Nm Shah 2018 Solutions for Class 11 Commerce Economics Chapter 1 Organisation Of Data are provided here with simple stepbystep explanations. These solutions for Organisation Of Data are extremely popular among Class 11 Commerce students for Economics Organisation Of Data Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Nm Shah 2018 Book of Class 11 Commerce Economics Chapter 1 are provided here for you for free. You will also love the adfree experience on Meritnation’s Nm Shah 2018 Solutions. All Nm Shah 2018 Solutions for class Class 11 Commerce Economics are prepared by experts and are 100% accurate.
Page No 128:
Question 1:
The frequency distribution of marks obtained by students in a class test is given below:
Marks  :  010  1020  2030  3040  4050 
No. of Students  :  3  10  14  10  3 
Answer:
Histogram is a joining rectangular diagram with equal class interval of size 10.
Page No 128:
Question 2:
Present the data given in the table below in the form of a Histogram :
Midpoints  :  115  125  135  145  155  165  175  185  195 
Frequency  :  6  25  48  72  116  60  38  22  3 
Answer:
The twodimensional diagrams that depict the frequency distribution of a continuous series by the means of rectangles are called histograms.
In order to construct a histogram, we first require the class intervals corresponding to the various midpoints, which is calculated using the following formula.
The value obtained is then added to the mid point to obtain the upper limit and subtracted from the midpoint to obtain the lower limit.
For the given data, the class interval is calculated by the following value of adjustment.
$\mathrm{Value}\mathrm{of}\mathrm{Adjustment}=\frac{125115}{2}=5$
Thus, we add and subtract 5 to each midpoint to obtain the class interval.
For instance:
The lower limit of first class = 115 – 5 = 110
Upper limit of first class = 115 + 5 = 120.
Thus, the first class interval is (110120). Similarly, we can calculate the remaining class intervals.
Midpoints  Class Interval  Frequency 
115 125 135 145 155 165 175 185 195 
110 − 120 120 − 130 130 − 140 140 − 150 150 − 160 160 − 170 170 − 180 180 − 190 190 − 200 
6 25 48 72 116 60 38 22 3 
Page No 128:
Question 3:
Make a frequency Polygon and Histogram using the given data:
Marks obtained  :  1020  2030  3040  4050  5060  6070 
No. of Students  :  5  12  15  22  14  4 
Answer:
Page No 128:
Question 4:
Draw Histogram from the following data:
Marks obtained  :  1020  2030  3040  4050  5070  70100 
No. of Students  :  6  10  15  10  6  3 
Answer:
The data is given in the form of unequal class interval. So, we will first make appropriate adjustment in the frequencies to make the class intervals equal. The general formula for the adjustment of the frequency is as follows.
$\mathrm{Adjusted}\mathrm{Frequency}=\frac{\mathrm{Required}\mathrm{class}\mathrm{interval}\times \mathrm{Frequency}}{\mathrm{Actual}\mathrm{class}\mathrm{interval}}$
Marks  No. of Students  Adjusted Frequency 
10−20  6  − 
20−30  10  − 
30−40  15  − 
40−50  10  − 
50−70  6  $\frac{10\times 6}{20}=3$ 
70−100  3  $\frac{10\times 3}{30}=1$ 
Page No 128:
Question 5:
In a certain colony a sample of 40 households was selected . The data on daily income for this sample are given as follows:
200  120  350  550  400  140  350  85 
180  110  110  600  350  500  450  200 
170  90  170  800  190  700  630  170 
210  185  250  120  180  350  110  250 
430  140  200  400  200  400  210  300 
(b) Show that the area under the polygon is equal to the area under the histogram.
Answer:
Income  Tally Marks  Frequency  Area for each class 
0 − 100  2  100 × 2 = 200  
100 − 200  14  100 × 14 = 1400  
200 − 300  8  100 × 8 = 800  
300 − 400  5  100 × 5 = 500  
400 − 500  5  100 × 5 = 500  
500 − 600  2  100 × 2 = 200  
600 − 700  2  100 × 2 = 200  
700 − 800  1  100 × 1 = 100  
800 − 900  1  100 × 1 = 100  
40  Total area =4000 
(b) The area under a histogram and under a frequency polygon is the same (i.e equal to 4000) because of the fact that we extend the first class interval to the left by half the size of class interval as the starting point of the frequency polygon. Similarly, the last class interval is extended to the right by the same amount as the end point of the frequency polygon. This ensures that the area that was excluded while joining the midpoints is included in the frequency polygon such that the area under the frequency polygon and the area of histogram is the same.
Page No 128:
Question 6:
Present the data given in the table below in Histogram:
Marks  :  2529  3034  3539  4044  4549  5054  5559 
Frequency  :  4  5  23  31  10  8  5 
Answer:
Before proceeding to construct histogram, we first need to convert the given inclusive series into an exclusive series using the following formula.
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.
$\mathrm{Here},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{adjustment}=\frac{3029}{2}=0.5$
Therefore, we add 0.5 to the upper limit and subtract 0.5 from the lower limit of each class.
Marks  Frequency 
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 
4 5 23 31 10 8 5 
Page No 128:
Question 7:
A survey showed that the average daily expenditures (in rupees) of 30 households in a city were:
11, 12, 14, 16, 16, 17, 18, 18, 20, 20, 20, 21, 21, 22, 22, 23, 23, 24, 25, 25, 26, 27, 28, 28, 31, 32, 32, 33, 36, 36.
