Advertisements
Advertisements
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.
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
Given below are the seats won by different political parties in the polling outcome of a state assembly elections:-
| Political Party | A | B | C | D | E | F |
| Seats Won | 75 | 55 | 37 | 29 | 10 | 37 |
- Draw a bar graph to represent the polling results.
- Which political party won the maximum number of seats?
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?
Read the bar graph given in Fig. 23.19 and answer the following questions:
(i) What information is given by the bar graph?
(ii) In which years the areas under the sugarcane crop were the maximum and the minimum?
(iii) State whether true or false:
The area under the sugarcane crop in the year 1982 - 83 is three times that of the year 1950 - 51
Read the bar graph given in Fig. 23.20 and answer the fol1owing questions:
(i) What information is given by the bar graph?
(ii) What was the expenditure on health and family planning in the year 1982-83?
(iii) In which year is the increase in expenditure maximum over the expenditure in previous year? What is the maximum increase?
The following data gives the number (in thousands) of applicants registered with an
| Year | 1995 | 1996 | 1997 | 1998 | 1999 | 2000 |
| Number of applicants registered (in thousands) | 18 | 20 | 24 | 28 | 30 | 34 |
Construct a bar graph to represent the above data.
The following table shows the interest paid by a company (in lakhs):
| Year | 1995-96 | 1996-97 | 1997-98 | 1998-99 | 1999-2000 |
| Interest (in lakhs of rupees | 20 | 25 | 15 | 18 | 30 |
Draw the bar graph to represent the above information.
The distribution of heights (in cm) of 96 children is given below. Construct a histogram and a frequency polygon on the same axes.
| Height (in cm): | 124 to 128 |
128 to 132 |
132 to 136 |
136 to 140 |
140 to 144 |
144 to 148 |
148 to 152 |
152 to 156 |
156 to 160 |
160 to 164 |
| No. of Children: | 5 | 8 | 17 | 24 | 16 | 12 | 6 | 4 | 3 | 1 |
In the 'less than' type of ogive the cumulative frequency is plotted against
A histogram is a pictorial representation of the grouped data in which class intervals and frequency are respectively taken along
The percentage of marks obtained, in different subjects by Ashok Sharma (in an examination) is given below. Draw a bar graph to represent it.
| English | Hindi | Maths | Science | Social Studies |
| 85 | 60 | 35 | 50 | 70 |
