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प्रश्न
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.
उत्तर
See the given table, the class intervals are of unequal width.
So, we calculate the adjusted frequency for each class.
Now, minimum size = 20 – 10 = 10
The formula of adjusted frequencies are:
Adjusted frequency = `("Minimum class" - "size")/("Class" - "size") xx` Frequency of the class
Now, the modified table for frequency distribution is given by:
Marks | Numbers of students (Frequency) |
Adjusted frequency |
10 – 20 | 6 | `10/10 xx 6 = 6` |
20 – 30 | 17 | `10/10 xx 17 = 17` |
30 – 50 | 15 | `10/20 xx 15 = 15/2 = 7.5` |
50 – 70 | 16 | `10/20 xx 16 = 16/2 = 8` |
70 – 100 | 26 | `10/30 xx 26 = 26/3 = 8.67` |
Now, let’s construct rectangles with class limits as base and respective adjusted frequencies as height
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संबंधित प्रश्न
The following table gives the distribution of students of two sections according to the mark obtained by them:-
Section A | Section B | ||
Marks | Frequency | Marks | Frequency |
0 - 10 | 3 | 0 - 10 | 5 |
10 - 20 | 9 | 10 - 20 | 19 |
20 - 30 | 17 | 20 - 30 | 15 |
30 - 40 | 12 | 30 - 40 | 10 |
40 - 50 | 9 | 40 - 50 | 1 |
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
The following data gives the number (in thousands) of applicants registered with an
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Number of applicants registered (in thousands) | 18 | 20 | 24 | 28 | 30 | 34 |
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Year | 1992 | 1993 | 1994 | 1995 | 1996 |
Loan (in crores of rupees) |
28 | 33 | 55 | 55 | 80 |
(i) Represent the above data with the help of a bar graph.
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Which one of the following is not the graphical representation of statistical data:
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32 | 16 | 24 | 20 | 8 |
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Class interval (km/h) | Frequency |
30 – 40 | 3 |
40 – 50 | 6 |
50 – 60 | 25 |
60 – 70 | 65 |
70 – 80 | 50 |
80 – 90 | 28 |
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30 – 40 | 3 |
40 – 50 | 6 |
50 – 60 | 25 |
60 – 70 | 65 |
70 – 80 | 50 |
80 – 90 | 28 |
90 – 100 | 14 |
Draw the frequency polygon representing the above data without drawing the histogram.