)

The sum of deviation of items from median is zero.

(ii)

An average alone is not enough to compare series.

(iii)

Arithmetic mean is a positional value.

(iv)

Upper quartile is the lowest value of top 25% of items.

(v)

Median is unduly affected by extreme observations.

(i)

The sum of deviation of items from median is zero.

●

False: This statement is not true. The sum of deviations from the median is not necessarily zero. This property holds true for the arithmetic mean, not the median. The median is the middle value of a dataset and is not calculated based on the sum of deviations.

(ii)

An average alone is not enough to compare series.

●

True: This statement is considered true. While averages provide a summary measure of data, they do not give a complete picture. Additional information, such as the distribution of data, variability, and presence of outliers, is also important for comprehensive comparison and analysis of series.

(iii)

Arithmetic mean is a positional value.

●

False: This statement is false. The arithmetic mean is not a positional value; it is a measure calculated by summing all the values in a data set and then dividing by the number of values. Positional values are values that depend on the position within a sorted list, such as the median or quartiles.

(iv)

Upper quartile is the lowest value of the top 25% of items.

●

True: This statement is true. The upper quartile, or the third quartile ({Q_3}), marks the boundary of the highest 25% of values in a dataset. It is the median of the upper half of the dataset.

(v)

Median is unduly affected by extreme observations.

●

False: This statement is false. One of the advantages of the median is that it is not unduly affected by extreme values (or outliers). Unlike the mean, the median provides a central value that is more resistant to skewness in the data.

4. 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 ₹)

(in ₹)

0-10

10-20

20-30

30-40

40-50

50-60

Number of retail shops

12

18

27

–

17

6

Let’s denote the number of retail shops as {f} (frequency)

Let us assume the missing frequency as {f_i}

(a) To find the missing frequency:

Let’s now calculate the mid-values ({m}) and the product of mid-values and frequency i.e., {fm}.

Profit

per

Retail Shop

(in ₹)

per

Retail Shop

(in ₹)

No. of

Retail Shops

({f})

Retail Shops

({f})

Mid-Value

({m})

({m})

{fm}

(2) × (3)

(2) × (3)

(1)

(2)

(3)

(4)

0-10

12

5

60

10-20

18

15

270

20-30

27

25

675

30-40

{f_i}

35

{35f_i}

40-50

17

45

765

50-60

6

55

330

{∑f = 80 + f_i}

{∑fm = 2100 + 35f_i}

Now, the arithmetic mean is given as

{\overline{X} = \dfrac{∑fm}{∑f}}

{⇒ 28 = \dfrac{2100 + 35f_i}{80 + f_i}}

{⇒ 28 × (80 + f_i) = 2100 + 35f_i}

{⇒ 2240 + 28f_i = 2100 + 35f_i}

Rearranging the terms, we have,

{2240 - 1200 = 35f_i - 28f_i}

{⇒ 140 = 7f_i}

{⇒ 7f_i = 140}

{⇒ f_i = \dfrac{140}{7}}

{⇒ f_i = 20}

(b) To find the median of the series:

To find the median, let’s calculate the cumulative frequency {c.f.}. Also, note that, going forward we’ll use the missing frequency value that is found out in the previous calculation.

Profit

per

Retail Shop

(in ₹)

per

Retail Shop

(in ₹)

No. of

Retail Shops

({f})

Retail Shops

({f})

Cumulative

Frequency

({c.f.})

Frequency

({c.f.})

(1)

(2)

(3)

0-10

12

12

10-20

18

30

{(c.f)}

{(c.f)}

20-30

{(L = 20)}

{(L = 20)}

27

{(f)}

{(f)}

57

30-40

20

77

40-50

17

94

50-60

6

100

{∑f = N = 100}

Now, the median is given as

Median

{= \text{Value of } \left(\dfrac{N}{2}\right)^{th} \text{ item}}

{= \text{Value of } \left(\dfrac{100}{2}\right)^{th} \text{ item}}

= Value of 50th item

The 50th item will be in the class interval whose cumulative frequency is 57 i.e., in the class interval 20-30

