Marks obtained by 200 students in an examination are given below:

Marks	Number of students
0 - 10	5
10 - 20	10
20 - 30	14
30 - 40	21
40 - 50	25
50 - 60	34
60 - 70	36
70 - 80	27
80 - 90	16
90 - 100	12

Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm = 20 students on other axis. From the graph, find:

(i) the median

(ii) the upper-quartile

(iii) number of students scoring more than 65 marks

(iv) if 10 students qualify for merit-scholarship, find the minimum marks required to qualify.

Cumulative frequency distribution table :

Here, n = 200, which is even.

Steps of construction:

1. Take 1 cm along x-axis = 10 marks

2. Take 2 cm along y-axis = 20 students

3. Plot the point (0, 0) as ogive starts from x- axis representing lower limit of first class.

4. Plot the points (10, 5), (20, 15), (30, 29), (40, 50), (50, 75), (60, 109), (70, 145), (80, 172), (90, 188), (100, 200).

5. Joint the points by a free hand curve.

(i) To find the median :

Let A be the point on y-axis representing frequency = $\dfrac{\text{n}}{2} = \dfrac{200}{2}$ = 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 57.5.

Hence, the median is 57.5.

(ii) By formula,

$\text{Upper quartile} = \Big(\dfrac{3\text{n}}{4}\Big)^ \text{th}\text{ term} \\[1em] \Rightarrow \text{Upper quartile} = \Big(\dfrac{(3 \times 200)}{4}\Big)^\text{th}\text{ term} \\[1em] \Rightarrow \text{Upper quartile} = \dfrac{600}{4}^\text{th}\text{ term} \\[1em] \Rightarrow \text{Upper quartile} = 150^\text{ th}\text{ term} \\[1em]$

Draw a line parallel to x-axis from point R (Number of students) = 150, touching the graph at point Q. From point Q draw a line parallel to y-axis touching x-axis at point N.

From graph,

N = 72

Hence, the upper quartile is 72.

(iii) Total marks = 100.

Let E be the point on x-axis representing marks = 65.

Through E draw a vertical line to meet the ogive at B. Through B, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 128.

Hence, 128 students score less than or equal to 65, so, students scoring more than 65 = 200 - 128 = 72.

Hence, the number of students who scored more than 65 marks is 72.

(iv) Given, 10 students qualify for merit-scholarship. This corresponds to 190th student. Since, 200 - 10 = 190

Draw a line parallel to x-axis from point F = 190, touching the graph at point G. From point G draw a line parallel to y-axis touching x-axis at point S.

From graph,

S = 91

Hence, the minimum marks required to qualify is 91.