Mathematics
Marks obtained by 200 students in an examination are given below:
| Marks | Number of students |
|---|---|
| 0 - 10 | 5 |
| 10 - 20 | 11 |
| 20 - 30 | 10 |
| 30 - 40 | 20 |
| 40 - 50 | 28 |
| 50 - 60 | 37 |
| 60 - 70 | 40 |
| 70 - 80 | 29 |
| 80 - 90 | 14 |
| 90 - 100 | 6 |
Draw an ogive for the given distribution taking 1 cm = 10 marks on one axis and 2 cm = 20 students on other axis.
Using the graph, determine :
(i) The median marks
(ii) The number of students who failed, if minimum marks required to pass is 40.
(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.
Measures of Central Tendency
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Answer
The cumulative frequency distribution table is as follows:
| Marks | Number of students | Cumulative frequency |
|---|---|---|
| 0 - 10 | 5 | 5 |
| 10 - 20 | 11 | 16 (5 + 11) |
| 20 - 30 | 10 | 26 (16 + 10) |
| 30 - 40 | 20 | 46 (26 + 20) |
| 40 - 50 | 28 | 74 (46 + 28) |
| 50 - 60 | 37 | 111 (74 + 37) |
| 60 - 70 | 40 | 151 (111 + 40) |
| 70 - 80 | 29 | 180 (151 + 29) |
| 80 - 90 | 14 | 194 (180 + 14) |
| 90 - 100 | 6 | 200 (194 + 6) |
Steps of construction :
Take 1 cm along x- axis = 10 marks
Take 2 cm along y- axis = 20 students
Plot the point (0, 0) as ogive starts from x- axis representing lower limit of first class.
Plot the points (10, 5), (20, 16), (30, 26), (40, 46), (50, 74), (60, 111), (70, 151), (80, 180), (90, 194), (100, 200).
Joint the points by a free hand curve.

(i) Here, n (no, of students) = 200.
To find the median :
Let A be the point on y-axis representing frequency = = 100.
Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the points M represents 57.
Hence, the median marks is 57.
(ii) Minimum passing marks = 40
Let T be the point on x-axis representing marks = 40
Through T, draw a vertical line to meet the ogive at S. Through S, draw a horizontal line to meet the y-axis at D.
D = 46
No. of students who scored less than 40 marks = 46
Hence, the no. of students who failed if the minimum marks required to pass is 40 = 46.
(iii) Total marks = 100.
Given, scoring 85 and more marks is considered as grade one.
Let E = 85.
Through E draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C.
C = 188.
Hence, 188 students score less than or equal to 85, so, students scoring more than 85 = 200 - 188 = 12.
Hence, the number of students who scored more than 85 marks in the examination is 12.
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Related Questions
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(ii) the upper-quartile
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