Now, the median is calculated as

{\text{Median } = L + \dfrac{\dfrac{N}{2} - c.f.}{f}} × h

We know that

{L}

=

Lower limit of the median class

=

20

{c.f.}

=

Cumulative Frequency of the class preceding the median class

=

30

{f}

=

frequency of the median class

=

27

{h}

=

magnitude of the median class interval

=

10

Substituting these value, we have,

Median

{= L + \left(\dfrac{\dfrac{N}{2} - c.f.}{f}\right) × h}

{= 20 + \left(\dfrac{\dfrac{100}{2} - 30}{27}\right) × 10}

{= 20 + \left(\dfrac{50 - 30}{27}\right) × 10}

{= 20 + \left(\dfrac{20}{27}\right) × 10}

{= 20 + \dfrac{200}{27}}

= 20 + 7.41

= 27.41

∴ The value of the missing frequency is 20 and value of the median is ₹ 27.41

5. 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 ₹)

(in ₹)

120

150

180

200

250

300

220

350

370

260

Workers

Daily Income

(in ₹)

{(X)}

(in ₹)

{(X)}

A

120

B

150

C

180

D

200

E

250

F

300

G

220

H

350

I

370

J

260

{∑X = 2400}

In this case, the total number of workers is given as

{N = 10}

Now, the formula for the arithmetic mean is given as

{\overline{X} = \dfrac{∑X}{N}}

Substituting the values, we have

{\overline{X}}

{= \dfrac{2400}{10}}

= 240

∴ The arithmetic mean is ₹ 240

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

Income

(in ₹)

(in ₹)

Number

of

families

of

families

More than 75

150

More than 85

140

More than 95

115

More than 105

95

More than 115

70

More than 125

60

More than 135

40

More than 145

25

Note that in the problem the number of families whose income is more than a given value is given. If we carefully analyze the data, we see that the cumulative frequencies ({c.f.}) are given. So, from these we find the actual frequencies {(f)}.

From the total number of families whose income is more than 75 if we subtract the number of families whose income is more than 85, we get the families whose income is between 75 and 85 i.e., we get the frequency of the class interval 75-85.

Applying this to all the given data,

Income

Class Interval

Class Interval

No.of

Families

{c.f.}

Families

{c.f.}

Frequency

{(f)}

{(f)}

Mid-Value

{(m)}

{(m)}

{fm}

(3) × (4)

(3) × (4)

(1)

(2)

(3)

(4)

(5)

75-85

150

150 – 140 = 10

80

800

85-95

140

140 – 115 = 25

90

2250

95-105

115

115 – 95 = 20

100

2000

105-115

95

95 – 70 = 25

110

2750

115-125

70

70 – 60 = 10

120

1200

125-135

60

60 – 40 = 20

130

2600

135-145

40

40 – 25 = 15

140

2100

145-155

25

25

150

3750

{∑f = 150}

{∑fm = 17,450}

Now, the arithmetic mean {A.M.} is calculated as below:

{A.M.}

{= \dfrac{∑fm}{∑f}}

{= \dfrac{17450}{150}}

= 116.33

So, the average income or the arithmetic mean of the families’ income is ₹ 116.33

7. 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)

of

Land Holdings

(in acres)

Less than 100

100–200

200–300

300–400

400 and above.

Number of families

40

89

148

64

39

Let’s first find the cumulative frequency.

Size of

Land Holdings

(in acres)

Land Holdings

(in acres)

No. of

Families

{(f)}

Families

{(f)}

Cumulative

Frequency

{(c.f.)}

Frequency

{(c.f.)}

0-100

40

40

100-100

89

129

{(c.f.)}

{(c.f.)}

200-300

{(L = 200)}

{(L = 200)}

148

{(f)}

{(f)}

277

300-400

64

341

400-500

39

380

{∑f = 380}

So, we have {∑f = N = 380}

Now, the median class can be found as

Median

{= \text{Value of }\left(\dfrac{N}{2}\right)^{th}\text{item}}

{= \text{Value of }\left(\dfrac{380}{2}\right)^{th}\text{item}}

= Value of 190th item

So, the 190th item lies in the class whose cumulative frequencey is 277 i.e., the class interval 200-300

