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Question
Attempt this question on a graph paper. The table shows the distribution of marks gained by a group of 400 students in an examination.
Marks (Less than ) |
10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 |
No.of student | 5 | 10 | 30 | 60 | 105 | 180 | 270 | 355 | 390 | 400 |
Using scaie of 2cm to represent 10 marks and 2 cm to represent 50 student, plot these point and draw a smooth curve though the point
Estimate from the graph :
(1)the median marks
(2)the quartile marks.
Solution 1
Marks (less than) | No.of students |
10 | 5 |
20 | 10 |
30 | 30 |
40 | 60 |
50 | 105 |
60 | 180 |
70 | 270 |
80 | 355 |
90 | 390 |
100 | 400 |
Number of terms=400
∴ Median= `400/2=200^("th")` term
Through 200th term mark draw a line parallel to the x-axis which meets the curve at A. draw a prependiculer to x-axis which meet it at B value of B is the median = 62
Lower Quartile = `Q_1=400/4=100^("th")` term=49
Upper Quartile= `400xx3/4=300^("th")` term=74
Solution 2
By plotting the points (10, 5), (20, 10), (30, 30), (40, 60), (50, 105), (60, 180), (70, 270), (80, 335), (90, 390) and (100, 400), we get the ogive for the given frequency table, as shown in the figure.
Scale: 2 cm to represent 10 marks
2 cm to represent 50 students.
(i) To find the median, we shall draw a horizontal line at c.f. = `"N"/(2) = (400)/(2)` = 200. Intersecting the ogive at the point (200, 63).
Hence the median is 63.
(ii) To find the lower quartile, we shall construct a horizontal line at c.f. `"N"/(2) = (400)/(2)` = 100, intersecting the ogive at the point (49, 100). Hence, 49 is the lower line at c.f. `(3"N")/(4) = (3 xx 400)/(4)` = 300.
Intersecting the ogive at the point (300, 72). Hence, the upper quartile mark is 72.
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