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प्रश्न
The following table gives the distribution of IQ's (intelligence quotients) of 60 pupils of class V in a school:
| IQ's: | 125.5 to 13.25 |
118.5 to 125.5 |
111.5 to 118.5 |
104.5 to 111.5 |
97.5 to 104.5 |
90.5 to 97.5 |
83.5 to 90.5 |
76.5 to 83.5 |
69.5 to 76.5 |
62.5 to 69.5 |
| No. of pupils: |
1 | 3 | 4 | 6 | 10 | 12 | 15 | 5 | 3 | 1 |
Draw a frequency polygon for the above data.
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उत्तर १
We first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively.
The given data is a continuous grouped frequency distribution with equal class-intervals. To draw the frequency polygon of the given data without using histogram, obtain the class-limits of the class intervals. Obtain the class-limits of two class-intervals of 0 frequencies, i.e. on the horizontal axis, one adjacent to the first, on its left and one adjacent to the last, on its right. These class-intervals are known as imagined class-intervals. Then plot the frequencies against class-limits.
The following table is useful to draw the frequency polygon of the given data.
| Class - Intervals | Class - Marks | Frequency |
| 55.5-62.5 | 59 | 0 |
| 62.5-69.5 | ||
We represent class marks on X-axis on a suitable scale and the frequencies on Y-axis on a suitable scale.
To obtain the frequency polygon we plot the points (66, 1), (73, 3), (80, 5), (87, 15), (94, 12), (101, 10), (108, 6), (115, 4), (122, 3), (129, 1).
Now we join the plotted points by line segments. The end points (66, 1) and (129, 1) are joined to the mid points (59, 0) and ( 136, 0) respectively of imagined class intervals to obtain the frequency polygon.
उत्तर २
We first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively.
The given data is a continuous grouped frequency distribution with equal class-intervals. To draw the frequency polygon of the given data without using histogram, obtain the class-limits of the class intervals. Obtain the class-limits of two class-intervals of 0 frequencies, i.e. on the horizontal axis, one adjacent to the first, on its left and one adjacent to the last, on its right. These class-intervals are known as imagined class-intervals. Then plot the frequencies against class-limits.
The following table is useful to draw the frequency polygon of the given data.
| Class - Intervals | Class - Marks | Frequency |
| 55.5-62.5 | 59 | 0 |
| 62.5-69.5 | 66 | 1 |
| 69.5 - 76.5 | 73 | 3 |
| 76.5 - 83.5 | 80 | 5 |
| 83.5 - 90.5 | 87 | 15 |
| 90.5 - 97.5 | 94 | 12 |
| 97.5-104.5 | 101 | 10 |
| 104-111.5 | 108 | 6 |
| 111.5 - 118.5 | 115 | 4 |
| 118.5-125.5 | 122 | 3 |
| 125.5-132.5 | 129 | 1 |
| 132.5-139.5 | 136 | 0 |
We represent class marks on X-axis on a suitable scale and the frequencies on Y-axis on a suitable scale.
To obtain the frequency polygon we plot the points (66, 1), (73, 3), (80, 5), (87, 15), (94, 12), (101, 10), (108, 6), (115, 4), (122, 3), (129, 1).
Now we join the plotted points by line segments. The end points (66, 1) and (129, 1) are joined to the mid points (59, 0) and ( 136, 0) respectively of imagined class intervals to obtain the frequency polygon.
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संबंधित प्रश्न
Read the following bar graph(Fig. 23.15) and answer the following questions:
(i) What information is given by the bar graph?
(ii) What was the production of a student in the year 1980 - 81?
(iii) What is the minimum and maximum productions of cement and corresponding years?
The following data gives the amount of manure (in thousand tonnes) manufactured by a company during some years:
| Year | 1992 | 1993 | 1994 | 1995 | 1996 | 1997 |
| Manure (in thousand tonnes) |
15 | 35 | 45 | 30 | 40 | 20 |
(i) Represent the above data with the help of a bar graph.
(ii) Indicate with the help of the bar graph the year in which the amount of manufactured by the company was maximum.
(iii) Choose the correct alternative:
The consecutive years during which there was maximum decrease in manure production are:
(a) 1994 and 1995
(b) 1992 and 1993
(c) 1996 and 1997
(d) 1995 and 1996
The following data gives the value (in crores of rupees) of the Indian export of cotton textiles for different years:
| Years | 1982 | 1983-1984 | 1984-1985 | 1985-1986 | 1986-1987 |
| Value of Export of Cotton Textiles (in crores of rupees) |
300 | 325 | 475 | 450 | 550 |
Represent the above data with the help of a bar graph. Indicate with the help of a bar graph the year in which the rate of increase in exports is maximum over the preceding year.
Draw frequency polygons for each of the following frequency distribution:
(a) using histogram
(b) without using histogram
|
C.I |
10 - 30 |
30 - 50 |
50 - 70 | 70 - 90 | 90 - 110 | 110 - 130 | 130 - 150 |
| ƒ | 4 | 7 | 5 | 9 | 5 | 6 | 4 |
For the following table, draw a bar-graph
| A | B | C | D | E | F |
| 230 | 400 | 350 | 200 | 380 | 160 |
Read the following bar graph and answer the following questions:
a. What information is given by the graph?
b. Which state is the largest producer of wheat?
c. Which state is the largest producer of sugar?
d. Which state has total production of wheat and sugar as its maximum?
e. Which state has the total production of wheat and sugar minimum?
The frequency distribution has been represented graphically as follows:
| Marks | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 100 |
| Number of Students | 10 | 15 | 20 | 25 |
Do you think this representation is correct? Why?
Is it correct to say that in a histogram, the area of each rectangle is proportional to the class size of the corresponding class interval? If not, correct the statement.
In the following figure, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table.
The marks obtained (out of 100) by a class of 80 students are given below:
| Marks | Number of students |
| 10 – 20 | 6 |
| 20 – 30 | 17 |
| 30 – 50 | 15 |
| 50 – 70 | 16 |
| 70 – 100 | 26 |
Construct a histogram to represent the data above.