Now, the median is calculated as

{\text{Median } = L + \left(\dfrac{\dfrac{N}{2} - c.f.}{f}\right) × h}

We know that

{L}

=

Lower limit of the median class

=

200

{c.f.}

=

Cumulative Frequency of the class preceding the median class

=

129

{f}

=

frequency of the median class

=

148

{h}

=

magnitude of the median class intervalue

=

100

Substituting these value, we have,

Median

{= L + \left(\dfrac{\dfrac{N}{2} - c.f.}{f}\right) × h}

{= 200 + \left(\dfrac{\dfrac{380}{2} - 129}{148}\right) × 100}

{= 200 + \left(\dfrac{190 - 129}{148}\right) × 100}

{= 200 + \left(\dfrac{61}{148}\right) × 100}

= 200 + 41.22

= 241.22

∴ The median size of the land holdings is 241.22 acres.

8. 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 ₹)

(in ₹)

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.)

Indirectly, this problem is asking us to compute median, lower quartile and upper quartile. So, let’s first calculate the cumulative frequencies.

Also note that the class intervals are adjusted to 9.5-14.5, 14.5-19.5, etc., to help in eliminating biases and inaccuracies that can occur due to the exclusive nature of the original intervals and also to ensure a more accurate and continuous representation of the data.

(a) To find the highest income of lowest 50% workers i.e., median.

Daily Income

(in ₹)

(in ₹)

No.of

Workers

{(f)}

Workers

{(f)}

Cumulative

Frequency

{(c.f.)}

Frequency

{(c.f.)}

9.5-14.5

5

5

14.5-19.5

10

15

19.5-24.5

15

30

{(c.f.)}

{(c.f.)}

24.5-29.5

{(L = 24.5)}

{(L = 24.5)}

20

{(f)}

{(f)}

50

29.5-34.5

10

60

34.5-39.5

5

65

{∑f = 65}

So, we have {∑f = N = 65}

So,

Median

{= \text{Value of}\left(\dfrac{N}{2}\right)^{th}\text{item}}

{= \text{Value of}\left(\dfrac{65}{2}\right)^{th}\text{item}}

= Value of 37.5th item

Now, the 37.5th item lies in the class whose cumulative frequency is 50 i.e., the class interval 24.5-29.5

Now, the median is calculated as

{\text{Median } = L + \left(\dfrac{\dfrac{N}{2} - c.f.}{f}\right) × h}

We know that

{L}

=

Lower limit of the median class

=

24.5

{c.f.}

=

Cumulative Frequency of the class preceding the median class

=

30

{f}

=

frequency of the median class

=

20

{h}

=

magnitude of the median class interval

=

5

Substituting these value, we have,

Median

{= L + \left(\dfrac{\dfrac{N}{2} - c.f.}{f}\right) × h}

{= 24.5 + \left(\dfrac{\dfrac{65}{2} - 30}{20}\right) × 5}

{= 24.5 + \left(\dfrac{32.5 - 30}{20}\right) × 5}

{= 24.5 + \left(\dfrac{2.5}{20}\right) × 5}

= 24.5 + 0.625

= 25.13

∴ The highest income of lowest 50% workers is ₹ 25.13

(b) To find the minimum income earned by the top 25% workers i.e., lower quartile.

Note: Note that the cumulative frequency table is provided again below, to highlight the lower quartile related values (for your reference). This need not be done while actually solving the problem.

Daily Income

(in ₹)

(in ₹)

No.of

Workers

{(f)}

Workers

{(f)}

Cumulative

Frequency

{(c.f.)}

Frequency

{(c.f.)}

9.5-14.5

5

5

14.5-19.5

10

15

{(c.f.)}

{(c.f.)}

19.5-24.5

{(L = 19.5)}

{(L = 19.5)}

15

{(f)}

{(f)}

30

24.5-29.5

20

50

29.5-34.5

10

60

34.5-39.5

5

65

{∑f = 65}

So, we have {∑f = N = 65}

So,

Lower Quartile {Q_1}

{= \text{Value of}\left(\dfrac{N}{4}\right)^{th}\text{item}}

{= \text{Value of}\left(\dfrac{65}{4}\right)^{th}\text{item}}

= Value of 16.25th item

Now, the 16.25th item lies in the class whose cumulative frequency is 30 i.e., the class interval 19.5-24.5

