40 students enter for a game of shot put competition. The distance thrown (in metres) is recorded below.

Distance (in m)	Number of students
12 - 13	3
13 - 14	9
14 - 15	12
15 - 16	9
16 - 17	4
17 - 18	2
18 - 19	1

Use a graph paper to draw an ogive for the above distribution.

Use a scale of 2 cm = 1 m on one axis and 2 cm = 5 students on the other axis. Hence using your graph, find

(i) the median

(ii) Upper quartile

(iii) Number of students who cover a distance which is above 16 $\dfrac{1}{2}$ m.

Cumulative frequency distribution table :

Steps of construction:

1. Since, the scale on x-axis starts at 12, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 12.

2. Take 2 cm along x-axis = 1 m.

3. Take 2 cm along y-axis = 5 students.

4. Plot the point (12, 0) as ogive starts from x-axis representing lower limit of first class.

5. Plot the points (13, 3), (14, 12), (15, 24), (16, 33), (17, 37), (18, 39) and (19, 40).

6. Join the points by a free hand curve.

(i) The total number of students is N = 40. The median position is found at $\dfrac{\text{N}}{2} = \dfrac{40}{2}$ = 20.

Draw a line parallel to x-axis from point A (number of students) = 20, touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at point C.

From graph, C = 14.6

Hence, the median = 14.6 m.

(ii) Here, n = 40, which is even.

By formula,

Upper quartile = $\dfrac{3\text{N}}{4} = \dfrac{3 \times 40}{4} = \dfrac{120}{4}$ = 30.

Draw a line parallel to x-axis from point J (number of students) = 30, touching the graph at point K. From point K draw a line parallel to y-axis touching x-axis at point L.

From graph, L = 15.6

Hence, the upper quartile = 15.6 m.

(iii) Draw a line parallel to y-axis from point D (Distance) = $16\dfrac{1}{2}$ m = 16.5 m, touching the graph at point E. From point E draw a line parallel to x-axis touching y-axis at point F.

From graph, F = 35.

It means there are 35 students who cover a distance either less or equal to $16\dfrac{1}{2}$ m.

Number of student who cover a distance which is above $16\dfrac{1}{2}$ m = 40 - 35 = 5.

Hence, number of students who cover a distance above $16\dfrac{1}{2}$ m = 5.