(a) Prepare a frequency distribution using class intervals:
1014,  1519,  2024,  2529,  3034  and  3539. 
(b) What percent of the households spend more than ₹ 29 each day?
(c) Draw a frequency histogram for the above data.
Answer:
(a)
Class Interval  Tally Marks  Frequency 
10 − 14 15 − 19 20 − 24 25 − 29 30 − 34 35 − 39 
3 5 10 6 4 2 

30 
(b) Households that spend more than 29 each day$=\frac{6}{30}\times 100=20\%$
(c) To construct histogram, we first need to convert the given inclusive series into an exclusive series using the following formula.
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.
$\mathrm{Here},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{adjustment}=\frac{1514}{2}=0.5$
Therefore, we add 0.5 to the upper limit and subtract 0.5 from the lower limit of each class.
Class Interval  Frequency 
9.5 − 14.5 14.5 − 19.5 19.5 − 24.5 24.5 − 29.5 29.5 − 34.5 34.5 − 39.5 
3 5 10 6 4 2 
30 
Page No 128:
Question 8:
Draw Histogram from a given data relating to monthly pocket money allowance of the students of class XII in a school:
Size of classes( in ₹ )  05  510  1015  1520  2025  2530  3035  3540 
No of students  5  10  15  20  25  15  10  5 
Answer:
Page No 129:
Question 9:
Draw a Histogram and Frequency Polygon of the following information.
Wages in Rupees  :  7580  8085  8590  9095  95100  100105  105110  110115 
No. of workers  :  9  12  15  11  20  20  11  2 
Answer:
Page No 129:
Question 10:
Draw ogive$\u2014$(a) less than, and (b) more than of the folowing data:
Weekly wages of No of workers  :  100105 
105110 
110115  115120  120125  125130  130135 
Workers (₹)  :  200  210  230  320  350  520  410 
Answer:
In order to construct the ogives, we first need to calculate the less than and the more than cumulative frequencies as as follows.
Weekly Wages  Cumulative Frequency 
Less than 105 Less than 110 Less than 115 Less than 120 Less than 125 Less than 130 Less than 135 
200 200 + 210 = 410 410 + 230 = 640 640 + 320 = 960 960 + 350 = 1410 1410 + 520 = 1930 1930 + 410 = 2240 
Weekly Wages  Cumulative Frequency 
More than 100 More than 105 More than 110 More than 115 More than 120 More than 125 More than 130 
2240 2240$$200 = 2040 2040$$210 = 1830 1830$$230 = 1600 1600$$320 = 1280 1280$$350 = 930 930$$520 = 410 
Page No 129:
Question 11:
Prepare a less than ogive from the following data:
Class  :  06  612  1218  1824  2430  3036 
Frequency  :  4  8  15  20  12  6 
Answer:
In order to prepare a less than ogive, we first need to calculate the less than cumulative frequency distribution as follows:
Class  Cumulative Frequency 
Less than 6 Less than 12 Less than 18 Less than 24 Less than 30 Less than 36 
4 4 + 8 = 12 12 + 15 = 27 27 + 20 = 47 47 + 12 = 59 59 + 6 = 65 
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.
Page No 129:
Question 12:
From the following frequency distribution prepare a 'less than ogive'.
Capital (₹ in lakhs)  :  010  1020  2030  3040  4050  5060  6070  7080 
No. of companies  :  2  3  7  11  15  7  2  3 
Answer:
For constructing a less than ogive, we convert the frequency distribution into a less than cumulative frequency distribution as follows:
Capital  Cumulative Frequency 
Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 
2 2 + 3 = 5 5 + 7 = 12 12 + 11 = 23 23 + 15 = 38 38 + 7 = 45 45 + 2 = 47 47 + 3 = 50 
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.
Page No 129:
Question 13:
Arrange the following information on a timeseries graph:
Year  :  200910  201011  201112  201213  201314  201415  201516 
NDP(₹ in '000 crores)  :  35  36  37  40  41  44  44 
Answer:
Time Series Graph
Page No 129:
Question 14:
Plot the following data of annual profits of a firm on a time series graph.
Year  :  2012  2013  2014  2015  2016  2017 
Profits(₹ in thousand)  :  60  72  75  65  80  95 
Answer:
Time Series Graph
Page No 129:
Question 15:
Prepare a suitable graph from the following data relating to export and import.
Year  201011  201112  201213  201314  201415  201516  201617 
Exports (₹ in crores)  1505  2265  2070  1805  1632  1527  1845 
Imports( ₹ in crores)  1005  1145  1980  1335  1547  1435  1740 
Answer:
The given data can be presented in the form of a time series graph as follows:
Here, the smallest value is 1,005 which is far higher than zero. Therefore, in this case we use a false base line starting from 1,000.
Page No 129:
Question 16:
Present graphically the following sales of Delhi branch of USHA FANS.
Year  :  2008  2009  2010  2011  2012  2013  2014  2015  2016 
Sales (₹ in '000) 
:  13  15  12  19  25  31  29  27  35 
Answer:
The given data can be presented in the form of a time series graph as follows:
Page No 129:
Question 17:
Prepare a graph showing total cost and total production of a scooter manufacturing company.
Year  :  2012  2013  2014  2015  2016 
Production (in units)  :  8500  9990  11700  13300  15600 
Total cost (₹ in lakh)  :  24  29  34  45  49 
Answer:
The given data can be presented in the form of a time series graph (i.e multiple yaxis graph) as follows:
View NCERT Solutions for all chapters of Class 13