Now, the lower quartile ({Q_1}) is calculated as

{Q_1 = L + \left(\dfrac{\dfrac{N}{4} - c.f.}{f}\right) × h}

We know that

{L}

=

Lower limit of the lower quartile i.e. {Q_1} class

=

19.5

{c.f.}

=

Cumulative Frequency of the class preceding the lower quartile class

=

15

{f}

=

frequency of the lower quartile class

=

15

{h}

=

magnitude of the lower quartile class interval

=

5

Substituting these value, we have,

{Q_1}

{= L + \left(\dfrac{\dfrac{N}{4} - c.f.}{f}\right) × h}

{= 19.5 + \left(\dfrac{\dfrac{65}{4} - 15}{15}\right) × 5}

{= 19.5 + \left(\dfrac{16.25 - 15}{15}\right) × 5}

{= 19.5 + \left(\dfrac{1.25}{15}\right) × 5}

= 19.5 + 0.42

= 19.92

∴ The minimum income earned by the top 25% workers is ₹ 19.92

(c) To find the maximum income earned by the lowest 25% workers i.e., upper quartile.

Note: Note that the cumulative frequency table is provided again below, to highlight the upper quartile related values (for your reference). This need not be done while actually solving the problem.

Daily Income

(in ₹)

(in ₹)

No.of

Workers

{(f)}

Workers

{(f)}

Cumulative

Frequency

{(c.f.)}

Frequency

{(c.f.)}

9.5-14.5

5

5

14.5-19.5

10

15

19.5-24.5

15

30

{(c.f.)}

{(c.f.)}

24.5-29.5

{(L = 24.5)}

{(L = 24.5)}

20

{(f)}

{(f)}

50

29.5-34.5

10

60

34.5-39.5

5

65

{∑f = 65}

So, we have {∑f = N = 65}

So,

Upper Quartile {Q_3}

{= \text{Value of}\left(\dfrac{3N}{4}\right)^{th}\text{item}}

{= \text{Value of}\left(\dfrac{3 × 65}{4}\right)^{th}\text{item}}

{= \text{Value of}\left(\dfrac{195}{4}\right)^{th}\text{item}}

= Value of 48.75th item

Now, the 48.75th item lies in the class whose cumulative frequency is 50 i.e., the class interval 24.5-29.5

Now, the upper quartile ({Q_3}) is calculated as

{Q_3 = L + \left(\dfrac{\dfrac{3N}{4} - c.f.}{f}\right) × h}

We know that

{L}

=

Lower limit of the first quartile i.e. {Q_3} class

=

24.5

{c.f.}

=

Cumulative Frequency of the class preceding the first quartile class

=

30

{f}

=

frequency of the first quartile class

=

20

{h}

=

magnitude of the first quartile class interval

=

5

Substituting these value, we have,

{Q_3}

{= L + \left(\dfrac{\dfrac{3N}{4} - c.f.}{f}\right) × h}

{= 24.5 + \left(\dfrac{\dfrac{3 × 65}{4} - 30}{20}\right) × 5}

{= 24.5 + \left(\dfrac{\dfrac{195}{4} - 30}{20}\right) × 5}

{= 24.5 + \left(\dfrac{48.75 - 30}{20}\right) × 5}

{= 24.5 + \left(\dfrac{18.75}{20}\right) × 5}

= 24.5 + 4.69

= 29.19

∴ The maximum income earned by the lowest 25% workers is ₹ 29.19

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

Production yield

(kg. per hectare)

(kg. per hectare)

50–53

53–56

56–59

59–62

62–65

65–68

68–71

71–74

74–77

Number

of

farms

of

farms

3

8

14

30

36

28

16

10

5

To calculate the Mean

Note to Students/Learners:Let’s use the Step Deviation Method to solve this problem. As you’re aware, this method is used when there’re large data sets or large numerical values or when it makes sense to further simplify the calculations. However, let’s use this method in this case, even though the data is small and not that complex to ensure that it fits our learning objective. We solved all the earlier problems using the Direct Method. So, to get you acquainted with the Step Deviation method, let’s use it for solving this problem.

Also note that in this case, the mid-value {A = 63.5} is considered as the Assumed Mean.

