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Page No 128:

Question 1:

The frequency distribution of marks obtained by students in a class test is given below:

Marks : 0-10 10-20 20-30 30-40 40-50
No. of Students : 3 10 14 10 3
Draw a histogram to represent the frequency distribution  of marks . Comment on the shape of the histogram.
 

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 :

Mid-points : 115 125 135 145 155 165 175 185 195
Frequency : 6 25 48 72 116 60 38 22 3

Answer:

The two-dimensional 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 mid-points, 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 mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.
Value of Adjustment = 125 -115 2 = 5

Thus, we add and subtract 5 to each mid-point 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 (110-120). Similarly, we can calculate the remaining class intervals.

Mid-points 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 : 10-20 20-30 30-40 40-50 50-60 60-70
No. of Students : 5 12 15 22 14 4

Answer:

    

Page No 128:

Question 4:

Draw Histogram from the following data:

Marks obtained : 10-20 20-30 30-40 40-50 50-70 70-100
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.
Adjusted Frequency=Required class interval × FrequencyActual class interval
 

Marks No. of Students Adjusted Frequency
10−20 6
20−30 10
30−40 15
40−50 10
50−70 6 10×620=3
70−100 3 10×330=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
(a) Construct a Histogram and a frequency Polygon.
(b) Show that the area under the polygon is equal to the area under the histogram.

Answer:

 
(a)
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 mid-points 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 : 25-29 30-34 35-39 40-44 45-49 50-54 55-59
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.
Here, the value of adjustment = 30 - 292 = 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:

10-14, 15-19, 20-24, 25-29, 30-34 and 35-39.

(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=630×100=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.
Here, the value of adjustment = 15 - 142 = 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 ₹ ) 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40
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 : 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115
No. of workers : 9 12 15 11 20 20 11 2

Answer:

Page No 129:

Question 10:

Draw ogive(a) less than, and (b) more than of the folowing data:

Weekly wages of No of workers  :  100-105

105-110               
110-115 115-120 120-125 125-130 130-135
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 : 0-6 6-12 12-18 18-24 24-30 30-36
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) : 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
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 time-series graph:

Year : 2009-10 2010-11 2011-12 2012-13 2013-14 2014-15 2015-16
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 2010-11 2011-12 2012-13 2013-14 2014-15 2015-16 2016-17
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 y-axis graph) as follows:



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