Tr Jain & Vk Ohri (2017) for Class 11 Science Economics Chapter 4 - Organisation Of Data

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Data of weekly expenditure on food
 

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i) In the form of a frequency distribution, the given data can be arranged as follows.


ii) With cumulative frequencies the given data can be arranged as follows.

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Individual series is simply the arrangement of the given data in ascending (or, descending order)
 

Discrete series:


Continuous series:


Cumulative frequency series

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In the given data, the lowest value is 15, while the highest value is 48. Accordingly, we take (15 - 20) as the initial class interval.

Weight (kg)	Tally Bars	No. of Students
15−20		2
20−25		1
25−30		2
30−35		2
35−40		4
40−45		4
45−50		5
Total		$\sum f=20$

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The given information can be summarised as follows.
 

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In the given question, the size of the items refers to the number of alphabets in the word. Accordingly, in the given statement there are eight items with size equal to 2, namely, in, no, to, is, an, as, be, is, in. Similarly, there are 5 items with size equal to three, namely, the, may, the, all, for.  In the similar manner, we can find the frequency corresponding to the various items. 
The frequency distribution table is prepared in the following manner.
 

Size of Item	Tally Bars	Frequency
2		9
3		5
4		2
5		2
6		1
7		3
8		3
9		3
10		2
11		3
Total		$\sum f=33$

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i)


ii) Number of families spending more than Rs 29 per day = 4+2 = 6
Thus, there are 6 families that spend more than Rs 29 per day.

Percentage of families spending more than Rs 29
$$\begin{gathered} =\frac{\mathrm{Number} \mathrm{of} \mathrm{families} \mathrm{spending} \mathrm{more} \mathrm{than} \mathrm{Rs} 29}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{families}}\times 100 \\ = \frac{6}{30}\times 100=20\% \\ \mathrm{Thus}, 20\% \mathrm{of} \mathrm{the} \mathrm{families} \mathrm{spend} \mathrm{more} \mathrm{than} \mathrm{Rs} 29 \mathrm{per} \mathrm{day}. \end{gathered}$$

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The class interval can be calculated from the mid-points using the following adjustment formula. 

The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.

Thus, we add and subtract 5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 5 – 5 = 0

Upper limit of first class = 5 + 5 = 10

Thus, the first class interval is (0 − 10). Similarly, we can calculate the remaining class intervals.
 

Answer:

The class interval can be calculated from the mid-points using the following adjustment formula. 

The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.
$\mathrm{Value} \mathrm{of} \mathrm{adjustment} = \frac{17 - 12}{2}= 2.5$

Thus, we add and subtract 2.5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 12 – 2.5 = 9.5

Upper limit of first class = 12 + 2.5 = 14.5

Thus, the first class interval is (9.5 − 14.5). Similarly, we can calculate the remaining class intervals.