Production

Yield

(kg/hectare)

Yield

(kg/hectare)

No. of

Farms

{(f)}

Farms

{(f)}

Mid-Value

{X}

{X}

{d = X - A}

{d' = \dfrac{d}{c}}

{fd'}

50-53

3

51.5

-12

-4

-12

53-56

8

54.5

-9

-3

-24

56-59

14

57.5

-6

-2

-28

59-62

30

60.5

-3

-1

-30

62-65

36

63.5

{A}

{A}

0

0

0

65-68

28

66.5

+3

+1

28

68-71

16

69.5

+6

+2

32

71-74

10

72.5

+9

+3

30

74-77

5

75.5

+12

+4

20

{∑f = 150}

{∑fd' = 16}

Now, the arithmetic mean {\overline{X}} is calculated as

{\overline{X}}

{= A + \dfrac{∑fd'}{∑f} × c}

{= 63.5 + \dfrac{16}{150} × 3}

= 63.5 + 0.32

= 63.82

So, the average production yield is 63.82 kg/hectare

To calculate the median:

To calculate the median, let’s find out the cumulative frequencies.

Production

Yield

(kg/hectare)

Yield

(kg/hectare)

No. of

Farms

{(f)}

Farms

{(f)}

Cumulative

Frequency

{(c.f.)}

Frequency

{(c.f.)}

50-53

3

3

53-56

8

11

56-59

14

25

59-62

30

55

{(c.f)}

{(c.f)}

62-65

{(L = 62)}

{(L = 62)}

36

{(f)}

{(f)}

91

65-68

28

119

68-71

16

135

71-74

10

145

74-77

5

150

{∑f = 150}

The median is given as

Median

= Value of {\left(\dfrac{N}{2}\right)^{th}} item

= Value of {\left(\dfrac{150}{2}\right)^{th}} item

= Value of 75th item

The 75th item will be in the class interval whose cumulative frequency is 91 i.e., in the class interval 62-65

Now, the median is calculated as

{\text{Median } = L + \dfrac{\dfrac{N}{2} - c.f.}{f}} × h

We know that

{L}

=

Lower limit of the median class

=

62

{c.f.}

=

Cumulative Frequency of the class preceding the median class

=

55

{f}

=

frequency of the median class

=

36

{h}

=

magnitude of the median class interval

=

3

Substituting these value, we have,

Median

{= L + \left(\dfrac{\dfrac{N}{2} - c.f.}{f}\right) × h}

{= 62 + \left(\dfrac{\dfrac{150}{2} - 55}{36}\right) × 3}

{= 62 + \left(\dfrac{75 - 55}{36}\right) × 3}

{= 62 + \left(\dfrac{20}{36}\right) × 3}

= 62 + 1.67

= 63.67

∴ The median of the production yield is 63.67 kg/hectare

To calculate the mode:

If we carefully analyze the data in the given table, we see that the modal class interval is 62.65 as it has the highest value of frequence i.e., 36.

Production

Yield

(kg/hectare)

Yield

(kg/hectare)

Frequency

{(f)}

{(f)}

50-53

3

53-56

8

56-59

14

59-62

30

{(f_0)}

{(f_0)}

62-65

{(L = 62)}

{(L = 62)}

36

{(f_1)}

{(f_1)}

65-68

28

{(f_2)}

{(f_2)}

68-71

16

71-74

10

74-77

5

Now, the mode is calculated as

{\text{Mode } = L + \dfrac{D_1}{D_1 + D_2} × h}

We know that

{L}

=

Lower limit of the modal class

=

62

{D_1}

=

Difference between the frequency of the modal class and the frequency of the class preceding the modal class (ignoring signs)

=

{f_1 - f_0}

=

36 – 30

=

6

{D_2}

=

Difference between the frequency of the modal class and the frequency of the class succeeding the modal class (ignoring signs)

=

{f_1 - f_2}

=

36 – 28

=

8

{h}

=

magnitude of the modal class interval

=

65 – 62

=

3

Substituting these value, we have,

Median

{= L + \left(\dfrac{D_1}{D_1 + D2}\right) × h}

{= 62 + \left(\dfrac{6}{6 + 8}\right) × 3}

{= 62 + \left(\dfrac{6}{14}\right) × 3}

= 62 + 1.29

= 63.29

∴ The mode of the production yield is 63.29 kg/hectare